Class 9 Maths - Statistics

The expenditure of a family on different heads in a month is given below:

Head Food Education Clothing House rent Others Saving
Expenditure (rs) $4000$ $2500$ $1000$ $3500$ $2500$ $1500$

Draw a bar graph to represent the data above.

$\textbf{The bar graph for given data is shown Above.}$

$\textbf{Step-1:We draw a bar graph of this data in the following steps}$

$\text{1.We represent the expenditure on the vertical axis. Since, the}$

$\text{maximum expenditure is Rs. 4000, wecan choose the scale as 1 unit = Rs. 500}$

$\text{2.To represent our first head i.e.,food, we draw a rectangular bar }$

$\text{with width 1 unit and height 8 units}$

$\text{3.Similarly, other heads are represented on the same graph.}$

$\textbf{Hence,Above graph is required bar graph}$

2153366_1795076_ans_636c314bae7a4b0f9f86b87fd5258ba5.png

The bar graph given below represents the circulation of newspapers (dailies) in a town in six languages (the figures are approximated to hundreds).

Study the bar graph and answer the following questions:
(a) Find the total number of newspapers read in Hindi, Punjabi, Urdu, Marathi and Tamil.
(b) Find the excess number of newspapers read in Hindi than those in English.
(c) Name the language in which the least number of newspapers are read.
(d) Write the total circulation of newspapers in the town.

(a) Total number of newspaper read in Punjabi, Hindi, Marathi, Urdu, and Tamil

$800+400+200+300+100=1800$

(b) Number of newspaper read in Hindi =

$800$

Number of newspaper read in English =

$500$

Therefore,

$800-500=300$ Extra newspaper in Hindi than in English

(d) Total number of newspaper in the town

$800+500+400+300+200+100=2300$

the following histograms shows the frequency distribution of teaching experiences of $30$ teachers in various schools.
(a) What is the class width ?
(b) How many teachers are having the maximum teaching experience and how many have the least teaching experience /
(c) How many teachers have teaching experience of $10$ to $20$ years?

(a)

From the above histogram , we can conclude that

$5-0=5$ is the class width.

(b)

From the above histogram, we can conclude that two teachers are

$15-20$ years that is maximum experience holder.

And five teachers are minimum experience holder that is of

$0-5$ years.

(c)

Total number of

$10-20$ years experienced teacher is

$7+2=9$

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
What information does the above bar graph give ?

The bar graph shows the number of shirts sold from Monday to Saturday. With maximum sell on Saturday

Form the frequency distribution table from the following data:
Marks (out of 90) Number of candidates
More than or equal to 80 4
More than or equal to 70 6
More than or equal to 60 11
More than or equal to 50 17
More than or equal to 40 23
More than or equal to 30 27
More than or equal to 20 30
More than or equal to 10 32
More than or equal to 0 34

The frequency distribution table for the given data is:

Class Interval	No of students
0-10	34-32=2
10-20	32-30=2
20-30	30-27=3
30-40	27-23=4
40-50	23-17=6
50-60	17-11=6
60-70	11-6=5
70-80	6-4=2
80-90	4

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
On which day were the maximum number of shirts sold ? How many shirts were sold on that day ?

From the graph it can be observed that a maximum of

$60$ shirts are sold on Saturday.

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
What is the scale chosen on the horizontal line representing number of shirts ?

1 unit length = 5 shirts

Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class:
Marks
Below 20 Below 40 Below 60 Below 80 Below 100
Number of students 17 22 29 37 50
Form the frequency distribution table for the data.

Marks	No of students
0-20	12
20-40	22-17=5
40-60	29-22=5
60-80	37-29=8
80-100	50-37=13

The frequency distributed table for given data.

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
How many shirts were sold on Thursday?

$35$ shirts were sold on Thursday

Fill in the blanks:
The number of times a particular observation occurs in a data is called ....... of the observation.

The number of times a particular observation occurs in a data is called frequency of the observation.

The following table shows the cumulative frequency distribution of marks of 800 students in an examination:
Marks Number of students
Below 10 10
Below 20 50
Below 30 130
Below 40 270
Below 50 440
Below 60 570
Below 70 670
Below 80 740
Below 90 780
Below 100 800
Construct a frequency distribution table for the data above.

Class interval	No of students
0-10	10
10-20	50-10=40
20-30	130-50=80
30-40	270-130=140
40-50	440-270=170
50-60	570-440=130
60-70	670-570=100
70-80	740-670=100
80-90	780-740=40
90-100	800-780=20

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
On which day were the minimum number of shirts sold?

Minimum number of shirts sold Tuesday

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:
Name the subject(s) in which she got $80$ or more marks.

Gurmeet obtained

$80$ or more marks in Science and mathematics.
2004835_1800533_ans_eb8e715200d04a48a18177c65b525f30.png

Given below a graph:
Read the bar graph carefully and answer the following question:
What is the information given by the bar graph?

Representation of the expenditures in different purposes is shown in the given bar graph.

Fill in the blanks:
In a bar graph, the bars are of uniform...........

In a bar graph, the bars are of uniform width.

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:
What is the scale of this bar graph ?

Scale: 1 unit length = 10 marks obtained.
2004826_1800530_ans_47a4305464e84372a49c3325f0248dd7.png

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:
In which subject she obtained minimum marks ?

Gurmeet obtained minimum marks in Hindi.
2004833_1800532_ans_7ed535aa1f234ab4bde913ec7f04314f.png

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:
In which subject Gurmeet obtained maximum marks ?

Gurmeet obtained maximum marks in mathematics.
2004828_1800531_ans_991899111c984558b6771b47a8467a72.png

The weight of $29$ patients in a hospital were recorded as follows:
Weight (in $kg$ )
$50-55$ $55-60$ $60-65$ $65-70$ $70-75$ $75-80$
Number of patients $7$ $4$ $4$ $9$ $2$ $3$
Draw a histogram to represent this data visually.

Weight (in $kg$ )	$50-55$	$55-60$	$60-65$	$65-70$	$70-75$	$75-80$
Number of patients	$7$	$4$	$4$	$9$	$2$	$3$

The histogram for the given data is shown below:

1715376_1809359_ans_4f7d2c8634314becbe2cba2b30934120.png

Given below a graph:
Read the bar graph carefully and answer the following question:
On which item the expenditure is minimum?

The expenditure on clothing is minimum

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:
Answer the following question:
Numbers of workers whose weakly wages are more than or equal to $Rs. 400$ but less than $Rs. 450$

The weekly wages of workers in a factory is shown in the given histogram:
Number of workers whose weekly wages are more than or equal to

$Rs. 400$ but less than

$Rs. 450$

$=18+10=28$ .

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:
Answer the following question:
How many workers get more than $Rs. 425$ ?

The weekly wages of workers in a factory is shown in the given histogram:
Total no. of workers more than

$Rs 125=10+20+4=34$

$34$ workers are getting more than

$Rs. 425$

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.
Answer the following:
For how many hours did the maximum number of students watch $T.V$ ?

From the given histogram:
Maximum number of students watch

$T.V$ for

$4-5$ hours

Given below a graph:
Read the bar graph carefully and answer the following question:
State whether true or false:
Expenditure on education is twice the expenditure on clothing.

Yes, the expenditure on education is twice the expenditure on clothing.

In a study diabetic patients, the following data was obtained:
Age (in years) $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
$70-80$
Number of patients $3$ $8$ $30$ $36$ $27$ $15$ $6$
Represents the above data by a histogram.

Age (in years)	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Number of patients	$3$	$8$	$30$	$36$	$27$	$15$	$6$

The histogram representing the above given data is as follows:
1715390_1809369_ans_3065eab789ef485ab2831d8a14c0d7e8.png

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:
Answer the following question:
What is the class having a minimum frequency?

The weekly wages of workers in a factory is shown in the given histogram:
The class having a minimum frequency is

$475-500$ that is

$4$ .

Given below a graph:
Read the bar graph carefully and answer the following question:
On which item the expenditure is maximum?

The expenditure on food is maximum.

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:
Answer the following question:
How many workers get less than $Rs. 475$ ?

The weekly wages of workers in a factory is shown in the given histogram:
Total no. of workers less than

$Rs.475=20+10+18+6=54$

$54$ workers are getting less than

$Rs. 475$

The following observations have been arranged in ascending order. If the median the data is $47.5$ , find the value of $x$ .
$17,21,23,29,39,40,x,50,51,54,59,67,91,93$

Given data

$17,21,23,29,39,40,x,50,51,54,59,67,91,93$

Here,

$n=14$ which is even

As the given data is arranged in descending order

Median

$\cfrac { 1 }{ 2 } \left[ (14/2)th\ term+(14/2+1)th\ term \right] =\cfrac { 1 }{ 2 } (7th\quad term+8th\quad term)\quad$

$\Rightarrow 47.5=\cfrac { 1 }{ 2 } (x+50)\Rightarrow 95=x+50\Rightarrow x=95-50=45$

Hence the value of

$x$ is

$45$ .

In a Science test given to a group of students, the marks scored by them (out of 100) are:
$41,39,52,48,54,62,46,52,40,96,42,40,98,60,52$
Find the mean and median of this data.

Given that, the marks scored by a group of students, out of

$100$ are:

$41,39,52,48,54,62,46,52,40,96,42,40,98,60,52$

On arranging the marks obtained by the students in ascending order, we get:

$39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98$

The total number of observations are

$15$ .

Hence,

$n=15$

We know that, Mean

$(\bar { x } )=\dfrac{\sum { { x }_{ i }}}{n}$

$\therefore \ \bar{x}=\dfrac{(39+40+40+41+42+46+48+52+52+52+54+60+62+96+98)}{15}$

$\Rightarrow \bar{x}=\dfrac{822}{15}$

$\therefore \ \bar{x}=54.8$

Also,

$n$ is odd

We know that for odd number of observations,

Median is the

$\left(\dfrac{n+1}{2}\right)^{th}$ observation.

Hence, the median for the given data is

$\left(\dfrac{15+1}{2}\right)^{th}$ observation, which is the

$8^{th}$ observation from the beginning.

$8^{th}$ observation in the data is

$52$ .

$\therefore \$ Median

$=52$

Hence, the mean of the given data is

$54.8$ and the median is

$52$ .

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.
Answer the following:
For how many students spent more than $2$ hours but less than $4$ in watching $T.V$ ?

From the given histogram:
No. of students spending more than 2 hours but less than 4 in watching T.V

$=8+22=30$

$30$ students spent more than

$2$ hours but less than

$4$ hours in watching

$T.V$

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.
Answer the following:
For how many students spent more than $5$ hours in watching $T.V$ ?

From the given histogram:
No. of students spending more than 5 hours in watching T.V

$=8+6=14$

$14$ students spent more than

$25$ hours in watching

$T.V$

The number of literate females in the group of $10$ to $40$ years in a town is shown in the histogram alongside.
Answer the following question:
What is the class size?

In the given histogram, the class size is

$5$

Rearrange the following raw data in descending order:
$5.3, 5.2, 5.1, 5.7, 5.6, 6.0, 5.5, 5.9, 5.8, 6.1, 5.5, 5.8, 5.7, 5.9$ and $5.4$ . The write the:
range of values

The given numbers in descending order are as follows:

$6.1, 6.0, 5.9, 5.9, 5.8, 5.7, 5.7, 5.6, 5.5, 5.5, 5.4, 5.3, 5.2, 5.1$
Range of values = Difference between highest value and lowest value
Hence,
Range of values = highest value - lowest value

$=6.1-5.1$

$=1.0$

The number of literate females in the group of $10$ to $40$ years in a town is shown in the histogram alongside.
Answer the following question:
In which age group is the number of literate females the highest?

$15-20$ age group, the literate females are highest.

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.
Answer the following:
How many students watched $T.V$ for less than $4$ hourse?

From the given histogram:
No. of students watching TV les than

$4$ hours

$= 4+8+22=34$

$34$ students watch

$T.V$ for less than

$4$ hours

The number of literate females in the group of $10$ to $40$ years in a town is shown in the histogram alongside.
Answer the following question:
Write the classes assuming all the classes are of equal width.

From the given histogram, the classes having equal width are

$10-15, 15-20,20-25,25-30,30-35,35-40$

The number of literate females in the group of $10$ to $40$ years in a town is shown in the histogram alongside.
Answer the following question:
In which age are the literate females the least?

$10-15$ age group, the literate females are least

The mean of $18,\ 28,\ x,\ 32,\ 14$ and $36$ is $23$ . Find the value of $x$ Sum of data

Given
Mean of

$18,\ 28,\ x,\ 32,\ 14$ and

$36$

$=23$
Hence, the value of

$x$ can be calculated as below

Mean

$=$

$\dfrac{\text {Sum of all terms}} {\text {total number of terms}}$

$23=\dfrac{(18+28+x+32+14+36) }{ 6}$

On further calculation, we get

$\Rightarrow 23 \times 6=x+128$

$\Rightarrow 138=x+128$

$\Rightarrow x=138-128$

$\Rightarrow x=10$

Hence, the value of

$x$ is

$10$

The bar graph drawn below shoes the number of tickets sold during a fair by $6$ students $A, B, C, D, E$ and $F$
Using the Bar graph, answer the following question:
Who sold the maximum number of tickets?

From the given graph, the student

$E$ sold the maximum number of tickets i.e.

$24$ tickets

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
On which day were the maximum number of patients examined?

From the given graph, the maximum number of patients is examined on Tuesday

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
On which day were the least number of patients examined?

From the given graph, the minimum number of patients is examined on Friday

The above bar graph shows the number of children, in various classes, in a school in Delhi.
Using the given bar graph, find:
How many more children are there in Class 5 compared to Class 6?

From the given graph,
Number of children in class $$

5$$ are $$120$$
And that in class $$

6$$ are $$90$$.
Difference of number of children in class

$5$ and class

$6=120-90$

$=30$

Hence, there are

$30$ children more in class

$5$ compared to class

$6$ .

The bar graph drawn below shoes the number of tickets sold during a fair by $6$ students $A, B, C, D, E$ and $F$
Using the Bar graph, answer the following question:
How many tickets were sold by $D, E$ and $F$ taken together?

From the graph, total number of tickets sold by the student

$D, E$ and

$F$ can be calculated as below
Tickets sold by

$D, E$ and

$F$ taken together = ( Tickets sold by

$D$ )+ ( Tickets sold by

$E$ )+( Tickets sold by

$F$ )

$=7+24+14$
We get,

$=45$
Therefore, total tickets sold by

$D, E$ and

$F$ together is

$45$ tickets

The bar graph drawn below shoes the number of tickets sold during a fair by $6$ students $A, B, C, D, E$ and $F$
Using the Bar graph, answer the following question:
How many tickets were sold by $A, B$ and $C$ taken together?

From the graph, total number of tickets sold by the student

$A, B$ and

$C$ can be calculated as below
Tickets sold by

$A, B$ and

$C$ taken together = ( Tickets sold by

$A$ )+ ( Tickets sold by

$B$ )+( Tickets sold by

$C$ )

$=16+9+20$
We get,

$=45$
Therefore, total tickets sold by

$A, B$ and

$C$ together is

$45$ tickets

The following bar graph shows the number of children, in various classes, in a school in Delhi.
Using the given bar graph, find:
the average number of children in a class

Average number of students in each class can be calculated as below
Average number of students in each class = ( Total number of students in classes )/ Number of classes

$=700/8$
We get,

$=87.5$

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
On which days were equal number of patients examined?

From the given graph, equal number of patients is examined on Sunday and Thursday

The bar graph drawn below shoes the number of tickets sold during a fair by $6$ students $A, B, C, D, E$ and $F$
Using the Bar graph, answer the following question:
Who sold the least number of tickets?

From the given graph, the student

$D$ sold the least number of tickets i.e.

$7$ tickets

The mean of the number $6, y, 7, x$ and $14$ is $8.$ Express $y$ in terms of $x$ .

Given,
No. of terms

$(n) = 5$ and mean

$= 8$
So, the sum of all terms

$= 5 \times 8 = 40$ ……. (i)
but sum of numbers

$= 6 + y + 7 + x + 14 = 27 + y + x$ ……. (ii)
On equating (i) and (ii), we get

$27 + y + x = 40$

$x + y = 13$
Hence,

$y = 13 – x$

If the mean of $6, 4, 7,$ $a$ and $10$ is $8$ . Find the value of $a$

Given,
No. of terms

$(n) = 5$
Mean

$= 8$
Sum of all terms =

$8 \times 5 = 40$ …… (i)
But, sum of numbers

$= 6 + 4 + 7 + a + 10 = 27 + a$ ….. (ii)
On equating (i) and (ii), we get

$27 + a = 40$
Thus,

$a = 13$

Find the actual lower and upper class limits and also the class marks of the classes:
$1.1-2.0,2.1-3.0$ and $3.1-4.0$

Case

$1$ : frequency

$1.1-2.0,$
The lower class limit is

$1.1$
The upper class limit is

$2.0$
Mid-value is,

$\dfrac{1.1+2.0}{2}=1.55$
Case

$2$ : frequency

$2.1-3.0,$
The lower class limit is

$2.1$
The upper class limit is

$3.0$
Mid-value is,

$\dfrac{2.1+3.0}{2}=2.55$
Case

$3$ : frequency

$3.1-4.0,$
The lower class limit is

$3.1$
The upper class limit is

$4.0$
Mid-value is,

$\dfrac{3.1+4.0}{2}=3.55$

Following data shows the weekly wages (in ) of $10$ workers in a factory. $3500, 4250, 4000, 4250, 4000, 3750, 4750, 4000, 4250$ and $4000$ .What is the range of wages (in )?

The frequency distribution table for weekly wages of

$10$ workers can be given as:
Range of wages

$= 4750 – 3500 = ₹\,1250$

Weekly wages (in Rs.)	Tally marks	Frequency
$3500$	\|	$1$
$3750$	\|	$1$
$4000$	\|\|\|\|	$4$
$4250$	\|\|\|	$3$
$4750$	\|	$1$
Total		$10$

Fill in the blanks:
The range of the data 7,13,6,25,18,20,16 is .......

25-6=19

Arrange the following data in descending order: $3.3, 3.2, 3.1, 3.7, 3.6, 4.0, 3.5, 3.9, 3.8, 4.1, 3.5, 3.8, 3.7, 3.9$ and $3.4$ .
Determine the range.

We can write the given data in descending order

$4.1, 4.0, 3.9, 3.9, 3.8, 3.7, 3.7, 3.6, 3.5, 3.5, 3.4, 3.3, 3.2, 3.1$

$Range = 4.1 – 3.1 = 1.0$

Uses the table given to find:
The actual class limits of the fourth class

Case

$1$ : frequency

$1.1-2.0,$
The lower class limit is

$1.1$
The upper class limit is

$2.0$
Mid-value is,

$\dfrac{1.1+2.0}{2}=1.55$
Case

$2$ : frequency

$2.1-3.0,$
The lower class limit is

$2.1$
The upper class limit is

$3.0$
Mid-value is,

$\dfrac{2.1+3.0}{2}=2.55$
Case

$3$ : frequency

$3.1-4.0,$
The lower class limit is

$3.1$
The upper class limit is

$4.0$
Mid-value is,

$\dfrac{3.1+4.0}{2}=3.55$

Consider the following numbers:
$68, 76, 63, 75, 93, 83, 70, 115, 82, 105, 90, 103, 92, 52, 99, 73, 75, 63, 77$ and $71$ .
What is the range of these numbers?

We know that

Range of these numbers = largest number – smallest number

So we get

$= 115 – 52$

$= 63$

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
What is the total number of patients examined in the week?

Total number of patients examined in a week can be calculated as given below
Total number of patients examined in a week

$=50+40+70+60+50+30$
We get,

$=300$ students
Hence,

$300$ patients are examined in a week

Hundred students from a certain locality use different modes of travelling to school as given below. Draw a bar graph.
Bus Car Rickshaw Bicycle Walk
$32$ $16$ $24$ $20$ $8$

1781048_1839490_ans_717c48fbd10a462b9260db0044ee7e6c.png

Fill in the blanks in the following table:
Class interval Frequency Cumulative frequency
$25-34$ __ $15$
$35-44$ _ $28$
$45-54$ $21$
$55-64$ $16$

$65-74$ $73$
$75-84$ $12$

Class interval	Frequency	Cumulative frequency
$25-34$	$15$	$15$
$35-44$	$13$	$28$
$45-54$	$21$	$49$
$55-64$	$16$	$65$
$65-74$	$8$	$73$
$75-84$	$12$	$85$

Uses the table given to find:
The class boundary of the sixth class

The actual class limit of the fourth class

=54.5−59.5

In the following table, the investment made by $210$ families is shown. Present it in the form of a histogram.
Investment
(Thousand Rupees) $10 - 15$ $15 - 20$ $20 - 25$ $25 - 30$ $30 - 35$
No. of families $30$ $50$ $60$ $55$ $15$

The histogram for the given data is
1815215_1852745_ans_be10473551e5498293ebba495ac2cc45.png

Uses the table given to find:
The class mark of the third class

The class marks of the third class will be the average of the lower and the upper bound of the interval.
Hence, class mark will be:

$\dfrac{40+44}{2}=42$

Draw a histogram for the following frequency distribution.
Use of electric
(unit) $50-70$ $70-90$ $90-110$ $110-130$ $130-150$ $150-170$
No. of families $150$ $400$ $460$ $540$ $600$ $350$

The histogram for the given data is
1914583_1853136_ans_1ff471e1ed1d424fbc5aa8138965818d.png

Draw a histogram of the following data.
Height of student ( $\text{cm}$ ) $135 - 140$ $140 - 145$ $145 - 150$ $150 - 155$
No. of students $4$ $12$ $16$ $8$

The histogram of given data is shown above.
1815209_1852693_ans_cecfcad71cbe401b898f651636e85245.png

Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.
TIme (minutes) $60 - 80$ $80 - 100$ $100 - 120$ $120 - 140$ $140 - 160$
No.of students $14$ $20$ $24$ $22$ $16$

The histogram for the given data is
1815220_1852762_ans_0de72a2c0ab64e939a8da445b8199174.png

Classify the following information as primary or secondary data. Information of attendance of every student collected by visiting every class in a school.

Since the information is collected from every house. So, the information collected from every house in the village Nandpur regarding students not attending school is primary data.

Classify the following information as primary or secondary data. In the village Nandpur, the information collected from every house regarding students not attending school.

Since the information is collected from every house. So, the information collected from every house in the village Nandpur regarding students not attending school is primary data.

Classify the following information as primary or secondary data. The information of the heights of students was gathered from school records and sent to the head office, as it was to be sent urgently.

Since the information is collected from every house. So, the information collected from every house in the village Nandpur regarding students not attending school is primary data.

Classify the following information as primary or secondary data:
For a science project, information of trees gathered by visiting a forest.

We know that, primary data is data collected by the researchers directly from the main source, while secondary data is data collected from some other source that obtained the data from the main source or other source.

While gathering the information by visiting the forest, the data is being collected from the main source directly. Hence, it is primary data.

The mean of the following frequency table isBut the frequencies $f_{1}$ and $f_{2}$ in class 20 - 40 and 60 - 80 respectively are missing. Find the missing frequencies.
Classes 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Total
Frequency 17 $f_{1}$ 32 $f_{2}$ 19 $N = \sum f_{i} = 120$

Assumed mean = a = 50 and h = 20

$\sum f_{i} = 68 + f_{1} + f_{2} = 120$

$f_{1} + f_{2} = 120 - 68$

$f_{1} + f_{2} = 52 \rightarrow (1)$

$\mathbf{Mean} = \bar{x} = a + \dfrac{\sum f_{i}u_{i}}{\sum f_{i}} \times h$

$50 = 50 + \dfrac{(4 - f_{1} + f_{2})}{\not{120}_{6}} \times \not{20}$

$50 - 50 = \dfrac{4 - f_{1} + f_{2}}{6}$

$0 = 4 - f_{1} + f_{2}$

$f_{1} + f_{2} = 4 \rightarrow (2)$
Adding equation (1) and (2)

$f_{1} + f_{2} = 52$

$f_{1} - f_{2} = 4$

$2f_{1} = 56$

$f_{1} = \dfrac{56}2 = 28$
Put

$f_{1} = 28$ in equation (1)

$28 + f_{2} = 52$

$f_{2} = 52 - 28$

$f_{2} = 24$

$\therefore$ missing frerquencies

$f_{1}$ is

$28$ and

$f_{2}$ is

$24.$

In a city, $38$ people donated to an organisation working for differently abled persons. The amount in rupees were as follows : $101, 500, 401, 201, 301, 160, 210, 125, 175, 190, 450, 151, 101, 351, 251, 451, 151, 260, 360, 410,$ $150, 125, 161, 195, 351, 170, 225, 260, 290, 310, 360, 425, 420, 100, 105, 170, 250, 100$ .
(i) By taking classes $100-149, 150-199, 200-249...$ prepare a grouped frequency distribution table.
(ii) From the table, find the number of people who donated rupees $350$ or more.

The frequency distribution table so obtained is given below.

Number of people who donated

$Rs.350$ or more

$=4+4+3$

$=11$

Hence,

$11$ people have donated

$Rs.350$ or more.

1817410_1855423_ans_49b806818fdc4b989c3f58c97db77e5f.png

If the mean of the following data is $20.2$ , then find the value of p.
$x_i$ 10 15 20 25 30
$f_i$ 6 8 p 10 6

We have,

$x_i$	$f_i$	$$x_if_i$$
10	6	60
15	8	120
20	p	20p
25	10	250
30	6	180

If the median of the distribution given is 28.5, find the values of 'x' and 'y'
Class Interval Frequency
0 - 10 5
10 - 20 x
20 - 30 20
30 - 40 15
40 - 50 y
50 - 60 5
Total 60

Median is 28.5
Class interval which has median is = (20 - 30)

$l = 20, n = 60, f = 20, cf = 5 + a \times h = 10$

$\therefore \mathbf{Median} = l + \left [ \dfrac{\dfrac{n}{2} - cf}{f} \right ] \times h$

$28.5 = 20 + \left \{ \dfrac{30 - (5 + x)}{20} \right \} \times 10$

$8.5 = \dfrac{25 - x}{2}$

$17 = 25 - x$

$\therefore x = 8$

$5 + x + 20 + 15 + y + 5 = 60$

$x + y - 45 = 60$

$x + y = 15$

$y = 15 - x$

$y = 15 - 8$

$\therefore y = 7$

$\therefore x = 8, y = 7$

Find the class mark of the class interval 30 - 40.

Class mark

$= \dfrac{Upper \; limit + Lower \;limit}{2}$

$= \dfrac{30 + 40}{2} = \dfrac{70}{2} = 35$

If the mean of the following distribution is 6, find the value of P.
$x_{i}$ 2 4 6 10 P + 5
$f_{i}$ 3 2 3 1 2

$\bar{X} = 6$

$\mathbf{Mean} = \bar{x} = \dfrac{\sum f_{i}x_{i}}{\sum f_{i}}$

$6 = \dfrac{2P + 52}{11}$

$66 = 2P + 52$

$2P = 66 - 52$

$2P = 14$

$P = \dfrac{14}{2} = 7$

$P = 7$

1841163_1869446_ans_92f3d19845814a63b9c26edf62795488.PNG

If the classmark is $10$ and class width is $6$ then find the class.

We have, class mark

$=10$ and class width

$= 6$ .
Since,

$\dfrac{6}{2} = 3$
The lower class limit

$= 10 - 3 = 7$ and the upper class limit

$= 10 + 3 = 13$
So, the class is

$7-13$ .

The monthly maximum temperature of a city is given in degree celcius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
$29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3,$ $31.4, 30.3, 34.7, 35.0, 32.5, 33.5, 29.0, 29.5, 29.9, 33.2, 30.2$
From the table, answer the following questions.
(i) For how many days the maximum temperature was less than $34^\circ C$ ?
(ii) For how many days the maximum temperature was $34^\circ C$ or more than $34^\circ C$ ?

(a)

The class intervals for which upper limit is less than or equal to

$34^oC$ will represent that the maximum temperature is less than

$34^oC,$

$28-29\rightarrow 2$

$29-30\rightarrow 6$

$30-31\rightarrow 5$

$31-32\rightarrow 3$

$32-33\rightarrow 5$

$33-34\rightarrow 3$

Total

$\rightarrow 24$

Hence, in

$24$ days the maximum temperature was less than

$34^oC.$

(b)

The class intervals for which the lower limit is greater than or equal to

$34^oC$ will represent that the maximum temperature is greater than or equal to

$34^oC,$

$34-35\rightarrow 3$

$35-36\rightarrow 2$

$36-37\rightarrow 1$

Total

$\rightarrow 6$

Hence, in

$6$ days the maximum temperature was greater than or equal to

$34^oC.$

1817561_1855447_ans_87b675de9a994cb6a5ed775ec3a969d9.png

Prepare the frequency distribution table for the given set of scores . 39 , 16 , 30 , 37 53 , 15 , 16 , 60 , 58 , 26 , 28 , 19, 20 , 12 , 14 , 24 , 59 , 21, 57 , , 38 , 25 , 36 , 34 , 15 , 25 , 41 , 52 , 45 , 60 , 63 , 18 , 26 , 43 , 18 , 27 , 59 , 63 , 46 , 48 , 25 , 33 , 46 , 27 , 46 , 42 , 48 , 35 , 64 , 24 Take class intervals as (10 - 20 ), ( 20 , 30 ) ............. and answer the following .
What is the size of each class interval ? Find the midpoint of the class interval 30 - 40

Size of each class interval is

$10$ and midpoint of the class interval

$30 - 40$ is

$\dfrac{(30 + 40)}{2} = \dfrac{70}{2} = 35$
1955322_1870989_ans_4fbc30dffa7840bb966523b400cb0526.png

The mark scored by 40 candidates in an examination (out of 100 ) is given below .
75 , 65 , 57 , 50 , 32 , 54 , 67 , 75 ,88 , 80 , 42 , 40 , 41 , 34 , 61 , 42 , 46 , 68 , 52 , 43 , 49 , 59 , 49, 67 , 34 , 33 , 87 , 97 , 47 , 46 , 54 , 48 , 45 , 51 , 47 , 41 , 43 .
Prepare a frequency distribution table with the class size 10 . Take the class intervals as 30 - 39 , 40 - 49 and answer the following questions .
Which class intervals have highest and lowest frequency .

The class interval

$40 - 49$ has highest frequency and

$90 - 99$ has the lowest frequency .
1955308_1870974_ans_e6c3c9c267334cbab107619531504e74.png

The mark scored by 40 candidates in an examination (out of 100 ) is given below .
75 , 65 , 57 , 50 , 32 , 54 , 67 , 75 ,88 , 80 , 42 , 40 , 41 , 34 , 61 , 42 , 46 , 68 , 52 , 43 , 49 , 59 , 49, 67 , 34 , 33 , 87 , 97 , 47 , 46 , 54 , 48 , 45 , 51 , 47 , 41 , 43 .
Prepare a frequency distribution table with the class size 10 . Take the class intervals as 30 - 39 , 40 - 49 and answer the following questions .
What are the range of the given distribution ?

Range = Highest score - lowest score

$= 97 - 32$
Range

$= 65$

1955311_1870980_ans_52183a3ea26841bcbbb1182874239e7c.png

The mark scored by 40 candidates in an examination (out of 100 ) is given below .
75 , 65 , 57 , 50 , 32 , 54 , 67 , 75 ,88 , 80 , 42 , 40 , 41 , 34 , 61 , 42 , 46 , 68 , 52 , 43 , 49 , 59 , 49, 67 , 34 , 33 , 87 , 97 , 47 , 46 , 54 , 48 , 45 , 51 , 47 , 41 , 43 .
Prepare a frequency distribution table with the class size 10 . Take the class intervals as 30 - 39 , 40 - 49 and answer the following questions .
Writhe the upper and lower limits of the class interval 30 - 49

The lower limit of the class interval

$30 - 39$ is

$29. 5$ and upper limit is

$39.5$

1955309_1870977_ans_dfac576d61cf49ec90eb0ae22e81cdb2.png

The following are the marks scored in a unit test ( out of 25 ) . Prepare a frequency distribution table , taking class interval as 0 - 4 , 5 - 9 , 10 - 14 , 15 - 19 , 20 - 24.
21 , 14 , 3 , 7 , 23 , 18 , 24 , 16 , 18 , 17 , 18 , 21 , 23 , 19 , 12 , 14 , 11 . From the table
Find the class interval having maximum frequency .

The class interval

$15 - 19$ has maximum frequency.
1956496_1871153_ans_4fb9396ca6b142eca3e542263d722ace.png

Prepare a frequency distribution table for the scores given.
42 , 22 , 55 , 18 , 50 , 10 , 33 , 29 , 17
29 , 29 , 27 , 34 , 15 , 40 , 42 , 40 , 41
35 , 27 , 44 , 31 , 38 , 19 , 54 , 55 , 38
19 , 20 , 30 , 42 , 59 , 15 ; 27 , 23
40 , 32 , 28 , 51
Take the class intervals as 10 - 10 , 20 - 30 , 30 - 40 , 40 - 50 , 50 - 60 . From the frequency distribution table answer the following questions .
What does the frequency , corresponding to the class intervals 20 - 30 indicates ?

The largest number of scores lie in the class intervals 20 - 30
1956674_1871137_ans_164d376a78484a04b12c8726e4f76d3c.png

Prepare the frequency distribution table for the given set of scores . 39 , 16 , 30 , 37 53 , 15 , 16 , 60 , 58 , 26 , 28 , 19, 20 , 12 , 14 , 24 , 59 , 21, 57 , , 38 , 25 , 36 , 34 , 15 , 25 , 41 , 52 , 45 , 60 , 63 , 18 , 26 , 43 , 18 , 27 , 59 , 63 , 46 , 48 , 25 , 33 , 46 , 27 , 46 , 42 , 48 , 35 , 64 , 24 Take class intervals as (10 - 20 ), ( 20 , 30 ) ............. and answer the following .
What is the range of the given set of scores ?

minimum score = 12 ,

maximum score = 64 ,

Range = maximum score - minimum score

= 64 - 12

= 52

1956435_1870991_ans_585b62ed51ec4b9083d61895909914e6.png

Prepare a frequency distribution table for the scores given.
42 , 22 , 55 , 18 , 50 , 10 , 33 , 29 , 17
29 , 29 , 27 , 34 , 15 , 40 , 42 , 40 , 41
35 , 27 , 44 , 31 , 38 , 19 , 54 , 55 , 38
19 , 20 , 30 , 42 , 59 , 15 ; 27 , 23
40 , 32 , 28 , 51
Take the class intervals as 10 - 10 , 20 - 30 , 30 - 40 , 40 - 50 , 50 - 60 . From the frequency distribution table answer the following questions .
Find the range of the scores.

The range

$= 59 - 10 = 49$
1956489_1871141_ans_66e3e8f350dc4c1496a5def5d084ccae.png

The following are the marks scored in a unit test ( out of 25 ) . Prepare a frequency distribution table , taking class interval as 0 - 4 , 5 - 9 , 10 - 14 , 15 - 19 , 20 - 24.
21 , 14 , 3 , 7 , 23 , 18 , 24 , 16 , 18 , 17 , 18 , 21 , 23 , 19 , 12 , 14 , 11 . From the table
Find the range of the scores.

Range = 24 - 3 = 21
1956516_1871154_ans_5b6ba4baca6d4d27aef4467ca62c39de.png

Draw a histogram to represent the following frequency distribution:
Class interval Frequency
$20 - 25$ $5$
$25 - 30$ $10$
$30 - 35$ $18$
$35 - 40$ $14$
$40 - 45$ $12$

1955398_1870993_ans_9fd3ef72e06f4217a0259270f8c9812e.png

If median of the following distribution is 27, then find the values of x and y.
Class-interval Frequency
0-10 5
10-20 $x$
20-30 20
30-40 14
40-50 $y$
50-60 8
Total 68

Given: N = 68

$\Rightarrow 47 + x + y = 68$

$\Rightarrow x + y = 68 - 47$

$x + y = 21$ ............(i)
Now

$\frac{N}{2} = \frac{68}{2} = 34$
Median of distribution is 27 which lies in class-interval 20-30.

$\therefore$ Median class 20-30.

$l = 20, c = 5 + x, f = 20$ and

$h = 10$

$\therefore Median = l + \left (\frac{\frac{N}{2} - C}{f} \right )$

$\Rightarrow 27 = 20 + \left ( \frac{34 - (5+x)}{20} \right ) \times 10$

$= 20 + \frac{34 - 5 - x}{2} = \frac{40 + 29 - x}{2}$

$\Rightarrow 69 -x = 2 \times 27$

$\Rightarrow x = 69 -54 = 15$ Putting value of x in eqn (i)

$15+y=21$

$\Rightarrow y = 21 - 15 = 6$
Thus,

$x = 15, y = 6$
1866518_1879250_ans_a2ecf64a44664f38a090172242560122.PNG

If mean of following data is 21.5, then find K :
Class 0-10 10-20 20-30 30-40 40-50
Frequency 6 4 3 K 2

$\because Arithmetic mean = \frac{\sum f_{i}x_{i}}{\sum f_{i}}$

$21.5 = \frac{255 + 35K}{15 + K}$

$\Rightarrow 322.5 - 255 = 35K - 21.5k$

$\Rightarrow 13.5k = 67.5$

$K= \frac{67.5}{13.5} = 5$

$\therefore K = 5$

1866285_1879029_ans_d27e60bef03246e7b973a2068c0f756b.PNG

Draw a histogram for the following frequency distribution .

1955585_1871156_ans_3a4d3b3f3a7b403a86eeb8518d25e010.png

Class Interval $0-20$ $20-40$ $40-60$ $60-80$ $80 -100$
Frequency $5$ $6$ $12$ $4$ $3$
Construct histogram for the following frequency table:

Take

$OX$ and

$OY$ as x-axis and y-axis respectively on the graph paper and

$O$ as the origin. By taking suitable scale at x-axis (

$20$ class interval

$= 1 cm$ ) and y-axis (

$2$ frequency

$= 1 cm$ ), draw graph for the given frequency table.
So, rectangles

$A, B,C, D$ and

$E$ are constructed.
1880784_1880129_ans_23eefd0d02714aaea24b05cd2aae1af6.png

Draw a histogram for the following frequency distribution

The given distribution is in the inclusive from.
d = lower limit of class - upper limit of a class before it

$= 11 - 10$

$d = 1 , \dfrac{d}{2} = \dfrac{1}{2} = 0 . 5$
1955586_1871158_ans_4b0eebd3f81e4c7184919e853e6e7b6e.png

Construct histogram for the following frequency table :

Class Interval $0-5$ $5-10$ $10-15$ $15-20$ $20-25$
Frequency $18$ $15$ $14$ $8$ $10$

We take

$OX$ and

$OY$ as x-axis and y-axis respectively on the graph paper and

$O$ as the origin. We represent class interval on the x-axis by choosing the scale as

$1 cm = 5$ units. We represent the frequency on y-axis by taking the scale as

$1 cm = 4$ .
We now draw graph for the given frequency table, so rectangles

$A, B, C, D$ and

$E$ are constructed.

1880782_1880128_ans_22b8565eac584cebb8adb5f3a44d65ec.png

The value of $\pi$ upto $50$ decimal places is given below.
$3.14159265358979323846264338327950288419716939937510$
(ii) What are the most and the least frequency occurring digits?

(ii) The most frequently occurred digits are

$3$ and

$9$ while

$0$ occurs least frequently.
1880757_1880112_ans_8c48d15626164272a9598614ed3d92e5.png

Find the range ad coefficient of range of the following table -
x 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5
f 4 5 6 3 2 1 3 5

Lower limit (S) = 4.5
Upper limit (L) = 11.5
Range = L - S

$11.5 - 4.5 = 7$
Coefficient of Range =

$\dfrac{L - S }{L + S}$

$= \dfrac{11.5 - 4.5}{11.5 + 4.5}$

$= \dfrac{7}{16} = 0.44$
Thus range is 7 and coefficient of range is 0.44

What are primary and secondary data Differentiate them?

When an investigator collects data himself with a definite plan or design in his/her mind is called primary data.
Data which are not originally collected rather obtained from published or unpublished source are known as secondary data.

1882567_1880052_ans_c7114faf73ca4d068d0fdd7dace7c7de.PNG

1882567_1880052_ans_c7114faf73ca4d068d0fdd7dace7c7de.PNG

Variate (x) 5 10 15 20 25 30
Frequency (f) 2 6 4 1 5 2
Construct a frequency polygon for the following frequency distribution.

The given distribution is ungrouped. So, by taking the scale

$1 cm = 5$ units along x-axis and

$1 cm = 1$ unit along y-axis respectively, we mark the variate and frequencies.
Now, the point

$(5, 2), (10, 6), (15, 4), (20, 1), (25, 5)$ , and

$(30, 2)$ are plotted and joined. Since, the value of the variate

$0$ comes before the first variate, so we join the first point

$(5, 2)$ to the origin

$(0, 0)$ . Similarly,

$35$ comes ahead than the last variate, so we join the last point

$(30, 2)$ to the point

$(35, 0)$ .
Thus, the required frequency polygon is obtained
1883150_1880136_ans_7276f0ee2f714b7fbb0c159f843d8033.png

Construct a frequency polygon for the following frequency distribution.

To draw a frequency polygon without a histogram, first we find the class-mark of the classes.
1883152_1880138_ans_339c5ab315c94655bf8cc5024831a31e.PNG

1883152_1880138_ans_339c5ab315c94655bf8cc5024831a31e.PNG

Class interval $5 - 9$ $10 - 14$ $15 - 19$ $20 - 24$
Frequency $3$ $5$ $4$ $2$
Construct histogram for the following frequency distribution.

The given frequency distribution is grouped but not continuous. So, first we will make them regular.
Lower limit of

$2^{\text{nd}}$ class interval

$= 10$ ;
Upper limit of

$1^{\text{st}}$ class interval

$= 9$
Difference

$(h) = 10 - 9 = 1$

$\implies \dfrac{h}{2} = \dfrac{1}{2} = 0.5$
We subtract

$0.5$ from the lower limits and add

$0.5$ in the upper limits to make the class intervals regular.
Here, we take

$OX$ and

$OY$ as the x-axis and y-axis respectively on a graph paper with

$O$ as the origin.

We take a suitable scale on both x-axis

$(5$ units

$= 1 \text{ cm})$ and y-axis

$(2$ units

$= 1 \text{ cm})$ .

Then we draw the graph for the given frequency distribution and obtains the shown histogram

1882644_1880131_ans_3de07ae6ee20429eb4bdba2ae0913714.png

Mid values 8 14 20 26 32
Frequency 10 15 25 9 6
Construct histogram from the following distribution.

Here, the given frequency distribution is ungrouped and mid values of the distribution are given. So, with the help of class marks, we make continuous grouped frequency distribution.

1880793_1880132_ans_a724e239ca604276b749ed6c2629205e.png

1880793_1880132_ans_a724e239ca604276b749ed6c2629205e.png

Class Interval $3-6$ $6-12$ $12-13$ $13-14$ $14-15$
Frequency $150$ $420$ $100$ $110$ $50$
Construct histogram for the following frequency table:

Let us represent the class interval along x-axis and the frequency along y-axis. Since, the intervals are not of equal width, therefore we have to construct the rectangles with the classes as their corresponding heights will be as follows :

$Height = \frac{Frequency}{Width of the corresponding interval}$
(i)

$\frac{150}{3} = 50$
(ii)

$\frac{420}{6} =70$
(iii)

$\frac{100}{1} = 100$
(iv)

$\frac{110}{1} = 110$
(v)

$\frac{50}{1} = 50$

1880788_1880130_ans_9aacbb36abaf47f3b4870d6313d2dff9.png

What is an individual series ?

In an individual series, the original value of each individual item or limit is written separately. In such series, frequency of each item or value is one.

The following frequency distribution table shows marks obtained by 180 students in Mathematics examination :
Marks Number of Students
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50 25
$x$
30
2 $x$
65
Find the value of $x$ .
Also draw a histogram representing the above information.

1896570_1911702_ans_89aad4e751ba44258c946bc357e1a095.jpeg

How many types of series are there on the basis of quantitative/numerical classification?

On the basis of quantitative or numerical classification, series can be divided into three parts namely
(i) Individual Series
(ii) Discrete Series
(iii) Continuous Series

Make class intervals from the following mid-values :

Mid-value is the average value of the upper and lower limits of a class.

1963005_1880262_ans_1b47accd3883496aa03d27b4e2138a5e.png

Fill in the blanks in the following table :

The cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors. The last value will always be equal to the total for all observations, since all frequencies will already have been added to the previous total.
1963018_1880276_ans_038020ae65ec42fa9c6178a0e890b717.png

1963018_1880276_ans_038020ae65ec42fa9c6178a0e890b717.png

What is exclusive series ?

In exclusive series, the upper limit of the previous class is the same as the lower limit of the next class. In exclusive series, the upper limit of a class are not included in that class.
For example : In the class

$5 - 10, 10$ is not included. Similarly in

$10 - 20, 20$ is not included.

What is inclusive series ?

In inclusive series, both the limits of a class are included in that class.

For example : In

$0 - 9, 0$ and

$9$ both are included. Similarly in

$10 - 19, 10$ and

$19$ both are included.

The mean of five numbers is $27$ . If one number is excluded, then the new mean is $25$ . Find the excluded number.

Let five numbers are

$x_1, x_2, x_3, x_4, x_5$

$\bar{x} = \frac{x_1 + x_2 + x_3 + x_4 + x_5}{5}$

$\Rightarrow 27 = \frac{x_1 + x_2 + x_3 + x_4 + x_5}{5}$

$\Rightarrow 5 \times 27 = x_1 + x_2 + x_3 + x_4 + x_5$

$\Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 = 135$ ...(i)

Suppose

$x_5$ is excluded, then new mean becomes

$25$

i.e.

$\frac{x_1 + x_2 + x_3 + x_4 }{4} = 25$

$\Rightarrow = x_1 + x_2 + x_3 + x_4 = 100$

Subtracting (ii) from (i), we get the excluded number,

$x_5 = 135 - 100 = 35$

Change the following frequency distribution in exclusive method.

To convert from inclusive to exclusive

1. First find the difference between the upper limit of a class interval and the lower limit of the next class interval.

2. Half the difference is added to the upper limit of each class interval and remaining half is deducted from the lower limit of each class-interval.

Step 1 :

$9010-9000=10$

Step 2 :

$10/2=5$

Now, on subtracting 5 from lower limit of each class and adding 5 to upper limit of each class, we get the above table.

1963015_1880275_ans_f167ec8ae5c74720925dae691da130e5.png

Find the missing frequencies in the following frequency distribution, if the mean of the distribution is $1.46$ .

${\textbf{Step 1: Make the Equation}}$

${\text{Given that }}\sum {{f_i}} = 200$

$\therefore {\text{ }}200 = 86 + {f_1} + {f_2}$

$\Rightarrow {f_1} + {f_2} = 200 - 86$

$\Rightarrow {f_1} + {f_2} = 114{\text{ - - - - - - (1)}}$

${\text{Also, mean }}\left( {\mathop x\limits^ - } \right){\text{ is given as 1}}{\text{.46}}$

${\text{So, }}\therefore {\text{1}}{\text{.46 = }}\dfrac{{140 + {f_1} + 2{f_2}}}{{86 + {f_1} + {f_2}}}$ $\left[\because \textbf{Mean}=\boldsymbol{\dfrac{\sum f_i x_i}{\sum f_i}}\right]$

$\Rightarrow 1.46 = \dfrac{{140 + {f_1} + 2{f_2}}}{{86 + 114}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\textbf{ - - - [from (1)]}}$

$\Rightarrow 1.46 = \dfrac{{140 + {f_1} + 2{f_2}}}{{200}}$

$\Rightarrow 292 = 140 + {f_1} + 2{f_2}$

$\Rightarrow {f_1} + 2{f_2} = 152\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\textbf{ - - - - (2)}}$

${\textbf{Step -2: Find }} \boldsymbol{f_1\ \&\ f_2}\textbf{ from equation (1) and (2)}$

${\text{Subtract (1) from (2) we get}}$

${\text{ }}{f_1} + 2{f_2} - {\text{ }}{f_1} - {f_2} = 152 - 114$

$\Rightarrow {\text{ }}{f_2} = 38$

${\text{Put the value of }}{f_2}{\text{ in (1) we get}}$

${f_1} + {f_2} = 114$

$\Rightarrow {f_1} + 38 = 114$

$\Rightarrow {f_1} = 76$

${\textbf{ Hence, the value of }}{f_1}{\textbf{ and }}{f_2}{\textbf{ is }}{f_1} = 76{\textbf{ and }}{f_2} = 38.{\textbf{ }}$

The Patel family flies from their home town, H, to Kiruna , K , in Lapland.
The Patel family has six suitcases.
The number of items in each suitcase is shown below.
$15$ $16$ $16$ $18$ $19$ $21$
Find the range.

Number of items in each suitcase are arranged in ascending order as,

$15,\quad 16,\quad 16,\quad 18,\quad 19,\quad 21$

Range of the series is difference between Largest and lowest value in given series

$\therefore Range=21-15=6$

Calculate the mean from the following frequency distribution:
X 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89
No. of Persons 32 42 40 56 20 6 2 2

Solution is given by:
2157599_2031927_ans_da38b962b2ea459ea88487bc8be2292f.PNG

Find the class mark of class interval 100-120.

we know that, The class midpoint (or class mark) is a specific point in the center of the class in a frequency distribution table and average of the upper and lower class limits.

$\text{We have given interval is 100 - 120.}$

$\text{Here upper limit is 120 and lower limit is 100.}$

$\therefore$

$\text{Mid point (Class mark) =}$

$\dfrac{100 + 120}{2}$

$\boldsymbol{\left[\because \text{Class mark =} \dfrac{\text{upper + lower}}{2}\right]}$

$= \dfrac{220}{2}$

$= 110$

$\textbf{Hence, answer is 110.}$

Find the class mark of the class 31-40.

Class mark = (Upper limit + Lower limit)/2 = (31 + 40)/2 = 71/2 = 35.5.

Prepare a frequency distribution by inclusive method, taking class interval of 7 from the following data: 28,17,15,22,29,21,23,27,18,12,7,2,9,4,6,1,8,3,10,5,20,16,12,8,4,33,27,21,15,9,3,36,27, 18,9,2,4,6,32,31,29,18,14,13,15,11,9,7,1,5,37,32,28,26,24,20,19,25,19,20

2169893_2037244_ans_c9eb7e29db2a42bc88a339950383ec31.jpg

Convert the normal distribution into more than and less than series:
Class 0-9 10-19 20-29 30-39
Frequency 12 20 22 18

More than series:

Class	Frequency
More than 0	72
More than 10	72 - 12 = 60
More than 20	60 - 20 = 40
More than 30	40 - 22 = 18

Less than series:

Class	Frequency
Less than 9	12
Less than 19	12 + 20 = 32
Less than 29	32 + 22 = 54
Less than 39	54 + 18 = 72

Find out the mean from the following data:
X 12 15 18 21 24 27 30
Y 6 8 10 12 14 16 18

2143225_2023428_ans_423bfa9994c94a41930c929a0131b3d6.png

Prepare a frequency table taking class intervals 20-24, 25-29, 30-34 and so on from the following data:
21 20 55 39 48 46 36 54 42 30
29 42 32 40 34 31 35 37 52 44
54 36 41 33 23 39 28 44 45 38
39 45 37 33 51 53 52 46 43 47
51 26 52 48 25 34 37 33 36 27

Table is given by:
2157594_2031922_ans_9f169401802d4b66b6a6915e88755249.jpeg

Calculate mean from the following data:
Size
Frequency
20
14
25
18
30
33
35
30
40
20
45
15
50
13
55
7

The following distribution gives the mass of $48$ objects measured to the nearest gram. Draw a histogram to illustrate the data.
Mass in (gms) $10-19$ $20-24$ $25-34$ $35-49$ $50-54$
No. of objects $6$ $4$ $12$ $18$ $8$

Re-arrange the class intervals

$9.5-19.5,19.5-24.5,24.5-34.5,34.5-49.5,49.5-54.5$

1174910_619672_ans_a78baba4071748ba878c043bf6e9f472.png

The median of the following data isFind the values of p and q, if the sum of all the frequencies is 90.
Marks Frequency
20-30
30-40
40-50
50-60
60-70
70-80
80-90
p
15
25
20
q
8
10

1229685_1322152_ans_3a0a5a3c6ea84d3cae8522da433313b4.PNG

In the order to study the qualities of apples and grade them, 40 apples were randomly chose and weighed the weights (in gram) are given her. Write the upper and lower limits of the class intervals that you will form and prepare a frequency table .
160,145,122,167,130,137,142,174, 150,155,157,144,130,140,154,150,128,132,140,168,
170,167,171,163,137,170,160,161,135,140,139,166,161,171,159,169,141,150,154,152

2042672_1146856_ans_c957522925404e2a9b263bab402bccac.jpg

The following are the daily expeditures (in $Rs$ ) of $30$ student in a school.
Ependiture (in $Rs$ )
No. of students
$10-15$ $3$
$15-20$ $5$
$20-25$ $10$
$25-30$ $6$
$30-35$ $4$
$35-40$ $2$
Total $30$
How many student spend daily between $20$ and $25$ rupees?

$10$ students spend daily between 20 to 25 Rupees

If $x _ { i }$ 's are the mid-points of the class interval of grouped data, $f _ { i }$ 's are the corresponding frequencies and $\overline { x }$ is the mean, then find $\Sigma \left( f _ { i } x _ { i } - \overline { x } \right)$

1976091_1332912_ans_eb896c500898410d8b6d15b0c22d7061.jpg

In the following frequency distribution of $165$ observations, the mode is $34.5$ . Find the missing frequencies $a$ and $b$ .
Class $5-14$ $14-23$ $23-32$ $32-41$ $41-50$ $50-59$ $59-68$
Frequency $5$ $11$ $a$ $53$ $b$ $16$ $10$

$\Rightarrow$ Here, modal class

$=32-41$

$\Rightarrow$

$l=32,\,f_0=a,\,f_1=53,\,f_2=b$ and

$h=9$

$\Rightarrow$

$Mode=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h$

$\Rightarrow$

$34.5=32+\dfrac{53-a}{2\times 53-a-b}\times 9$

$\Rightarrow$

$\dfrac{2.5}{9}=\dfrac{53-a}{106-a-b}$

$\Rightarrow$

$265-2.5a-2.5b=477-9a$

$\Rightarrow$

$7.5a-2.5b=212$

$\Rightarrow$

$75a-25b=2120$

$\therefore$

$13a-5b=424$ ------ ( 1 )

As total observation are

$165$

$\Rightarrow$ So,

$5+11+a+53+b+16+10=165$

$\Rightarrow$

$95+a+b=165$

$\therefore$

$a+b=70$ -------- ( 2 )

$\therefore$

$b=70-a$

Substitute

$b=70-a$ in equation ( 1 )

$\Rightarrow$

$13a-5(70-a)=424$

$\Rightarrow$

$13a-350+5a=424$

$\Rightarrow$

$18a=424+350$

$\Rightarrow$

$18a=774$

$\therefore$

$a=43$

$\Rightarrow$

$b=70-43=27$

$\therefore$ Frequency

$a=43$ and

$b=27$

The average frequency is $18$ . Find the missing frequency $f$ .
Class Interval Frequency
$11-13$ $7$
$13-15$ $6$
$15-17$ $9$
$17-19$ $13$
$19-21$ $f$
$21-23$ $5$
$23-25$ $4$

$\begin{array}{l} \dfrac { { 7+6+9+13+f+5+4 } }{ 7 } =18 \\ 44+f=18\times 7 \\ f=126-44=82 \end{array}$

The following are the daily expeditures (in $Rs$ ) of $30$ student in a school.
Ependiture (in $Rs$ )
No. of students
$10-15$ $3$
$15-20$ $5$
$20-25$ $10$
$25-30$ $6$
$30-35$ $4$
$35-40$ $2$
Total $30$
Find the class-marks of $20-25$ and $25-30$ .

Class markes of

$20-25$ is

$22.5$ and

$25-30$ is

$27.5$

The mean of the following frequency distribution is 62.8 and the sum of all frequencies isCompute the missing frequencies ${ f }_{ 1 }$ and ${ f }_{ 2 }$ .
Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 Total
Frequency 5 ${ f }_{ 1 }$ 10 ${ f }_{ 2 }$ 7 8 50

Given mean of frequency distribution is

$62.8$

& the sum of all frequencies is

$50$

Class	mid point of class $(x_i)$	frequency $(f_i)$	$f_i x_i$
$0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$	$10$ $30$ $50$ $70$ $90$ $110$	$5$ $f_1$ $10$ $f_2$ $7$ $8$	$50$ $30 f_1$ $500$ $70 f_2$ $630$ $880$
		$50$	$2060+30f_1+70 f_2$

$\therefore \sum f_i = 50$

$30 + f_1 + f_2 = 50 \Rightarrow f_1 + f_2 = 50 - 30 = 20$

$f_1 + f_2 = 20$ ___(1)

mean

$\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{2060 + 30 f_1 + 70 f_2}{50}$

$62.8 \times 50 = 2060 + 30 f_1 + 70 f_2$

$30 f_1 + 70 f_2 = 3140 - 2060=1080$

$10 (3f_1 + 7f_2) = 1080$

$3f_1 + 7f_2 = 108$ __(2)

$f_1 = 9 \Rightarrow (1) \Rightarrow f_2 = 20 - f_1 = 20 - 9$

$f_2 = 11$

$\therefore f_1 = 9 , f_2 = 11$

The marks obtained by $40$ students of a class in an examination are given:
$8, 47, 22, 31, 17, 13, 38, 26, 3, 34, 29, 11, 22, 7, 15, 24, 38, 31, 21, 35, 42, 24,$ $45, 23, 21, 27, 29, 49, 25, 48, 21, 15, 18, 27, 19, 45, 14, 34, 37, 34$ .

Prepare a frequency distribution table with equal class intervals, starting from $0-10$ [where $10$ is not included].

Given that the marks obtained by

$40$ students of a class in an examination are

$8, 47, 22, 31, 17, 13, 38, 26, 3, 34, 29, 11, 22, 7, 15, 24, 38, 31, 21, 35, 42, 24,$

$45, 23, 21, 27, 29, 49, 25, 48, 21, 15, 18, 27, 19, 45, 14, 34, 37, 34$ .

the frequency distribution table with equal class intervals, starting from

$0-10$ [where

$10$ is not included].

frequency table

class intervals	Frequency
$0-10$ $10-20$ $20-30$ $30-40$ $40-50$	$3$ $8$ $14$ $9$ $6$
Total	$40$

The final marks in mathematics of 30 students are as follows:
$53,61,48,60,78,68,55,100,67,90$
$75,88,77,37,84,58,60,48,62,56$
$44,58,52,64,98,59,70,39,50,60$
Now answer the following:
How many have scored less than 50?

Given

final marks in mathematics of 30 students are as follows

$53,61,48,60,78,68,55,100,67,90$

$75, 88,77,37,84,58,60,48,62,56$

$44, 58,52, 64,98,59,70,39,50,60$

the observations arrange in ascending order

$37,39,44,48,48,50,52,53,55,56,58,58,59,60,60,60,$

$61,62,64,67,68,70,75,77,78,84,88,90,98,100$ .

The observations between

$50$ and

$60$ have not actually appeared are

$51, 54, 57$

$\rightarrow$ No of have marks scored less then

$50$ is

$5$

$(37,39,44,48,48)$

The final marks in mathematics of 30 students are as follows:
$53,61,48,60,78,68,55,100,67,90$
$75,88,77,37,84,58,60,48,62,56$
$44,58,52,64,98,59,70,39,50,60$
Arrange these marks in the ascending order, 30 to 39 one group, 40 to 49 second group, etc.

Final marks in mathematics of 30 students are as follows

$53, 61,48,60,78,68,55,100,67,90$

$75,88,77,37,84,58,60,48,62,56$

$44,58,52,64,98,59,70,39,50,60$

the observations arrange in ascending order

$37,39,44,48,48,50,52,53,55,56,58,58,59,60,60,60,$

$61,62,64,67,68,70,75,77,78,84,88,90,98,100$

Group	Ascending order
$30-39$ $40-49$ $50-59$ $60-69$ $70-79$ $80-89$ $89- 99$ $100 - 109$	$37,39$ $44,48, 48$ $50,52,53,55,56,58,58,59$ $60,60,60,61,62,64,67,68$ $70,75,77,78$ $84,88$ $90,98$ $100$

the mean of the following data is 50. Find the missing frequencies $f_{1}$ and $f_{2}$
C.L 0-20 20-40 40-60 60-80 80-100 Total
Frequency 17 $f_{1}$ 321 $f_{2}$ 19 120

Now we can prepare a table for calculating missing frequencies

Class limits	frequency (f)	midpoint (x)	fx
$0-20$ $20-40$ $40-60$ $60-80$ $80-100$	$17$ $f_1$ $321$ $f_2$ $19$	$10$ $30$ $50$ $70$ $90$	$170$ $30 f_1$ $16,050$ $70f_2$ $1710$
	$\sum f = 120$		$\sum fx = 17930+30f_1+70f_2$

$\therefore 357 + f_1 + f_2 = 120$

$f_1 + f_2 = -237$ __(1)

we know mean

$= \dfrac{\sum fx}{\sum f} = \dfrac{17930 + 30 f_1 + 70 f_2}{120}$

$50 \times 120 = 17930 + 30 f_1 + 70 f_2$

$30 f_1 + 70 f_2 = 17930 - 6000 = 11930$

$30 f_1 + 70 f_2 = 11930$ __(2)

$(1) \times 70 \Rightarrow$

$70 f_1 + 70 f_2 = -16590$

$30 f_1 + 70 f_2 = 11930$

$\underline{-\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,}$

$40 f_1 = -28520$

$f_1 = -713$

$f_1 + f_2 = -237$

$f_2 = -237 + 713 = 476$

The class size of a distribution is 25 and the first class-interval is $200-224$ . There are seven class-intervals.
Write the class-marks of each interval.

class-marks of each interval are

$212,237,262,287,312,337,362$
1669541_1670841_ans_1eec561fefad46f3b3928cec6edac13c.png

then mean of the following frequency distribution is 57.6 and he sum of all the frequencies isFind the values of p and q.
Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 7 p 12 q 8 5

1894565_1484059_ans_9f89c7b7951746cbb850adf95b1967c2.jpg

Construct a frequency polygon for the given data on a graph sheet of paper
Height (in cm) 102 - 108 112 - 118 122 - 128 132 - 138 142 - 148 152 - 158
No. of pupils 8 12 5 16 35 23

1486544_1535254_ans_3aa180bae0064ad9be368a465521efc0.jpg

The class marks of frequency distribution are $47,52,57,62,67,72,77,82,87,92,97,102$ . Find the class size and class unit

1486562_1562903_ans_b9642d1cc5d240c391e1faee043996b7.jpg

The mean of the following distribution is 48 and sum of all the frequencies isFind the missing frequencies x and y.
Class 20-30 30-40 40-50 50-60 60-70
Frequency 8 6 x 11 y

2150837_1542079_ans_52f593e62239484280971a30f0880f72.jpg

The weight of new born babies (in kg) in a hospital on a particular day are as follows:
$2.3,2.2,2.1,2.7,2.6,3.0,2.5,2.9,2.8,3.1,2.5,2.8,2.7,2.9,2.4$
Rearrange the weights in weights in descending order.

1659624_1670819_ans_3126dca68989489f832da0f297131bbb.jpg

The monthly wages of $30$ workers in a factory are given below:
$830,835,890,810,835,836,869,845,898,890,820,860,832,833,855,$
$845,804,808,812,840,885,835,836,878,840,868,890,806,840,890$ .
Represent the data in the form of a frequency distribution with class size $10$ .

The monthly wages of

$30$ workers in a factory are

$830,835,890,810,835,836,869,845,898,890,820,860,832,833,855,$

$845,804,808,812,840,885,835,836,878,840,868,890,806,840,890$ .

Frequency distribution with class size -

$10$

$800 - 810$ (here

$810$ not included)

Class interval	frequency
$800-810$ $810-820$ $820-830$ $830-840$ $840-850$ $850-860$ $860-870$ $870-880$ $880-890$ $890-900$	$3$ $2$ $1$ $8$ $5$ $1$ $3$ $1$ $1$ $5$
Total	$30$

The monthly profits (in Rs.) of 100 shops are distributed as follows:
Profits per shop: 0-50 50-100 100-50 150-200 200-250 250-300
No. of shops: 12 18 27 20 17 6
Draw a histogram for the data and show the frequency polygon for it.

To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-intyervals and the frequencies of the class-intervals respectively.

The given data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights. The scale for horizontal axis may not be same as the scale for vertical axis. Let us take one vertical division is equal to

$3$ shops.

The heights of the different rectangles are as following

$1.$ The height of the rectangle corresponding to the class-interval

$0-50$ is

$\dfrac{12}{3}=4$ big divisions.

$2.$ The height of the rectangle corresponding to the class-interval

$50-100$ is

$\dfrac{18}{3}=6$ big divisions.

$3.$ The height of the rectangle corresponding to the class-interval

$100-150$ is

$\dfrac{27}{3}=9$ big divisions.

$4.$ The height of the rectangle corresponding to the class-interval

$150-200$ is

$\dfrac{20}{3}=6.67$ big divisions.

$5.$ The height of the rectangle corresponding to the class-interval

$200-250$ is

$\dfrac{17}{3}=5.67$ big divisions.

$6.$ The height of the rectangle corresponding to the class-interval

$250-300$ is

$\dfrac{6}{3}=2$ big divisions.

For frequency polygon, first we will obtain the class marks as given in the following table.

Profits per shop	Class Marks	Number of shops
$0-50$	$25$	$12$
$50-100$	$75$	$18$
$100-150$	$125$	$27$
$150-200$	$175$	$20$
$200-250$	$225$	$17$
$250-300$	$275$	$6$

We plot the points

$(25, 12), (75, 18), (125, 27), (175, 20), (225, 17)$ and

$(275, 6)$ .

Now, we join the plotted points by line segments . The end points

$(25, 12)$ and

$(275, 6)$ are joined to the mid-points

$(−25, 0)$ and

$(325, 0)$ respectively of imagined class-intervals to obtain the frequency polygon.

1482798_1671078_ans_08740c355eec4244a43dd2e05fde2340.png

The monthly wages of $30$ workers in a factory are given below:
$830, 835, 890, 810, 835,836, 869, 845, 898, 890,$
$820, 860, 832, 833, 855, 845, 804, 808, 812,840,$
$885, 835, 836, 878, 840, 868, 890, 806, 840, 890$ .
Represent the data in the form of a frequency distribution with class size $10$ .

Given that

the monthly wages of

$30$ workers in a factory are

$830, 835, 890, 810, 835,836, 869, 845, 898, 890,$

$820, 860, 832, 833, 855, 845, 804, 808, 812,840,$

$885, 835, 836, 878, 840, 868, 890, 806, 840, 890$ .

Frequency distribution with class size

$10$

$800-810$ (there 810 not inclcded)

Class interval	Frequency
800 - 810	3
810 - 820	2
820 - 830	1
830 - 840	8
840 - 850	5
850 - 860	1
860 - 870	3
870 - 880	1
880 - 890	1
890 - 900	5
Total	30

Define the following terms :
Class limits

Class limits

Corresponding to a class internal, the class limits may be defined as the minimum value and the maximum value the class interval may contain.

The minimum value is known as the lower class limit (LCL) and the maximum value is known as the upper class limit (UCL)

Ex.

Suppose class interval

$10 - 14$

Class internal	Frequency
10 - 14	8
15 - 19	3
20 - 24	2

Here

$10$ is lower limit

$14$ is upper limit

Define the following terms :
Class size

Difference between the true upper limit and true lower limit of a class interval is called the class size.

$\to$ class size remains the same for all class intervals

for the class internal

$10-20$

Class size

$= 20 - 10$

$=10$

The daily minimum temperatures in degrees Celsius recorded in a certain Arctic region are as follows:
$-12.5,-10.8,-18.6,-8.4,-10.8,-4.2,-4.8,-6.7,-13.2,-11.8,-2.3,1.2,2.6,0,2.4,0,3.2,2.7,$
$3.4,0,-2.4,-2.4,0,3.2,2.7,3.4,0,-2.4,-5.8,-8.9,-14.6,-12.3,-11.5,-7.8,-2.9.$
Represent them as frequency distribution table taking $-19.9$ to $-15$ as the first class interval.

From given data,

Frequency distribution table

Class interval	Frequency
$(-19.9)\ - \ (-15)$ $(-14)\ - \ (-9.1)$ $(-9.2)\ - \ (-4.3)$ $(-4.2)\ -\ (0.7)$ $(0.6)\ -\ (5.5)$	$1$ $8$ $6$ $11$ $9$
Total	$35$

The daily maximum temperatures (in degree celsius) recorded in a certain city during the month of November are as follows:
$25.8,24.5,25.6,20.7,21.8,20.5,20.6,20.9,22.3,22.7,23.1,22.8,22.9,21.7,21.3,$
$20.5,20.9,23.1,22.4,21.5,22.7,22.8,22.0,23.9,24.7,22.8,23.8,24.6,23.9,21.1$
Represent them as a frequency distribution table with class size $1^{\circ}C$ .

The daily maximum temperatures (in degree celsius) recorded in a certain city during the month of November are as follows:

$25.8,24.5,25.6,20.7,21.8,20.5,20.6,20.9,22.3,22.7,23.1,22.8,22.9,21.7,21.3,$

$20.5,20.9,23.1,22.4,21.5,22.7,22.8,22.0,23.9,24.7,22.8,23.8,24.6,23.9,21.1$

Frequency distribution table with class size

$1^{\circ} C$ .

Class interval	Frequency
$20-21$ $21-22$ $22-23$ $23-24$ $24-25$ $25-26$	$6$ $5$ $9$ $5$ $3$ $2$
Total	$30$

Explain the reading and interpretation of bar graphs.

Reading and interpretation of Bar graph:

The first step in reading a bargraph is to know what does it represent. For this we read the caption which is generally written at the bottom of horizontal line x-axis and adjacent to vertical line y-axis.

After this we understand the scale of the graph in order to know the precise value of the given data.

After reading the bar graph, one must drawn conclusions from it. Drawing various conclusions from a given bar graph is referred as the interpretation of the bar graph.

The heights (in cm) of $30$ students of class VIII are given below:
$155, 158, 154, 158, 160, 148, 149, 150, 153, 159, 161, 148, 157, 153, 157,$
$162, 159, 151, 154, 156, 152, 156, 160, 152, 147, 155, 163, 155, 157, 153$ .
Prepare a frequency distribution table with $160-164$ as one of the class intervals.

Given that

The heights of 30 students of class

$IX$ are given below

$155, 158, 154, 158, 160, 148, 149, 150, 153, 159, 161, 148, 157, 153, 157,$

$162, 159, 151, 154, 156, 152, 156, 160, 152, 147, 155, 163, 155, 157, 153$ .

Frequency distribution table

Class interval	Frequency
145 - 149	4
150 - 154	8
155 - 159	13
160 - 164	5
165 - 169	-
Total	30

Draw, in the same diagram, a histogram and a frequency polygon to represent the following data which shows the monthly cost of living index of a city in a period of $2$ years:
Cost of living index: $440-460$ $460-480$ $480-500$ $500-520$ $520-540$ $540-560$ $560-580$ $580-600$
No. of months: $2$ $4$ $3$ $5$ $3$ $2$ $1$ $4$

The histogram and frequency polygon of the given data is as follows:

Refer Fig.

1669384_1671041_ans_3434b35611e24d0580996ba052d9567a.png

Draw a frequency polygon of the following distribution of the students obtaining marks in Economics.
Marks 0-10 10-20 20-30 30-40 40-50 50-60
Students 5 12 15 22 14 4

What is the class interval ( $h$ ) to calculate median?

Median class is

$70-74$ or

$69.5-74.5$ in continuous classes.

Class interval is given by,

$h=Upper\quad Limit-Lower\quad Limit$

$\therefore h=74.5-69.5$

$\therefore h=5$

The following data shows the average age of men in various countries in a certain year:
Country India Nepal China Pakistan U.K. U.S.A.
Average age (in years) 55 52 60 50 70 75
Represent the above information by a bar graph.

Given data shows the average age of man in various countries in a certain year

Country India Nepal China Pakistan U.K U.S.A

Average use(in years)

$56$

$52$

$60$

$50$

$70$

$75$

We can construct the bar graph as shown above by taking the countries on

$x-axis$ and the average age on the

$y-axis$ .

The height of each country bar corresponds to the average age, as given in the data. Constructing such bars of each country gives us the required bar graph.

1480913_1675551_ans_e6afc4399cb84bcfa7598e5fcba9d74f.png

Draw a histogram for the following data:
Class interval

$600-640$
$640-680$

$680-720$

$720-760$

$760-800$

$800-840$

Frequency

$18$

$45$
$153$

$288$

$171$

$63$

1617655_1775142_ans_e53360ca3dc64cdbb524d32108c1ce68.png

The following table gives the route length (in thousand kilometres) of the Indian Railways in some of the years:
Year $1960-1961$ $1970-1971$ $1980-1981$ $1990-1991$ $2000-2001$
Route length (in thousand kilometres) $56$ $60$ $61$ $74$ $98$
Represent the above data with the help of a bar graph.

The bar graph shown below is representing the route length (in thousand kilometers) of the Indian Railways in some of the years.

1480974_1675520_ans_1b7f97abd9a9463797b78c6593ab85f9.png

The following table shows the interest paid by a company (in lakhs):
Year 1995-96 1996-97 1997-98 1998-99 1999-2000
Interest (in lakhs of rupees) 20 25 15 18 30
Draw the bar graph to represent the above information.

$Y-$ axis represents interest in lakh and

$x-$ axis represents years
1480683_1677200_ans_c20496a4e12b4c4a91293f533ea06048.png

The bar graph given below shows the marks of students of a class in a particular subject:
Study the bar graph and answer the following questions:
(a) If 40 is the pass mark, then how many students have failed?
(b) How many students got marks from 50 to 69?
(c) How many students scored 90 marks and above?
(d) If students who scored marks above 80 are given merits then how many merit holders are there?
(e) What is the strength of the class?

Observing the bar graph,

(a)

$4$ students have failed and obtained less than

$40$ marks

$(30-39)$

(b)

$7$ and

$11$ students have obtained the marks

$(50-59)$ and

$(60-69)$ respectively.

Therefore,

$7 + 11=18$

$18$ students have obtained

$60$ to

$69$ marks.

(c)

$4$ students secured

$90$ and above marks.

(d)

$6$ students got

$80$ to

$89$ marks and

$4$ students got

$90$ to

$92$ .

Hence,

$6+4=10$ students are merit holders.

(e) Total number of students in the class

$=4+2+7+11+8+6+4=42$

Represent the following data in the form of a frequency distribution:
$52, 56, 72, 68, 52, 68, 52, 68, 52, 60, 56, 72, 56, 60, 64, 56, 48, 48, 64$ and $64$

The required frequency table for the given data is as follows:
Marks Tally marks Frequency

$^{48}{||}2$

$^{52}{||||}4$

$^{56}{||||}4$

$^{60}{||}2$

$^{64}{|||}3$

$^{68}{|||}3$

$^{72}{||}2$
Total

$20$

Define:
frequency of an observation

Frequency of an observation : The number of times a particular observation occurs is known as its frequency

The value of $\pi$ up to 35 decimal places is given below:
$14159265358979323846264338327950288$
Make a frequency distribution of the digits $0$ to $9$ after the decimal point.

Frequency Distribution Table for digits after decimal point in the value of Pi.

Digits	Frequency after decimal point
$0$	$1$
$1$	$2$
$2$	$5$
$3$	$6$
$4$	$3$
$5$	$4$
$6$	$3$
$7$	$2$
$8$	$5$
$9$	$4$
Total $=$	$35$

Convert the given frequency distribution into a continuous grouped frequency distribution:
Class Interval
Frequency
150-153
7
154-157
7
158-161
15
162-165
10
166-169
5
170-173
6
In which intervals would 153.5 and 157.5 be included?

Answer:-

To convert inclusive class interval into an exclusive class interval:
(1) Find the difference between the upper limit of the class interval and the lower limit of the next class interval.

For example, the upper limit of the class interval 150-153 is 153. The lower limit of the next class interval 154-157 is 154. The difference is 154-153=1

(2) Subtract half of the difference from the lower limit of the class interval and add half of that difference to the upper limit of a class interval.

Half of 1 is 0.5. Hence add 0.5 to the upper limit and subtract 0.5 from the lower limit for each class interval.

Thus, we get:-

Class Interval	Frequency
149.5-153.5	7
153.5-157.5	7
157.5-161.5	15
161.5-165.5	10
165.5-169.5	5
169.5-173.5	6

153.5 is included in the class interval 153.5 - 157.5 and 157.5 is included in the class interval 157.5 - 161.5.

The scores (out of $100$ ) obtained by $33$ students in a mathematics test are as follows:
$69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84\\ 66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71\\ 81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69$
Represent this data in the form of a frequency distribution.

Arrange the given scores in ascending order:

$48,48,48,58,58,58,64,64,64,64,66 ,66,66,66 ,66,66,66,69,69,69,69,69,69,71 ,71,71 ,73,73 ,81 ,83 ,83 ,84 ,84$

Scores	No. of Students
$48$	$3$
$58$	$3$
$64$	$4$
$66$	$7$
$69$	$6$
$71$	$3$
$73$	$2$
$81$	$1$
$83$	$2$
$84$	$2$
Total $=$	$33$

The following are the marks (out of 100) of 60 students in mathematics.16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75,31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62,52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4,Construct a grouped frequency distribution table with width 10 of each class starting from 0 - 9.

Answer:-

Arranging the following data in ascending order:

4, 5, 7, 7, 13, 15, 16, 16, 17, 17, 19, 24, 25, 26, 27, 28, 30, 31, 34, 34, 35, 35, 36, 36, 36, 42, 44, 45, 47, 48,

51, 52, 52, 54, 55, 55, 56, 56, 61, 62, 62, 63,

68, 70, 72, 72, 72, 74, 75, 75, 78, 80,, 81, 85, 86, 86, 92, 95, 97.

Marks of 60 students (class Interval)	Tally Marks	Frequency
0-9	$\|\|\|\|$	4
10-19		7
20-29		5
30-39		10
40-49		5
50-59		8
60-69		5
70-79		8
80-89		5
90-99		3

Total frequency = 60

The table shows a frequency distribution of the life time of $400$ radio tubes tested at a company.
Life time (hours) Number of tubes Life time (hours) Number of tubes
$300-399$ $14$ $800-899$ $62$
$400-499$ $46$ $900-999$ $48$
$500-599$ $58$ $1000-1099$ $22$
$600-699$ $76$ $1100-1199$ $6$
$700-799$ $68$
Total $400$
Column $II$ gives data for description given in column $I$ , match them correctly.

No.	Lifetime (hours)	No. of tubes
1	300-399	14
2	400-499	46
3	500-599	58
4	600-699	76
5	700-799	68
6	800-899	62
7	900-999	48
8	1000-1099	22
9	1100-1199	6
	Total	400

A) Upper limit of fifth class - 4)

$799$

B) Lower limit of eighth class - 3)

$1000$

C) Class marks of seventh class =

$\cfrac{upper limit + lower limit}{2}$

$\dfrac{900 + 999}{2}$ =

$949.5$

D) Clas Interval Size = Upper limit - lower limit + 1

$399 - 300 + 1$

$100$

The weekly wages (in Rs) of $30$ workers in a factory are:
$830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855,$
$845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840$
What is the frequency of $830-840$ interval?

Intervals Telly marked frequency

800-810 III 3

810-820 II 2

820-830 I 1

830-840 II/IIIIII 9

840-850 II/II 5

850-860 I 1

860-870 III 3

870-880 I 1

880-890 I 1

890-900 IIII 4

Total 30

The frequency of 830-840 interval is 9

The marks of $30$ students of a class, obtained in a test (out of $75$ ), are given below:
$42, 21, 50, 37, 42, 37, 38, 42, 49, 52, 38, 53, 57, 47, 29$
$59, 61, 33, 17, 17, 39, 44, 42, 39, 14, 7, 27, 19, 54, 51$ .
Form a frequency table with equal class intervals.

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

class intervals

No of students 1 4 3 7 7 7 1 0

The class intervals 30-40 ,40-50 and 50-60 have equal frequency.

Calculate the range from the following data.
Marks $26$ $38$ $54$ $65$ $72$ $88$
No. of students $5$ $10$ $15$ $20$ $25$ $30$

We observe that, highest score

$H = 88$ and lowest score

$L = 26$ .
Therefore, range

$= H - L = 88 - 26 = 62$ marks

Construct a grouped frequency distribution table with width 10 of each class, in such a way that one of the classes is $10 - 20$ (20 not included).

Class Interval	f
0-10	4
10-20	7
20-30	5
30-40	10
40-50	5
50-60	8
60-70	5
70-80	8
80-90	5
90-100	3
Total =	60

Given below are the marks obtained by $30$ students in an examination:
$08$
$17$
$33$
$41$
$47$
$23$
$20$
$34$
$09$
$18$
$42$
$14$
$30$
$19$
$29$
$11$
$36$
$48$
$40$
$24$
$22$
$02$
$16$
$21$
$15$
$32$
$47$
$44$
$33$
$01$

Taking class intervals $1- 10, 11 - 20 ........... ..41-50 ;$ make a frequency table for the above distribution.

$Marks$

$No. of students$

$0-10$

$4$

$10-20$

$8$

$20-30$

$6$

$30-40$

$6$

$40-50$

$6$

$30$

The following figures represent the amount of money spent in various years , Represent them with a suitable diagram .
Years
$1986-87$
$1987-88$
$1988-89$
$1989-90$
Amount ( in crores )
$45$
$70$
$95$
$90$

Define raw data.

The data obtained in the original form is called raw data.
Example: The marks obtained by $5$ students in an English test as given below:
$45, 25, 70, 60, 50$
The data in this form is called raw data.

A TV channel organized a SMs (Short Message Service) poll on prohibition on smoking, giving options like A- complete prohibition, B- prohibition in public places only, C- not necessary. SMS results in one hour were $A\ B\ A\ B\ C\ B\ A\ B\ B\ A\ C\ C\ B\ B\ A\ B\ B\ A\ B\ C\ B\ A\ B\ C\ B\ A\ B\ B\ A\ B\ B\ C\ B\ A\ B\ A\ B\ C\ B\ B\ A\ B\ C\ B\ B\ A\ B\ B\ A\ B\ B\ A\ B\ C\ B\ A\ B\ B\ A\ B\ C\ A\ B\ B\ A$
Represent the above data as grouped frequency distribution table. How many appropriate answers were received? What was the majority of people's opinion?

Option A B C

Frequency (f) 19 35 10

The total appopriate answers=64

Majority of peoples opinion=B

Convert data into grouped frequency distribution:
$3, 5, 1, 2, 3, 2, 2, 1, 6, 9, 5, 3, 9, 2$

First the arrange the data in order,

$1, 1, 2, 2, 2, 2, 3, 3, 3, 5, 5, 6, 9, 9$
The largest number

$= 9$
The smallest number

$= 1$
Range

$=$ largest number

$-$ smallest number

$= 9 - 1 = 8$
Hence, we shall calculate an approximate group size, by dividing the range by how many groups you would like.
If we want

$4$ groups, then Divide range with number of groups

$= \dfrac84 = 2$
The class interval width is

$2$ . Start from

$0$ .

So, the grouped data will be as follows.

Cass interval	Frequency
$0 -2$	$2$
$2 -4$	$7$
$4 -6$	$2$
$6 -8$	$1$
$8 - 10$	$2$

Convert the following data into grouped frequency table:
$12, 1, 2 , 3, 4, 4, 4, 5, 6, 3, 2, 1, 12, 4, 6, 7, 7, 4$

First the arrange the data in order

$1, 1, 2, 2, 2, 2, 3, 3, 3, 5, 5, 6, 9, 9$
The largest number

$= 9$
The smallest number

$= 1$
Range

$=$ largest number

$-$ smallest number

$= 9 - 1 = 8$
Hence, we shall calculate an approximate group size, by dividing the range by the number of groups we want.
If we want

$4$ groups, then Divide range with number of groups

$= \dfrac84 = 2$
The class interval width is

$2$ . Start from

$0$ .

So, the grouped data will be as follows.

Class interval	Frequency
$0 -2$	$2$
$2 -4$	$7$
$4 -6$	$2$
$6 -8$	$1$
$8 - 10$	$2$

Consider the following marks (out of $50$ ) scored in mathematics by $50$ students of $8th$ class:
$41, 31, 33, 32, 28, 31, 21, 10, 30, 22,33, 37, 12,05, 08, 15, 39, 26, 41, 46, 34, 22, 09$ .
$11, 16, 22, 25, 29, 31, 39, 23, 31, 21, 45, 47, 30, 22, 17, 36, 18, 20, 22, 44, 16, 24, 10, 27, 39, 28, 17$ .
Prepare a frequency distribution table

If we prepare a frequency distribution table for each observation, then the table would be too long. So for convenience, we make groups of observations like

$0 - 9, 10 - 19$ and so on. We obtain a frequency of distribution of the number of observations coming under each group. In this way, we prepare a frequency distribution table for the above data as below:

Forty candidates from $10th$ class of a school appear for a test. The number of questions (out of $60$ ) attempted by them in forty five minutes is given here.
$52, 42, 40, 36, 12, 28, 15, 37, 35, 22, 39, 50, 54, 39, 21, 34, 46, 31, 10, 09$ ,
$13, 24, 29, 31, 49, 58, 40, 44, 37, 28, 13, 16, 29, 36, 39, 41, 47, 55, 52, 09$ .
Prepare a frequency distribution table with the class size $10$ and answer the following:
(i) Which class has the highest frequency?
(ii) Which class has the lowest frequency?
(iii) Write the upper and lower limits of the class $(20 - 29)$ .
(iv) Which two classes have the same frequency?

Let us first prepare the frequency distribution table pertaining to this data.
Using this table, we can observe:
(i)

$(30-39)$ has the highest frequency i.e.

$11$ ;
(ii)

$(0-9)$ has the lowest frequency i.e.

$2$ ;
(iii) upper limit is

$29.5$ and lower limit is

$19.5$ of the class

$(20-29)$ ;
(iv)

$(10-19)$ and

$(50-59)$ have the same frequency.

The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
$\begin{matrix} 5 & 3 & 10 & 20 & 25 & 11 & 13 & 7 & 12 & 31 \\ 19 & 10 & 12 & 17 & 18 & 11 & 32 & 17 & 16 & 2 \\ 7 & 9 & 7 & 8 & 3 & 5 & 12 & 15 & 18 & 3 \\ 12 & 14 & 2 & 9 & 6 & 15 & 15 & 7 & 6 & 12 \end{matrix}$

Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as $0-5$ (5 not included). What main features do you observe from this tabular representation?

Mostly engineers who travel from residence to work place travel around

$20 \ km$ .

Classify the following data as primary or secondary data.
(1) Number of trains that passes through Thane railway station.
(2) Number of Barbie dolls available in different Barbie stores in Mumbai.
(3) Number of foodies visit the zomato website.
(4) Number of people visit in a museum during summer.
(5) Number of musicians in Mumbai city who knows both guitar and drum.

Primary data: If the investigation has definite object while collecting the information, then data is called as primary data.

Therefore, (3), (4), (5) are examples of primary data.

Secondary data: If information is gathered from source which already had information, then data is called secondary data.

Therefore, (1), (2) are examples of secondary data.

Find the upper limit of class $50-55$

In every Class interval, there is one Lower Class Limit and one Upper Class Limit

$50-55 ; 50$ is lower class limit and

$55$ is upper class limit

Given below are the marks of students in a less than cumulative frequency distribution table. Write the frequencies of the respective classes. Also write the Greater than cumulative frequencies. How many students marks are given in the table?
Class Interval (Marks) $1 - 10$ $11 - 20$ $21 - 30$ $31 - 40$ $41 - 50$
L.C.F. (No of students) $12$ $27$ $54$ $67$ $75$

$\Rightarrow$ Frequency table are as follows:

$Class\,Interval\,(Marks)$	$L.CF.(No\,of\,students$	$Frequency$
$1-10$	$12$	$12$
$11-20$	$27$	$27-12=15$
$21-30$	$54$	$54-27=27$
$31-40$	$67$	$67-54=13$
$41-50$	$75$	$75-67=8$

$\Rightarrow$ The following table is more than frequency table :

$Class\,Interval(Marks)$	$M.C.F(No\,of\,students)$
More than $1$	$12+63=75$
More than $11$	$15+48=63$
More than $21$	$27+21=48$
More than $31$	$13+8=21$
More than $41$	$8$

$\Rightarrow$ The total number of students are

$75$ .

The heights of $25$ children in centimetre are given below:
$174, 168, 110, 142, 156, 119, 110, 101, 190, 102, 190, 111, 172, 140,$
$136, 174, 128, 124, 136, 147, 168, 192, 101, 129, 114$ .
Prepare a frequency distribution table, taking the size of the class interval
as $20$ , and answer the following:
(i) Mention the class intervals of highest and lowest frequency.
(ii) What does the frequency $6$ corresponding to class interval $(160-180)$ indicate?
(iii) Find out the class mark (or midpoint) of $(140-160)$ .
(iv) What is the range of heights?

The frequency distribution for the given data is as follows:
Answers:
(i) Highest frequency:

$(100-120)$ ; lowest frequency:

$(140-160)$ . and

$(180-200)$ .
(ii) There are

$6$ children whose heights are in the range

$160 cm$ to

$180 cm$ .
(iii)

$\text{Class mark} = \dfrac{140+160}{2} = 150$

(iv)

$\text{Range = highest score - lowest score} \\\ \ \ \ \ \ = 192 -101 \\\ \ \ \ \ \ = 91$

Calculate the range from the following data.

Marks $0-5$ $5-10$ $10-15$ $15-20$ $20-25$
No. of students $5$ $12$ $23$ $8$ $2$

Lower limit of lower class,

$L = 0$

Higher limit of higher class,

$H = 25$

Hence, range

$= H - L$

$= 25 - 0$

$= 25$ marks

The electricity bills (in rupees) of $25$ houses in a locality are given below. Construct a grouped frequency distribution table with a class size of $75$ .
$170, 212, 252, 224, 310, 712, 412, 425, 322, 325, 192, 198, 230, 320, 412, 530, 602, 724, 370, 402, 317, 403, 405, 372, 413$

Electric bill in (Rs) Number of house (f)

class intervals

150-225 4

225-300 3

300-375 7

375-450 7

450-525 0

525-600 1

600-675 1

675-750 2

Draw the histogram for the following frequency distribution.
Class-Interval Frequency
$0 - 5$ $5$
$5 - 10$ $8$
$10 - 15$ $15$
$15 - 20$ $4$
$20 - 25$ $10$

The given distribution is in exclusive form. So, we can take the class-intervals as

$(0-5), (5-10)$ etc. along the

$x-axis$ and frequency along

$y-axis$ . Choosing a proper scale, we can construct a histogram as shown in the figure

Write the upper limit and the lower limit of the class $21 - 25$ .

Upper limit of the given class

$= 25$
Lower limit of the given class

$= 21$

Find the class mark of the class $35-39$ .

Class mark of a class

$a-b$ is given by

$\dfrac{a+b}{2}$

Hence, the class mark of given class

$35-39$ is

$=\dfrac{35+39}{2}=37$

Find the class mark of the classes $20 - 30$ and $30 - 40$

Class mark of class

$20 - 30 = \dfrac {20 + 30}{2} = 25$

and Class mark of class

$30 - 40 = \dfrac {30 + 40}{2} = 35$

Write the lower class limit of class $20-30$ .

Lower class limit of class

$20-30$ is

$20$ .

Write the class-size in each of the following :
$10-19,20-29,30-39$

Class size is

$9, 9, 9$

Hence, this is the answer.

Write the class size and class limits in each of the following:.
$104,114,124,134,144,154$ and $164$

Class size $=10$

As class size is the difference between upper limit and lower limit

EX : $114 - 104$

Class size = $10$

Class limit : the lowest and highest marks which are included in a class are called lower class limit and upper class limits of the class.

Ex : $104 -114$

lower class limit is 104 and upper class limit = $114$

Hence, this is the answer.

The marks of $30$ student of a class, obtained in a test (out of $75$ ), are given below:
$4,21,50,37,68,42,37,38,42,49,52,38,53,57,47,29,64, 29,63,33,17,17,39,44,42,7,27,19,54,51$

Form a frequency table with equal class intervals.

Class Interval	Number of Students
$0-15$	$2$
$15-30$	$7$
$30-45$	$10$
$45-60$	$8$
$60-75$	$3$

Total number of students in the class

$=30$

Therefore, according to the given data, the frequency table is thus obtained.

Write the class-size in each of the following :
i) $0-4,5-9,10-14$

$i) \ 0-4,5-9,10-14$

Class size

$=4,4,4.$

Hence, this is the answer.

Is it true to say that class-mark of a class interval is $1/2(lower+Upper limit)?$

$\Rightarrow$ Class mark is mid-point of class interval.

$\therefore$ Its is true that,

$Class\,mark=\dfrac{1}{2}(Lower\,limit+Upper\,limit)$

The following table gives the weekly wages (in Rs.) of workers in a factory:
Weekly wages (in Rs.) $50-55$ $55-60$ $60-65$ $65-70$ $70-75$ $75-80$ $80-85$ $85-90$
No. of workers $5$ $20$ $10$ $10$ $9$ $6$ $12$ $8$
Calculate the number of workers getting weekly wages below $Rs. 80$ .

$Weekly\,wages$ $(in\,Rs.)$	$No.\,of\,workers$
$50-55$	$5$
$55-60$	$20$
$60-65$	$10$
$65-70$	$10$
$70-75$	$9$
$75-80$	$6$
$Total$	$60$

We consider the class intervals till

$75-80$ only as we are looking for the wages under

$80$ .

$\therefore$ The number of workers getting weekly wages below

$Rs.80=60$

This is because

$5+20+10+10+9+6$ adds up to

$Rs60$

construct a L.C.F and G.C.F. curve for the following data.
Marks obtained 0-5 5-10 10-15 15-20 20-25
No. of students 10 3 5 2 2

We are going to form more than frequency table :

$Marks\,obtained$	$No.\,of\,students$ $(G.C.F.)$
More than $0$	$22$
More than $5$	$12$
More than $10$	$9$
More than $15$	$4$
More than $20$	$2$

We are going to form more than frequency table :

$Marks\,obtained$	$No.\,of\,students$ $(L.C.F)$
Less than $5$	$10$
Less than $10$	$13$
Less than $15$	$18$
Less than $20$	$20$
Less than $25$	$22$

1220133_1151598_ans_9d4a8ef2c9d549b88d2e896e7fbbea5e.png

There are following the weights of $30$ students of class $10$ of a secondary school:
$34,\ 34,\ 36,\ 37,\ 38,\ 33,\ 34,\ 35,\ 36,\ 37,\ 38,\ 33,\ 34,\ 35,\ 34,\ 33$ ,
$\ 35,\ 34,\ 38,\ 36,\ 35,\ 34,\ 35,\ 37,\ 38,\ 34,\ 35,\ 35,\ 37$
Represent the above data in the form of a frequency distribution table.

We are going to form frequency distribution table from given data :

$Weight$	$No.\,of\,students$
$33$	$3$
$34$	$8$
$35$	$7$
$36$	$3$
$37$	$4$
$38$	$4$

Write the class-size in each of the following :
$100-120,120-140,160-180$

Class size

$=20, 20, 20$

Hence, this is the answer.

Give the formula for class marks and class size of a data

$Class\, mark:$ Mid-value of each class interval is called its class mark.

$Class\, size:$ Difference between the true upper limit and true lower limit of a class interval is called the class size.

$\Rightarrow$

$Class\,mark=\dfrac{Upper\,limit+Lower\,limit}{2}$

$\Rightarrow$

$Class\,size=Upper\,limit-Lower\,limit$

Define primary and secondary data.

Primary data :

Data collected by the investigator himself/herself for a specific person is called as primary data. For example, data collected by a student for his/her research project.

Secondary data :

Data collected by someone else for some other purpose but being utilized by the investigator for another purpose is called secondary data.

Write the class-size in each of the following :
$5-5.01,5.01-5.02,5.02-5.03$

Class size

$=$ Higher limit -Lower limit of class

Class size

$=0.01$

The mid-value of a class interval isIf the class-size is 10, find the upper and lower limits of the class.

$\Rightarrow$ Let upper limit and lower limit be

$x$ and

$y$

$\Rightarrow$

$\dfrac{x+y}{2}=42$ [ Given ]

$\therefore$

$x+y=84$ ------ ( 1 )

Now, according to the question,

$\Rightarrow$

$x-y=10$ ------ ( 2 )

Adding equation ( 1 ) and ( 2 ) we get,

$2x=94$

$\therefore$

$x=47$

And

$47-y=10$

$\therefore$

$y=37$

$\therefore$ Required class interval

$=37-47$

Write the class-size in each of the following :
$0-0.25,0.25-0.50,0.50-0.75$

We know that,

$Class Size = Upper\ limit- lower \ limit \ of\ the\ class$

Therefore,

Class size for

$0-0.25, 0.25-0.50,0.5-0.75$

$\Rightarrow 0.25, 0.25, 0.25$

Hence, this is the answer.

Read the given instruction and answer the following questions:
$\left(i\right)$ What information is depicted in the histogram?
$\left(ii\right)$ There is a 'kink' before the class interval $125-130$ on the horizontal axis.What does it mean?

$(i)$ The y-axis (vertical axis) shows number of girls and x-axis (horizontal axis) shows height of the girls in a class.

$(ii)$ The kink shows that there are no irls of height

$0-125$ cms.

Find the class mark of $80-90$

Class mark of

$a-b$ is

$\dfrac{a+b}{2}$

Class mark of

$80-90$ is

$\dfrac{80+90}{2} = \dfrac{170}{2} = 85$

From the data given below find the mean .
Age in years Frequency
$0-5$ $3$
$6-10$ $4$
$11-15$ $5$
$16-20$ $8$

$x_i=\dfrac 52,\dfrac {15}2,\dfrac {25}2,\dfrac {35}2$

$f_i=3,4,5,8\\sum f_i=20$

$\sum x_if_i=\dfrac{15}{2}+30+\dfrac{125}{2}+140=240$

$\bar x=\dfrac{\sum x_if_i}{\sum f_i}=\dfrac{240}{20}=12$

The two consecutive class marks of a distribution are $52$ and $57$ . Find the class limit of the two intervals.

Class size

$=$ Difference between consecutive class marks

$=$

$57-52$

Lower class limit

$=52-\dfrac{5}{2}=49.5$

Upper class limit

$=52+\dfrac{5}{2}=54.5$

Arrange the following data in ascending order and prepare the frequency table:
$9,4,3,1,7,10,4,3,3,4,9,8,1,6,3,7,9,1,8,10$

Arranging given numbers in ascending order,

$1,1,1,3,3,3,3,4,4,4,6,7,7,8,8,9,9,9,10,10$

Now, we are going to form a frequency table :

$Numbers$	$Frequency$
$1$	$3$
$3$	$4$
$4$	$3$
$6$	$1$
$7$	$2$
$8$	$2$
$9$	$3$
$10$	$2$

The following table gives the distribution of $45$ students across the different levels of intelligent quotient. Draw the histogram for the data.

IQ Level 60-70 70-80 80-90 90-100 100-110 110-120 120-130
No. of Students 2 5 6 10 9 8 5

$\Rightarrow$ The above graph shows histogram of given frequency table.

$x-axis-$ IQ Level

$y-axis-$ Number of students.

1225934_1231334_ans_f30a823cc0134ce1af4194421f49c44a.png

Construct a frequency distribution table for the data on the weights (in kg) of $20$ students of a class using intervals $30 - 35, 35 - 40$ ans so on____
$40, 38, 33, 48, 60, 53, 31, 46, 34, 36, 49, 41, 55, 49, 65, 42, 44, 47, 38, 39$ .

Weight in kg No. of students

$30-35$

$3$

$35-40$

$5$

$40-45$

$3$

$45-50$

$5$

$50-55$

$2$

$55-60$

$1$

$60-65$

$1$

$a-b\Rightarrow a < W\leq b$

$W\rightarrow$ weight.

1111167_1208778_ans_e32e77f9bdd449ad9eb6afb62f41ba4a.jpg

The following is the distribution of weights (in kg) of $52$ persons:
Weight (in kg) persons
$30-40$ 10
$40-50$ 15
$50-60$ 17
$60-70$
6
$70-80$ 4
What is the class size?

Class size is the difference between the upper and lower limits of the class interval.

Therefore, Class size

$= 40-30=50-40$

$=10$

The following bar graph shows the number of children, in various classes, in a school in Delhi.
Using the given bar graph, find:
the total number of children from Class $1$ to Class $8$

Total number of students in Class

$1$ to

$8$ can be calculated as below
Total number of students

$=100+90+100+80+120+90+70+50$
We get,

$=700$ students
Hence, there are

$700$ students in class

$1$ to

$8$

What is the class size for the given data?
Marks Number of student
$30-35$ $5$
$35-40$ $8$
$40-45$ $15$
$45-50$ $20$
$50-55$ $5$
$55-60$ $4$
$60-65$ $4$
$65-70$ $2$
$70-75$ $3$

1207781_1506359_ans_922dab85e6414ba1b4398ab62d6d1f7e.jpg

If the mean of the following distribution is $6$ , find the value of $p$
$X$ $2$ $4$ $6$ $10$ $P+5$
$f$ $3$ $2$ $3$ $1$ $2$

$x_{i}$	$f_{i}$	$f_{i}x_{i}$
$2$	$3$	$6$
$4$	$2$	$8$
$6$	$3$	$18$
$10$	$1$	$10$
$P+5$	$2$	$2p+10$
Total	$\displaystyle\sum f_{i}=11$	$\displaystyle\sum f_{i}x_{i}=2p+52$

We have

$\displaystyle\sum f_{i}=11, \sum f_{i}x_{i}=2p+52, \bar{x}=6$

$\therefore$ Mean

$(\bar{x})=\dfrac{\displaystyle\sum f_{i}x_{i}}{\displaystyle\sum f_{i}}$

$\Rightarrow 6=\dfrac{2p+52}{11}\Rightarrow 66=2p+52$

$\Rightarrow 2p=14\Rightarrow p=7$

The weights of new born babies (in kg) in a hospital on a particular day are as follows:
$2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1, 2.5, 2.8, 2.7, 2.9, 2.4$
How many babies weigh below $2.5 kg$ ?

In Descending order: The lowest weight will come last and highest will come first.

$3.1, 3.0,2.9,2.9, 2.8, 2.8, 2.7, 2.7,2.6, 2.5,\underline{2.5}, 2.4,2.3, 2.2,2.1$

$2.1,2.2,2.3,2.4$ are the observations less than

$2.5$ .

$4$ babies weigh less than

$2.5$ kg.

Consider the following frequency distribution.
Class $0-5$ $6-11$ $12-17$ $18-23$ $24-29$
Frequency $13$ $10$ $18$ $8$ $11$
Find the upper limit of medians class.

Classes are not continuous, hence make them continuous by adding

$0.5$ to the upper limits and subtracting

$0.5$ from the lower limits.

$C.I$	Frequency	Cumulative Frequency
$0-5.5$	$13$	$13$
$5.5-11.5$	$10$	$23$
$11.5-17.5$	$15$	$38$
$17.5-23.5$	$08$	$46$
$23.5-29.5$	$11$	$57$
Total	$\sum f=57$

Class interval can't be negative hence the first

$C.I.$ is starting from

$0$ .
Now to find median we calculate

$\dfrac{\sum f}{2}=\dfrac{57}{2}=28.5$

$\therefore$ Median class

$11.5-17.5$
So, the upper limit is

$17.5$

Given above is the bar graph indicating the marks obtained out of $50$ in mathematics paper by $100$ students. Read the bar and answer the following questions:
What is the number of students belonging to the highest mark group?

Let us prepare a chart of the

$100$ students using the data from the bar graph.

Marks	Number of Students
$0 − 9$	$27$
$10 − 19$	$12$
$20 − 29$	$20$
$30 − 39$	$24$
$40 − 49$	$17$

Clearly, the highest mark group is

$40 − 49$ .

and the number of students in this marks group

$= 17$ .

Therefore, the number of students belonging to the highest mark group is

$17$ .

What is the number of persons in the age-group $60-65$ ?

From the graph,

The number of persons in the age group $10-15$ is $=1400$ .

The number of persons in the age group $20-25$ is $=1200$ .

The number of persons in the age group $30-35$ is $=1100$ .

The number of persons in the age group $40-45$ is $=1000$ .

The number of persons in the age group $50-55$ is $=900$ .

The number of persons in the age group $60-65$ is $=800$ .

The number of persons in the age group $70-75$ is $=300$ .

Therefore, the number of persons in the age-group

$60-65$ from the graph is

$800$ .

If median of the distribution given is $28.5$ , find the values of $x$ and $y$ .
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ Total
Frequency $5$ $X$ $20$ $15$ $y$ $5$ $60$

Here, median

$=28.5$ and

$n=60$
Now, we have

Class interval	Frequency $(f_i)$	Cumulative frequency $(cf)$
$0-10$	$5$	$5$
$10-20$	$x$	$5+x$
$20-30$	$20$	$25+x$
$30-40$	$15$	$40+x$
$40-50$	$y$	$40+x+y$
$50-60$	$5$	$45+x+y$
Total	$\displaystyle \sum f_i=60$

Since the median is given to be

$28.5$ , thus the median is

$20-30$

$\therefore \ \dfrac {n}{2}=30, I=20, h=10, cf=5+x$ and

$f=20$

Hence,

$x=8$ and

$y=7$ .

What is the total number of persons living in the town in the age-groups $50-55$ ?

From the graph,

The number of persons in the age group $10-15$ is $=1400$ .

The number of persons in the age group $20-25$ is $=1200$ .

The number of persons in the age group $30-35$ is $=1100$ .

The number of persons in the age group $40-45$ is $=1000$ .

The number of persons in the age group $50-55$ is $=900$ .

The number of persons in the age group $60-65$ is $=800$ .

The number of persons in the age group $70-75$ is $=300$ .

Then clearly, the total number of persons living in the town in the age-groups

$50-55$ is

$900$ .

The range of the data $6,8, 16, 22, 8, 20, 7, 25$ is _____

The correct fill up is 19.
Range of data= maximum value- minimum value =

$25-6=19.$

In the class interval $26 - 33, 33$ is known as_____

The correct fill up is Upper class limit.
The uppermost value is known as upper class limit.

In the class interval $20 - 30$ the lower class limit is ______

The correct fill up is 20
The lowermost value is known as lower class limit.

The bar graph given below represents the approximate length (in kilometers) of some National Highways in India. Study the bar graph and answer the following questions:
Length of which National Highway is about three times
the National Highway $10$ ?

$N.H\ 2$ is (

$1500$ km) about three-time of

$N.H\ 10$ (

$500$ km).

Find the class mark of classes $15.5-18.5$ and $50-75.$

Class marks

$=\dfrac{upper\ limit+lower\ limit}{2}$

$\therefore$ Class marks of

$15.5-18.5=\dfrac{18.5+15.5}{2}$

$=\dfrac{34}{2}=17$

Class mark of

$50-75=\dfrac{75+50}{2}$

$=\dfrac{125}{2}=62.5$ .

The sixth class interval for a grouped data whose first two class intervals are $10 -15$ and $15 -20$ is _______

The class size observed in first two intervals is

$5$ . So, the intervals will go as follows:

$10-15$

$15-20$

$20-25$

$25-30$

$30-35$

$35-40$

The correct fill up is

$35-40$ .
So, the sixth class interval will be

$= 35-40$

In the class intervals $10 -20 , 20 -30$ etc. respectively, $20$ lies in the class_______

The correct fill up is

$20-30$

The difference between the upper and lower limit of a class interval is called the ________ of the class interval.

The correct fill up is size
Size of the class interval is the difference between upper and lower limit of a class.

In a histogram, class intervals and frequencies are taken along axis and axis.

The correct fill up is

$x$ and

$y$ .
In a histogram,

$x-\text{axis}$ represents class intervals and

$y-\text{axis}$ represents frequencies.

The arithmetic mean of the following frequency distribution is $53$ . Find the value of p:
Class 0-20 20-40 40-60 60-80 80-100
Frequency 12 15 32 p 13

Class	Frequency $({ f }_{ i })$	Class Mark $({ x }_{ i })$ ${ x }_{ i }=\frac { \text{ Upper class limit+ Lower class limit} }{ 2 }$	${ x }_{ i }{ f }_{ i }$
0-20	12	10	120
20-40	15	30	450
40-60	32	50	1600
60-80	p	70	70p
80-100	13	90	1170
Total	$\sum { { f }_{ i } } =72+p$		$\sum { { f }_{ i } } { x }_{ i }=3340+70p$

Now,

$\overline { x } =\cfrac { \sum { { f }_{ i } } { x }_{ i } }{ \sum { { f }_{ i } } }$

$\Rightarrow$

$53=\cfrac { 1900+70p }{ 72+p }$

$\Rightarrow$ 53(72 + p) = 3340 +70p

$\Rightarrow$ 3816 + 53p = 3340 +70p

$\Rightarrow$ 53p -70p = 3340 - 3816

$\Rightarrow$ -17p=-476

$\Rightarrow$ p=28

The mean of the following frequency distribution is $62.8$ and the sum of all frequency is $50$ . Compute the missing frequency $f_{ 1}$ and $f_{ 2 }$ :
Class 0-20 20-40 40-60 60-80 80-100 100-120 Total
Frequency 5 $f_{ 1 }$ 10 $f_{ 2 }$
7 8 50

Class	Frequency $f_{ i}$	Class Mark $x_{ i}$	${ x }_{ i }{ f }_{ i }$
0-20	5	10	50
20-40	$f_{ 1}$	30	30 $f_{ 1}$
40-60	10	50	500
60-80	$f_{ 2}$	70	70 $f_{ 2}$
80-100	7	90	630
100-120	8	110	880
Total	$\sum { f_{ i } } =30+f_{ 1 }+f_{ 2 }$		$\sum { { f }_{ i }{ x }_{ i } } =2060+30f_{ 1 }+70f_{ 2 }$

Now,

$\quad \overline { x } =\cfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { f_{ i } } }$

$\Rightarrow$

$62.8 = \quad \cfrac { 2060+30f_{ 1 }+70f_{ 2 } }{ 50 }$

$\Rightarrow$

$3140= 2060+30f_{ 1 }+70f_{ 2 }$

$\Rightarrow$

$3140- 2060= 30f_{ 1 }+70f_{ 2 }$

$\Rightarrow$

$1080= 30f_{ 1 }+70f_{ 2 }$

$\Rightarrow$

$108= 3f_{ 1 }+7f_{ 2 }$ ......(i)
and

$30+f_{ 1 }+f_{ 2 } =50$

$\Rightarrow$

$f_{ 1 }+f_{ 2 } =50-30$

$\Rightarrow$

$f_{ 1 }+f_{ 2 } =20$

$\Rightarrow$

$f_{ 1 } =20-f_{ 2 }$ .......(ii)
Now, putting the value of

$f_{ 1 }$ in eq. (i),
we get

$3(20-f_{ 2 })+7f_{ 2 } = 108$

$\Rightarrow$

$60-3f_{ 2 }+7f_{ 2 }=108$

$\Rightarrow$

$4f_{ 2 }=108-60$

$\Rightarrow$

$4f_{ 2 }=48$

$\Rightarrow$

$f_{ 2 }=12$
Now, substitute the value of

$f_{ 2 }$ in eq. (ii),
we get

$f_{ 1 }=20-12$

$\Rightarrow$

$f_{ 1 }=8$

The mean of the following frequency distribution is $62.8$ . Find the missing frequency x:
Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 5 8 x 12 7 8

Class	Frequency $({ f }_{ i })$	Class Mark $({ x }_{ i })$ ${ x }_{ i }=\frac {\text{ Upper class limit+ Lower class limit} }{ 2 }$	${ x }_{ i }{ f }_{ i }$
0-20	5	10	50
20-40	8	30	240
40-60	x	50	50x
60-80	12	70	840
80-100	7	90	630
100-120	8	110	880
Total	$\sum { { f }_{ i } } =40+x$		$\sum { { f }_{ i } } { x }_{ i }=2640+50x$

Now,

$\overline { x } =\cfrac { \sum { { f }_{ i } } { x }_{ i } }{ \sum { { f }_{ i } } }$

$\Rightarrow$

$62.8=\cfrac { 2640+50x }{ 40+x }$

$\Rightarrow 62.8 (40+x) =2640+50x$

$\Rightarrow 2512+62.8x = 2640 +50x$

$\Rightarrow 62.8x -50x=2640 -2512$

$\Rightarrow 12.8x= 128$

$\Rightarrow x=10$

Marks scored by $30$ students of class $VI$ are as given below:
$38, 46, 33, 45, 63, 53, 40, 85, 52, 75, 60, 73, 62, 22, 69, 43, 45, 33, 47, 41, 29, 43, 37, 49, 83, 44, 55, 22, 35$ and $45$ .
State:
the range of marks

Range of marks = Difference between highest and lowest marks

$=85-22$

$=63$
Therefore, the range of marks is

$63$

The class size of the class interval $60 -68$ is $8$ .

Class size = upper class limit - lower class limit =

$68 -60 = 8$

If the mean of the following distribution is $5$ . Find the value of $f_{ 1 }$ :
Class 0-20 20-40 40-60 60-80 80-100
Frequency 17 28 32 $f_{ 1 }$
19

Class	Frequency $(f_{ i })$	Class Mark $(x_{ i } )$ ${ x }_{ i }=\frac { \text{Upper class limit+ Lower class limit} }{ 2\quad \quad \quad \quad \quad \quad \quad }$	${ x }_{ i }{ f }_{ i }$
0-20	17	10	170
20-40	28	30	840
40-60	32	50	1600
60-80	$f_{ 1 }$	70	70 $f_{ 1 }$
80-100	19	90	1710
Total	$\sum { f_{ i } } =96+{ f }_{ 1 }$		$\sum { \quad { x }_{ i }{ f }_{ i } } =4320+70f_{ 1 }$

Now,

$\overline { x } =\cfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { f_{ i } } }$

$\Rightarrow$

$5=\cfrac { 4320+70f_{ 1 } }{ 96+f_{ 1 } }$

$\Rightarrow$

$(96+f_{ 1 }) =4320 + 70f_{ 1 }$

$\Rightarrow$

$480+ 5f_{ 1 } =4320 + 70f_{ 1 }$

$\Rightarrow$

$5f_{ i }-70f_{ 1 } =4320 -480$

$\Rightarrow$

$-65f_{ i } =+3840$

$\Rightarrow$

$f_{ 1 } =-59.07$
This is not possible as frequency cannot be negative.

Fill in the blanks:
The class mark of class interval $22-29$ is.......

The class mark is

$22-29=\dfrac{22+29}{2}=\dfrac{51}{2}=25.5$

A student spends his pocket money on various items, as given below:
Books: $Rs. 380$ , Postage: $Rs. 30$ , Toilet items: $Rs. 240$ ,
Stationary: $Rs. 220$ and Entertainment : $Rs. 120$
Draw a bar graph to represent his expenses.

The required bar graph of the given data will be as shown below:

Here scale = Rs.50

on $x-axis$ : Items

on $y-axis$ : Expenses (in Rs).

2222281_1823655_ans_f31ed67135344658b073d924786c3b90.PNG

The following bar graph shows the number of children, in various classes, in a school in Delhi.
Using the given bar graph, find:
the total number of children from class $6$ to Class $8$

From the given graph, the number of students from Class

6 to Class

8 is as follows:
Class

6=90 students
Class

7=70 students
Class

8=50 students
hence, total number of students in Class

6 to Class

8 can be calculated as below:
Total students = Students in Class

6 to Class

=90+70+50
We get,

=210
Hence, total number of students in Class

6,7 and

8 are

210

The following bar graph shows the number of children, in various classes, in a school in Delhi.
Using the given bar graph, find:
the total number of children in each class.

From the given graph, the number of students in each class is as follows:
Class

$1=100$ students
Class

$2=90$ students
Class

$3=100$ students
Class

$4=80$ students

Using the data, given below, construct a frequency distribution table:
$46, 44, 42, 54, 52, 60, 50, 58, 56, 62, 50, 56, 54, 58$ and $48$ . Now answer the following:
What is the range of the numbers?

Marks Tally marks Frequency

$42\ |\ 1$

$44\ |\ 1$

$46\ |\ 1$

$48\ |\ 1$

$50\ ||\ 2$

$52\ |\ 1$

$54\ ||\ 2$

$56\ ||\ 2$

$58\ ||\ 2$

$60\ |\ 1$

$62\ |\ 1$
Total

$15$
Range of numbers = Highest number - Lowest number

$=62-42$

$=20$

In a study of number of accidents per day, the observations for $30$ days were obtained as follows:
$6\ 3\ 5\ 6\ 4\ 3\ 2\ 5\ 4\ 2$
$4\ 0\ 5\ 3\ 6\ 1\ 5\ 5\ 2\ 6$
$2\ 1\ 2\ 2\ 0\ 5\ 4\ 6\ 1\ 6$
Construct a suitable frequency distribution table.

The required frequency table is shown below:
No. of accidents Tally marks Frequency

1754503_1823517_ans_cedee89979e14c3c8d4e415367f776fd.png

1754503_1823517_ans_cedee89979e14c3c8d4e415367f776fd.png

Given below are the marks obtained by 30 students in an examination :
08 17 33 41 47 23 20 34
09 18 42 14 30 19 29 11
36 48 40 24 22 02 16 21
15 32 47 44 33 01
Taking class intervals 1 - 10, 11 - 20, ...... 41 - 50; make a frequency table for the above distribution.

Class-intervals
$50-60$
$60-70$
$70-80$
$80-90$
$90-100$
$100-110$
Frequency
$30$
$25$
$45$
$15$
$20$
$40$
Draw a histogram to represent this data:

$\Rightarrow$ The above graph represent histogram of given data.

$\Rightarrow$

$X-axis$ represents class-intervals and

$Y-axis$ represents frequencies.

1048368_180424_ans_ce14ac437ada45f8988b51d1a1ad705a.jpeg

The marks of $24$ candidates in the subject mathematics are given below:
45
48
15
23
30
35
40
11
29
0
3
12
48
50
18
30
15
30
11
42
23
2
3
44
The maximum marks are $50.$ Make a frequency distribution taking class intervals $0-10 , 10-20 , ............ .$

$subjects$

$candidates$

$0-10$

$4$

$10-20$

$6$

$20-30$

$3$

$30-40$

$4$

$40-50$

$7$

$24$

The marks of $24$ candidates in the subject mathematics are given below :
$45$ $48$ $15$ $23$ $30$ $35$ $40$ $11$
$29$ $00$ $03$ $12$ $48$ $50$ $18$ $30$
$15$ $30$ $11$ $42$ $23$ $02$ $03$ $44$
The maximum marks are $50.$ Make a frequency distribution taking class intervals $0 - 10, 10 - 20, ..... .$

The class marks of a distribution are $2.03$ , $2.23$ , $2.43$ , $2.63$ , $2.83$ , $3.03$ and $3.23$ Determine the class size and the class boundaries

Sol. Class size is

$0.20$ . Class boundaries are

$1.93-2.13$ ,

$2.13-2.33$ ,

$2.33-2.53$ ,

$2.53-2.73$ ,

$2.73-2.93$ ,

$2.93-3.13$ ,

$3.13-3.33$

Find the number of examinees who have obtained 600 and more than marks
Enter 0 if answer is 216, else 1.

Number of examinee having more than 600 marks would be = 54+67+45+47 = 216

Inluding all students in categories aboe 600.

Construct

the following table into cumulative frequency table.
Marks $100$ $50$ $60$
Students $4$ $5$ $10$

The cumulative frequency is calculated from a

frequency table, by adding each frequency to the total of the frequencies of

all data values before it in the data set.

Marks	Students	Cumulative curve(cf)
$100$	$4$	$4$
$50$	$5$	$4+5=9$
$60$	$10$	$9+10=19$

Convert the following data into grouped frequency distribution:
$3, 4, 5, 6, 6, 5, 4, 3, 2, 2, 2, 1, 7, 7, 8, 2$

First arrange the data in ascending order

$1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8$
The largest number

$= 8$
The smallest number

$= 1$
Range

$=$ largest number

$-$ smallest number

$= 8 - 1 = 7$
We shall calculate an approximate group size, by dividing the range by the number of groups we want.
If we want

$3$ groups, then Divide range with number of groups

$= \dfrac73 = 2.333$
The class width will be

$2$ , Starting from

$1$ .

Class interval	Frequency
$1 -3$	$7$
$4 -6$	$6$
$7 -9$	$3$

"A bird in hand is worth two in the bush" from this sentence select the vowels for frequency column. Form a ungrouped frequency distribution table.

English vowels

$= a, e, i, o, u$
The number of occurrences of

$a, e, i, o$ and

$u$ are

$2, 1, 4, 2$ and

$1$
So, the ungrouped frequency is drawn in the table.

The weight of $30$ students of a class ( $kg$ ) are as given:
$43, 45, 64, 42, 54, 78, 91, 90, 56, 37, 48, 90, 50, 48, 67$
$85, 30, 39, 49, 40, 50, 45, 68, 90, 50, 34, 56, 88, 43, 50$

Convert the data in the form of grouped data frequency distribution.

Assuming the class size

$= 50 - 40 = 10$
Lower class limit

$=$

$30-\cfrac{10}{2} = 25$
Upper class limit

$=$

$30+\cfrac{10}{2} = 60$
Constructing the grouped frequency distribution table:

Weight ( $kg$ )	Frequency
$25-35$	$2$
$35-45$	$8$
$45-55$	$8$
$55-65$	$3$
$65-75$	$2$
$75-85$	$3$
$85-95$	$4$
Total	$30$

Draw a ungrouped frequency distribution for the vowels used in the given sentence.
"Empty vessels make the most noise"

English vowels

$= a, e, i, o, u$
The number of occurrences of

$a, e, i, o$ and

$u$ are

$1, 5, 1, 2$ and

$0$ .
So, the ungrouped frequency is drawn in the table.

The height of $20$ students of a class ( $cm$ ) are as given: $132, 131, 135, 136, 134, 132, 131, 142, 142, 152, 151, 151, 148, 145, 131, 132, 142, 156, 155, 152$ . Arrange the data in the form of table of grouped data frequency distribution.

Minimum height

$= 131\:cm$

Maximum height

$= 156\: cm$

Range

$= 156 - 131 = 25$

Number of classes

$= 4$

Class range

$= \dfrac {25}{4} = 6.25$ approx

$6$

class

$1 = 131 + 6 = 137$

class

$2 = 137 + 6= 143$

class

$3 = 143 + 6 = 149$

class

$4 = 149 + 6 155$

class

$5 = 155 + 6 = 161$

Class interval Frequency

$131 - 137$

$9$

$137 - 143$

$3$

$143 - 149$

$2$

$149 - 155$

$4$

$155 - 161$

$2$

The following data represent the number of hours of playing per week ( $X$ ) for $25$ people.
$3, 4, 6, 1, 2, 6, 6, 4, 3, 1, 3, 6, 7, 8, 1, 5, 8, 9, 5, 9, 4, 3, 7, 1, 7$ .

Picture the following data into ungrouped frequency table.

Hours	Number of people
$0-2$	$5$
$3-5$	$9$
$6-8$	$9$
$9-11$	$2$

Sketch a ungrouped frequency distribution for the vowels used in the given sentence.
"An ounce of discretion is worth a pound of wit"

English vowels

$= a, e, i, o, u$
The number of occurrences of

$a, e, i, o$ and

$u$ are

$1, 2, 4, 5$ and

$2$ .
So, the ungrouped frequency is drawn in the table.

Form a ungrouped frequency distribution table for the vowels used in the given sentence.
"A rolling stone gathers no moss"

English vowels

$= a, e, i, o, u$
The number of occurrences of

$a, e, i, o$ and

$u$ are

$1, 2, 4, 5$ and

$2$
So, the ungrouped frequency is drawn in the table.

Interval
$10-15$
$15-20$
$20-25$
$25-30$
$30-35$
$x$
$2$ $1$
$4$
$6$
$5$
What is the frequency polygon for the following table?

X	Interval
$10-15$	$2$
$15-20$	$1$
$20-25$	$4$
$25-30$	$6$
$30-35$	$5$

1020594_463082_ans_0009bf39da6144a5a8aabcf287c7862e.png

Form a histogram for the following data.
X
3-5
6-8
9-11
12-14
15-17
frequency
12
18
25
15
10

$CI$	$Frequency$
$2.5-5.5$	$12$
$5.5-8.5$	$18$
$8.5-11.5$	$25$
$11.5-14.5$	$15$
$14.5-17.5$	$10$

1115639_462572_ans_cc11f4abf8734aafafd6d9dece19fa53.jpg

Sketch frequency polygon for the following table:
$x$
$1-3$
$3-5$
$5-7$
$7-9$
$9-11$
$11-13$
frequency
$2$
$1$
$4$
$2$
$3$
$6$

X	Frequency
$1-3$	$2$
$3-5$	$1$
$5-7$	$4$
$7-9$	$2$
$9-11$	$3$
$11-13$	$6$

1020593_463081_ans_64d494f0f3344c8ba942fa9f9dad159b.png

Draw frequency polygon for the following histogram.

The frequency polygon is formed by joining the midpoints of each rectangle in a histogram.

The blood groups of 30 students of Class VIII are recorded as follows:
$A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,$
$A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O$ .
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

Blood group	Number of students
$A$	$9$
$B$	$6$
$O$	$12$
$AB$	$3$
Total	$30$

Most common blood group -

$O$ (

$12$ students)
Rarest blood group -

$AB$ (

$3$ students)

The following table gives the life times of $400$ neon lamps:
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than $700$ hours?

(ii) Lifetime of more than

$700$ hours :

$700 - 800, 800 - 900$ &

$900 - 1000$

$\therefore$ Number of neon lamps

$= 74+62+48 = 184$

The value of $\pi$ upto 50 decimal places is given below:
$3.14159265358979323846264338327950288419716939937510$
(i) Make a frequency distribution of the digits from $0$ to $9$ after the decimal point.
(ii) What are the most and the least frequently occurring digits?

(i)

Digits	Frequency
$0$	$2$
$1$	$5$
$2$	$5$
$3$	$8$
$4$	$4$
$5$	$5$
$6$	$4$
$7$	$4$
$8$	$5$
$9$	$8$
Total	$50$

(ii) Least frequency

$\rightarrow 0 \rightarrow 2$ times
Maximum frequency

$\rightarrow 3$ &

$9 \rightarrow 8$ times

The relative humidity (in %) of a certain city for a month of 30 days was as follows:
$\begin{matrix} 98.1 & 98.6 & 99.2 & 90.3 & 86.5 & 95.3 & 92.9 & 96.3 & 94.2 & 95.1 \\ 89.2 & 92.3 & 97.1 & 93.5 & 92.7 & 95.1 & 97.2 & 93.3 & 95.2 & 97.3 \\ 96.2 & 92.1 & 84.9 & 90.2 & 95.7 & 98.3 & 97.3 & 96.1 & 92.1 & 89 \end{matrix}$
(i) Construct a grouped frequency distribution table with classes $84-86, 86-88$ , etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?

Relative Humidity (in $\%$ )	Frequency
$84-86$	$1$
$86-88$	$1$
$88-90$	$2$
$90-92$	$2$
$92-94$	$7$
$94-96$	$6$
$96-98$	$7$
$98-100$	$4$
$\text{Total}$	$30$

(i) The table above is the frequency distribution.

(ii) The data appears to be taken in the rainy season as the relative humidity is high.
(iii) Range of data

$=$ Maximum value - Minimum value

$=99.2-84.9=14.3$

The histogram below represents the scores obtained by $25$ students in a Mathematics mental test. Use the data to frame a frequency distribution table.

Frequency distribution table:

C.I.	Frequency
$0 - 10$	$2$
$10 -20$	$5$
$20 - 30$	$8$
$30- 40$	$4$
$40 - 50$	$6$

The following are the daily expeditures (in $Rs$ ) of $30$ student in a school.
Ependiture (in $Rs$ )
No. of students
$10-15$ $3$
$15-20$ $5$
$20-25$ $10$
$25-30$ $6$
$30-35$ $4$
$35-40$ $2$
Total $30$
What is the class-size

The class size is

$5$

$upper \ limit - lower \ limit = class \ size$

A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
$\begin{matrix} 0.03 & 0.08 & 0.08 & 0.09 & 0.04 & 0.17 \\ 0.16 & 0.05 & 0.02 & 0.06 & 0.18 & 0.20 \\ 0.11 & 0.08 & 0.12 & 0.13 & 0.22 & 0.07 \\ 0.08 & 0.01 & 0.10 & 0.06 & 0.09 & 0.18 \\ 0.11 & 0.07 & 0.05 & 0.07 & 0.01 & 0.04 \end{matrix}$

(i) Make a grouped frequency distribution table for this data with class intervals as $0.00-0.04, 0.04-0.08$ , and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than $0.11$ parts per million?

Concentration of sulphur dioxide (in ppm)	Frequency
0.00 - 0.04	4
0.04 - 0.08	9
0.08 - 0.12	9
0.12 - 0.16	2
0.16 - 0.20	4
0.20 - 0.24	2
Total	30

(i) The figure above is the grouped frequency distribution.

(ii) The number of days for which concentration of

$S{O}_{2}$ is more than

$0.11 \rightarrow 0.12 - 0.16, 0.16 - 0.20, 0.20 - 0.24$

$\therefore$ Total number of days

$= 2 + 4 + 2 = 8$ days.

A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
$\begin{matrix} 2.6 & 3.0 & 3.7 & 3.2 & 2.2 & 4.1 & 3.5 & 4.5 \\ 3.5 & 2.3 & 3.2 & 3.4 & 3.8 & 3.2 & 4.6 & 3.7 \\ 2.5 & 4.4 & 3.4 & 3.3 & 2.9 & 3.0 & 4.3 & 2.8 \\ 3.5 & 3.2 & 3.9 & 3.2 & 3.2 & 3.1 & 3.7 & 3.4 \\ 4.6 & 3.8 & 3.2 & 2.6 & 3.5 & 4.2 & 2.9 & 3.6 \end{matrix}$

Construct a grouped frequency distribution table for this data, using class intervals of size $0.5$ starting from the interval $2-2.5$ .

Life of batteries (in years)	Frequency
2.0 - 2.5	2
2.5 - 3.0	6
3.0 - 3.5	14
3.5 - 4.0	11
4.0 - 4.5	4
4.5 - 5.0	3
Total	40

Class size :

$0.5$

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
$\begin{matrix} 1 & 6 & 2 & 3 & 5 & 12 & 5 & 8 & 4 & 8 \\ 10 & 3 & 4 & 12 & 2 & 8 & 15 & 1 & 17 & 6 \\ 3 & 2 & 8 & 5 & 9 & 6 & 8 & 7 & 14 & 12 \end{matrix}$

(i) Make a grouped frequency distribution table for this data, taking class width $5$ and one of the class intervals as $5-10$ .
(ii) How many children watched television for 15 or more hours a week?

Number of hours	Frequency
0-5	10
5-10	13
10-15	5
15-20	2
Total	30

(i) Above table
(ii) Number of children who watched TV for

$15$ hours or more

$= 2$ children.

The length of $40$ leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:
(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are $153 mm$ long? Why?

Length (in mm)	New length (in mm)	No. of leaves
118 - 126	117.5 - 126.5	3
127 - 135	126.5 - 135.5	5
136 - 144	135.5 - 144.5	9
145 - 153	144.5 - 153.5	12
154 - 162	153.5 - 162.5	5
163 - 171	162.5 - 171.5	4
172 - 180	171.5 - 180.5	2

In order to make a histogram, we need to convert discontinuous classes to continuous classes.

Adjustment

$= \dfrac 12 (\mbox {lower limit of class - upper limit of previous class)}$

$= \dfrac 12 (127 - 126) = 0.5$

So,

$0.5$ needs to be subtracted from upper limit of previous class and lower limit of next class.

(ii) Frequency polygon
(iii) No. It is a range. The maximum number of leaves lies in between

$144.5 mm - 154.5 mm$ . Not necessary that it all occurs as

$153 mm$ length.

Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
$\begin{matrix} 0 & 1 & 2 & 2 & 1 & 2 & 3 & 1 & 3 & 0 \\ 1 & 3 & 1 & 1 & 2 & 2 & 0 & 1 & 2 & 1 \\ 3 & 0 & 0 & 1 & 1 & 2 & 3 & 2 & 2 & 0 \end{matrix}$
Prepare a frequency distribution table for the data given above.

Number of heads	Frequency
0	6
1	10
2	9
3	5
Total	30

The marks scored by students in Mathematics in a certain examination are given below:
Marks Scored Number of Students
$0 - 20$ $3$
$20 - 40$ $8$
$40 - 60$ $15$
$60 - 80$ $17$
$80 - 100$ $7$
Draw histogram for the above data.

Mark the classes

$0-20, 20-40,40-60,...$ on the

$X$ -axis and number of students on the

$Y$ -axis.

For each class, mark a horizontal line at the corresponding frequency (no. of students) on the

$Y$ -axis.

Drop vertical lines from the endpoints of the classes. The obtained graph is a histogram.

The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
$\begin{matrix} 161 & 150 & 154 & 165 & 168 & 161 & 154 & 162 & 150 & 151 \\ 162 & 164 & 171 & 165 & 158 & 154 & 156 & 172 & 160 & 170 \\ 153 & 159 & 161 & 170 & 162 & 165 & 166 & 168 & 165 & 164 \\ 154 & 152 & 153 & 156 & 158 & 162 & 160 & 161 & 173 & 166 \\ 161 & 159 & 162 & 167 & 168 & 159 & 158 & 153 & 154 & 159 \end{matrix}$

(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as $160-165, 165-170,$ etc.
(ii) What can you conclude about their heights from the table?

Height (in cm)	Frequency
$150-155$	$12$
$155-160$	$9$
$160-165$	$14$
$165-170$	$10$
$170-175$	$5$
Total	$50$

(i) The above table is the grouped frequency distribution.
(ii) Students with

$160-165 cm$ heights are maximum i.e.

$14$ students.

A random survey of the number of children of various age group playing in a park was found as follows:
Draw a histogram to represent the data above.

Age (in years)	Frequency	Width	Length of the rectangle
1 - 2	5	1	$\dfrac 51 \times 1 = 5$
2 - 3	3	1	$\dfrac 31 \times 1 = 3$
3 - 5	6	2	$\dfrac 62 \times 1 = 3$
5 - 7	12	2	$\dfrac {12}{2} \times 1 = 6$
7 - 10	9	3	$\dfrac 93 \times 1 = 3$
10 - 15	10	5	$\dfrac {10}{5} \times 1 = 2$
15 - 17	4	2	$\dfrac 42 \times 1 = 2$

Width is not constant. So, we have to first calculate the length of rectangle to draw histogram according to the width as shown in the above figure.

Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the medians is an appropriate measure of central tendency.

(i) While if we take the case of calculating the weight of students in a class, then we should use mean rather than median. In such cases taking median is not suitable.

(ii) If the score of students in a class are

$1, 2, 3, 4, 20$
So if we calculate the mean

$= \dfrac{1+2+3+4+20}{5} = \dfrac{30}{5} = 6$
Median

$= 3$
So, median is better or appropriate measure because

$20$ is much greater than other numbers and because of

$20$ the mean has come out to

$6.$

$\therefore$ Its better to take median than mean.

The marks scored by $40$ candidates in an examination (out of $100$ ) is given below:
$75, 65, 57, 50, 32, 54, 75, 67, 75, 88, 80, 42, 40, 41,34, 78, 43, 61, 42, 46$
$68, 52, 43, 49, 59, 49, 67, 34, 33, 87, 97, 47, 46, 54, 48, 45, 51, 47, 41, 43$ .

Prepare a frequency distribution table with the class size $10$ . Take the class intervals as $(30 - 39), (40 - 49), . . .$ and answer the following questions:
(i) Which class intervals have the highest and lowest frequency?
(ii) Write the upper and lower limits of the class interval $30-39$ .

$\text{Class Interval}$

$\text{Frequency}$

$\quad 30-39$

$\quad 40-49$

$\quad 50-59$

$\quad 60-69$

$\quad 70-79$

$\quad 80-89$

$\quad 90-99$

$\quad 4$

$\quad 16$

$\quad 7$

$\quad 5$

$\quad 4$

$\quad 3$

$\quad 1$

(i) Class interval $40-49$ and $90-99$ have highest and lowest frequency respectively.

(ii) $\text{Upper Limit}=39$

$\text{Lower Limit} =30$

Given below are the ages of $45$ people in a colony.
$33\ 8\ 7\ 25\ 31\ 26\ 5\ 50\ 25\ 48\ 56\ 33\ 28\ 22\ 15\ 62\ 59\ 16\ 14\ 19\ 24$ $35\ 26\ 9\ 12\ 46\ 15\ 42\ 63\ 32\ 5\ 22\ 11\ 42\ 23\ 52\ 48\ 62\ 10\ 24\ 43\ 51\ 37\ 48\ 36$
Construct grouped frequency distribution for the given data with $6$ class intervals.

$\Rightarrow$ In given data minimum value is

$5$ and maximum value is

$63$

$\Rightarrow$ Range

$=63-5=58$

$\Rightarrow$ Number of class interval

$=6$ class interval (Average)

$\Rightarrow$ Length of the class interval

$=\dfrac{58}{6}=9.6\approx 10$

$Class-interval$	$Frequency$
$0-9$	$5$
$10-19$	$8$
$20-29$	$10$
$30-39$	$7$
$40-49$	$7$
$50-59$	$5$
$60-69$	$3$
$Total$	$45$

Cumulative frequency table is given below. Which type of cumulative frequency is given. Try to build the frequencies of respective class intervals.
Runs $0 - 10$ $10 - 20$ $20 - 30$ $30 - 40$ $40 - 50$
No of cricketers $3$ $8$ $19$ $25$ $30$

The given cumulative frequency is a less than cumulative frequency distribution as it is starting from lowest to highest.

Frequency distribution table-

Runs	No. of cricketers (C. f.)	Frequency
$0-10$	$3$	$3$
$10-20$	$8$	$8 - 3 = 5$
$20-30$	$19$	$19 - 8 = 11$
$30-40$	$25$	$25 - 19 = 6$
$40-50$	$30$	$30 - 25 = 5$

Prepare the frequency distribution table for the given set of scores:
$39, 16, 30, 37, 53, 15, 16, 60, 58, 26, 28, 19, 20, 12, 14, 24,$ $59, 21, 57, 38, 25, 36, 24, 15, 25, 41, 52, 45, 60, 63, 18, 26,$ $43, 36, 18, 27, 59, 63, 46, 25, 33, 46, 42, 48, 35, 64, 24$ .Take class intervals as $(10 - 20), (20 - 30), ....$ and answer the following:
(i) What does the frequency corresponding to the third class interval mean?
(ii) What is the size of each class interval? Find the midpoint of the class interval $30 - 40$ .
(iii) What is the range of the given set of scores?

Class Interval

Frequency

10-20

20-30

30-40

40-50

50-60

60-70

$(i)$ For Cumulative Frequency $30-40$ frequency is $7$

$(ii)$ Size is $11$

Mid point for cumulative frequency $30-40$ is $35$ .

$(iii)$ Range $=70-10=60$

The following is the number of daily commuters who travel different distances every day. Draw the histogram.
Distance (in km) $20-40$ $40-50$ $50-60$ $60-75$ $75-100$
No. of commuters $80$ $120$ $70$ $60$ $40$

We shall calculate the length of each rectangle corresponding to the class intervals using frequency density and minimum class width of the given data.

Distance(Class interval)	Commuters	Class width	Length of rectangle $=\dfrac{f\times C}{width}$
$20-40$	$80$	$20$	$\dfrac{80}{20}\times 10=40$
$40-50$	$120$	$10$	$\dfrac{120}{10}\times 10=120$
$50-60$	$70$	$10$	$\dfrac{70}{10}\times 10=70$
$60-75$	$60$	$15$	$\dfrac{60}{15}\times 10=40$
$75-100$	$40$	$25$	$\dfrac{40}{25}\times 10=16$

Here distance is the class intervals and number of commuters is the frequency.
Now, taking

$1$ cm

$= 10$ on

$X$ -axis and

$1$ cm

$= 10$ on

$Y$ -axis, we can draw the histogram of varying width. We get the histogram of this kind.

Construct histogram of variable width for the following data.
$CI$ $25-29$ $30-35$ $36-40$ $41-50$ $51-56$ $57-60$
$f$ $10$ $24$ $15$ $20$ $12$ $16$

$\Rightarrow$ Adjusted frequency

$=\dfrac{Minimum\,class\,size}{Class\,size}\times frequency$

Here, Minimum class size

$=4$

$CI$	$f$	$Class-size$	$Adjusted\,frequency$
$24.5-29.5$	$10$	$5$	$8$
$29.5-35.5$	$24$	$6$	$16$
$35.5-40.5$	$15$	$5$	$12$
$40.5-50.5$	$20$	$10$	$8$
$50.5-56.5$	$12$	$6$	$8$
$56.5-60.5$	$16$	$4$	$16$

We are going to draw histogram between class interval and adjusted frequency.

1241671_559320_ans_c5e4d2fe58ec455596e1d2998a7aef52.jpeg

Draw a frequency polygon for the following frequency distribution.
$CI$ $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
$f$ $12$ $18$ $6$ $10$ $8$ $25$ $22$ $4$

Construct a histogram for the given data. Mark the mid point of the tops of rectangles of histogram and join these points by line segments. To complete the polygon, mark two classes

$(-10$ to

$0)$ and

$(80$ to

$90)$ . Join the end point of the line segments to the mid points of these two classes. The curve so obtained is a frequency polygon.

Marks of $30$ students in mathematics test are given in the adjacent grouped frequency distribution.
Sl. No Marks No of students
$1$ $0 - 5$ $5$
$2$ $5 - 10$ $7$
$3$ $10 - 15$ $10$
$4$ $15 - 20$ $6$
$5$ $20 - 25$ $2$
What are the individual marks of $2$ students in the fifth group?

$\Rightarrow$ The first column here represents the marks obtained in class interval form. The lowest number in a class interval is called the lower limit and the highest number is called the upper limit. This example is a case of continuous class intervals as the upper limit of one class is the lower limit of the following class.

$\Rightarrow$ In continuous cases, any observation corresponding to the extreme values of a class is always included in that class where it is the lower limit. For example, if we had a student who has scored

$5$ marks in the test, his marks would be included in the class interval

$5-10$ and not

$0-5.$

$\therefore$ Marks of

$2$ students who are in

$5^{th}$ group is between

$20-24$ .

Construct histogram of variable width for the following data.
$CI$ $0-10$ $10-15$ $15-20$ $20-30$ $30-40$ $40-60$ $60-70$
$f$ $20$ $15$ $10$ $25$ $5$ $30$ $50$

$\Rightarrow$ Here, we can see class width is not same, so we have to find adjacent frequency to draw histogram.

$\Rightarrow$ Adjacent frequency

$=\dfrac{Minimum\,class\,size}{Class\,size}\times Frequency$

$\Rightarrow$ Where, minimum class size is

$5$

$CI$	$Frequency$	$Class\, size$	$Adjacent\,frequency$
$0-10$	$20$	$10$	$10$
$10-15$	$15$	$5$	$15$
$15-20$	$10$	$5$	$10$
$20-30$	$25$	$10$	$12.5$
$30-40$	$5$	$10$	$2.5$
$40-60$	$30$	$20$	$7.5$
$60-70$	$50$	$10$	$25$

$\Rightarrow$

$X-axis$ contains

$CI$ and

$Y-axis$ contains adjacent frequency.

1049460_559322_ans_16562b5fb1594df9b6810b58923d4cb9.jpeg

Draw a frequency polygon of the following data:
$CI$ $10-19$ $20-29$ $30-39$ $40-49$ $50-59$ $60-69$ $70-79$ $80-90$
$f$ $3$ $7$ $5$ $11$ $6$ $8$ $4$ $5$

We can also draw a frequency polygon without drawing the histogram. We shall understand this.
Observe that the class intervals are given in inclusive form. They are to be converted in to exclusive form by adding the correction factor. The correction factor is

$0.5$ . This is subtracted from the lower limit and added to the upper limit. Converting the data into exclusive form we get:

CI	$9.5-19.5$	$19.5-29.5$	$29.5-39.5$	$39.5-49.5$	$49.5-59.5$	$59.5-69.5$	$69.5-79.5$	$79.5-89.5$
M.P.	$14.5$	$24.5$	$34.$	$44.5$	$54.5$	$64.5$	$74.5$	$84.5$
$f$	$3$	$7$	$5$	$11$	$6$	$8$	$4$	$5$

Now take a suitable scale for class intervals on X-axis and corresponding frequency on Y-axis. Plot the frequency against the midpoint of the class interval and join these points by line segments to get the frequency polygon.

The number of goals scored by a hockey team in $20$ matches is given here:
$4, 6, 3, 2, 2, 4, 1, 5, 3, 0, 4, 5, 4, 5, 4, 0, 4, 3, 6, 4$ .
Find the mean.

To find the mean, let us prepare a frequency distribution table first. We observe that some scores are repeated. So to find the sum of all scores, we have to multiply each score with its frequency and then find the sum.

Scores $(x_i)$	Frequency $(f_i)$	$f_i x_i$
$0$	$2$	$0$
$1$	$1$	$1$
$2$	$2$	$4$
$3$	$3$	$9$
$4$	$7$	$28$
$5$	$3$	$15$
$6$	$2$	$12$
	$\displaystyle \sum N =20$	$\displaystyle \sum f_i x_i =69$

To do this, let us denote the scores by

$x_i$ and frequency by

$f_i$ , then multiply

$f_i$ and

$x_i$ and add the product

$f_i x_i$ . Here

$\sum f_i x_i$ denotes the sum of all the product

$f_i \times x_i$ . Now the mean is

$\begin{aligned}{}\overline X &= \frac{{{\text{sum of the scores}}}}{{{\text{number of scores}}}}\\& = \frac{{\sum {{f_i}} {x_i}}}{N}\\ &= \frac{{69}}{{20}}\end{aligned}$

Thus, we get mean goals scored

$\overline {X} = 3.45.$

Calculate the range and coefficient of range from the following data:
The heights of $10$ children in $cm : 122, 144, 154, 101, 168, 118, 155, 133, 160, 140$

$X_{max}=168,\quad X_{min}=101$

So, Range

$= X_{max}-X_{min}=67$

Coefficient of range

$=\cfrac{X_{max}-X_{min}}{X_{max}+X_{min}}=\cfrac{168-101}{168+101}=\cfrac{67}{269}=0.24$

Three coins were tossed $30$ times simultaneously. Each time the number of heads occurring was noted down as follows;
$1\ 2\ 3\ 2\ 3\ 1\ 1\ 1\ 0\ 3\ 2\ 1\ 2\ 2\ 1\ 1\ 2\ 3\ 2\ 0\ 3\ 0\ 1\ 2\ 3\ 2\ 2\ 3\ 1\ 1$
Prepare a frequency distribution table for the data given above

1. To make the frequency distribution table, first write the categories in one column (number of heads).

2. Next, write the frequency in the next column. The frequency is just the total.

Number of Heads	Frequency
$0$	$3$
$1$	$10$
$2$	$10$
$3$	$7$

The frequency distribution of the age of persons (in years) attending trekking camp is given below:
Age (in years) Number of persons
$15-19$ $16$
$20-24$ $60$
$25-29$ $50$
$30-34$ $30$
Find the class mark of the class $20-24$ .

Classmark is the mid point of a given class interval

The mid-point of the class

$20-24=\cfrac{20+24}{2}$

$=\cfrac{44}{2}=22$
Therefore class mark of the class

$20-24$ is

$22$ .

The marks obtained by $36$ students in a test in statistics (out of $50$ ) are as below.
$3\ 5\ 1\ 15\ 35\ 45\ 23\ 31\ 40\ 21\ 13\ 15\ 20\ 47\ 48\ 42\ 34\ 43\ 45\ 33\ 37\ 11\ 13\ 27\ 18\ 12\ 37\ 39\ 38\ 16\ 13\ 18$ $5\ 41\ 47\ 43$
Construct a frequency distribution table with equal class intervals, one of them being $10-20$ ( $20$ is not included)

$\Rightarrow$ Frequency distribution table with equal class interval are given below :

$Marks$	$Frequency$
$0-10$	$4$
$10-20$	$10$
$20-30$	$4$
$30-40$	$8$
$40-50$	$10$

Weekly wages of $250$ workers in a factory are given in the following table. Construct the histogram and frequency polygon on the same graph for the data given.
Weekly wages $500 - 550$ $550 - 600$ $600 - 650$ $650 - 700$ $700 - 750$ $750 - 800$
No of workers $30$ $42$ $50$ $55$ $45$ $28$

$Class-interval$	$Frequency$	$Class-mark$
$450-500$	$0$	$475$
$500-550$	$30$	$525$
$550-600$	$42$	$575$
$600-650$	$50$	$625$
$650-700$	$55$	$675$
$700-750$	$45$	$725$
$750-800$	$28$	$775$
$800-850$	$0$	$825$

$\Rightarrow$ Co-ordinates of frequency polygon are

$(475,0),(525,30),(575,42),(625,50),(675,55),(725,45),(775,28)$ and

$(825,0)$ .

$\Rightarrow$ Above graph shows histogram and frequency of polygon.

1118029_564920_ans_34d44545ef9a4fd3b7c017debf457c2a.png

A company manufactures car batteries of a particular type. The life (in years) of 40 batteries were recorded as follows:
$2.6\ 3.0\ 3.7\ 3.2\ 2.2\ 4.1\ 3.5\ 4.5$
$3.5\ 2.3\ 3.2\ 3.4\ 3.8\ 3.2\ 4.6\ 3.7$
$2.5\ 4.4\ 3.4\ 3.3\ 2.9\ 3.0\ 4.3\ 2.8$
$3.5\ 3.2\ 3.9\ 3.2\ 3.2\ 3.1\ 3.7\ 3.4$
$4.6\ 3.8\ 3.2\ 2.6\ 3.5\ 4.2\ 2.9\ 3.6$
Construct a grouped frequency distribution table with exclusive classes for this data, using class intervals of size $0.5$ starting from the interval $2 - 2.5$

Life (in years)	Number of batteries
2.0-2.5	2
2.5-3.0	6
3.0-3.5	14
3.5-4.0	11
4.0-4.5	4
4.5-5.0	3

Given is the life of

$40$ batteries in years.

We know that a grouped frequency distribution table has class-intervals or groups and their corresponding frequencies.

It is given that the class size needs to be

$0.5$ and the class intervals must start with

$2-2.5$ .

Also,

$4.6$ is the highest value in the given data.

Hence, our class intervals will be

$2-2.5,\ 2.5-3,\ 3-3.5,\ 3.5-4,\ 4-4.5\text{ and }4.5-5$ .

Also, the classes are exclusive. It means the upper limit of a class is excluded from that class (and is included in the next class).

So, the grouped frequency distribution table for the given data will be as shown above.

Ages of $60$ teachers in primary schools of a Mandal are given in the following frequency distribution table. Construct the Frequency polygon and frequency curve for the data without using the histogram. (Use separate graph sheets)
Ages $24 - 28$ $28 - 32$ $32 - 36$ $36 - 40$ $40 - 44$ $44 - 48$
No of teachers $12$ $10$ $15$ $9$ $8$ $6$

$Ages$	$Class-mark$	$No.\,of\,teachers$
$20-24$	$22$	$0$
$24-28$	$26$	$12$
$28-32$	$30$	$10$
$32-36$	$34$	$15$
$36-40$	$38$	$9$
$40-44$	$42$	$8$
$44-48$	$46$	$6$
$48-52$	$50$	$0$

1241694_564923_ans_a66644afb58241c3ad2ad758cad96d15.png

The age (in years) of $360$ patients treated in the hospital on a particular day are given below.
Age in years $10-20$ $20-30$ $30-50$ $50-60$ $60-70$
No. of patients $80$ $50$ $80$ $120$ $30$
Draw a histogram for the above data.

Consider the class intervals that is age in years on the X – axis and number of patients on the Y- axis and draw rectangles which provides the required histogram.

1130184_619681_ans_5d88b16559c242089e86b3df134b7d7a.png

The following is the frequency distribution of waiting time at $A$ centre; draw histogram to represent the data.
Waiting time(in seconds)
Number of Customers
$0-30$
$5$
$30-60$
$50$
$60-90$
$64$
$90-120$
$23$
$120-150$
$15$

Given the distribution of number of customers in respective waiting times.

Draw a histogram for the following table which represent the marks obtained by $100$ students in an examination.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Number of students $5$ $10$ $15$ $20$ $25$ $12$ $8$ $5$

The class intervals are all equal with length of

$10$ marks. Let us donate these class intervals along the X-axis. Denote the number of students along the Y-axis, with appropriate scale. The histogram is given below.

The marks obtained by $50$ students in a Mathematics test with maximum mrks of $100$ are given as follows.
43, 88, 25, 93, 68, 81, 29, 41, 45, 87, 34, 50, 61, 75, 51, 96, 20, 13, 18, 35, 25, 77, 62, 98, 47, 36, 15, 40, 9, 25, 39, 60, 37, 50, 19, 86, 42, 29, 32, 61, 45, 68, 41, 87, 61, 44, 67, 30, 54, 8.
Prepare a frequency table for the above data using class interval.

Total number of values

$=50$
Range

$=$ Highest value

$-$ Lowest value

$=98-8=90$
Let us divide the given data into

$10$ classes.

$\therefore$ Length of the Class interval

$=\displaystyle\frac{Range}{\text{Number of class interval}}$

$=\displaystyle\frac{90}{10}=10$
The frequency table of the marks obtained by

$50$ students in a mathematics test is as follows.

Draw a histogram for the following data:
Marks No. of Students
$10-20$ $15$
$20-30$ $30$
$30-40$ $50$
$40-50$ $40$
$50-60$ $20$

The given table has marks obtained by

$15$ students.

We know that a histogram is a pictorial representation of a grouped frequency distribution using rectangular bars of width equal to the size of the groups and length equal to the corresponding frequency.

Here, the group size is

$10$ .

Hence, the width of each bar will be

$10$ .

Also, the height of each bar represents the number of students. For example,

$15$ students have got marks between

$10-20$ , so the height of the rectangular bar representing the class

$10-20$ will be

$15$ . Similarly, we will decide the height of each rectangular bar in the histogram.

Hence, by taking marks on the

$x-axis$ and number of students on the

$y-axis$ , we can construct a histogram as shown above.

2090986_599870_ans_09ad770a8dbd4b5e99d01e8e7ad02d3a.png

Construct a histogram and frequey polygon for the following data.
Class interval $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency $7$ $10$ $23$ $11$ $8$ $5$

1297924_618274_ans_0ae7f8de59c94cd9aace04a69d547e3b.jpg

Draw a Histogram for the following data :
Class Interval Frequency
$0-10$ $35$
$10-20$ $70$
$20-30$ $20$
$30-40$ $40$
$40-50$ $50$

The histogram for the given data is constructed on the graph shown.

Consider the following data.
$15, 17, 17, 20, 15, 18, 16, 25, 16, 15,$
$16, 18, 20, 28, 30, 27, 18, 18, 20, 25,$
$16, 16, 20, 28, 15, 18, 20, 20, 20, 25$ . Form a frequency table.

The frequency table is given above.

Given below are marks obtained by $20$ students in Math out of $25$ .
$21, 23, 19, 17, 12, 15, 15, 17, 17, 19, 23, 23, 21, 23, 25, 25, 21, 19, 19, 19$
Prepare a frequency distribution table for the given data.

Arrange the given data in ascending order

$12,15,15,17,17,17,19,19,19,19,19,21,21,21,23,23,23,23,25,25$

Frequency Distribution Table

Marks	12	15	17	19	21	23	25
No. of students	1	2	3	5	3	4	2

Calculate the frequency data of age of class students.

Age (x)	No. of students(f)	$(x\times f)$
10	4	40
12	15	180
14	24	336
15	30	450
13	12	156
11	9	99
9	11	99
	Total =105	Total=1360

$Mean=\dfrac{\sum f.x}{\sum f }$

$=\dfrac{1360}{105 }$

$=12.95\sim 13$

Create an Ungrouped Frequency Distribution table with the data from the survey, accomplished among the students of university, which answered the question of how many books they read per year. The answers are presented below:
7 8 9 5 4 6 7 3 4

No. of books (read per year)	$3$	$4$	$5$	$6$	$7$	$8$	$9$
No. of students	$1$	$2$	$1$	$1$	$2$	$1$	$1$

Find the median for the following frequency distribution
Height (in $cm$ )
$160-162$ $163-165$ $166-168$ $169-171$ $172-174$
Frequency $15$ $117$ $136$ $110$ $14$

REF. Image.

Total No. of students (N) = 400

$\dfrac{N}{2} = 200$

Median Class

$= 165.5-168.5$ [having nearest value

$> N/2$ ]

Median

$= l+[\dfrac{N/2-cf}{f}]h$

l = Lower limit of median class

cf = Cumulative frequency of class preceding median class

f = frequency of median class

h = height of class

Median

$= 165.5+[\dfrac{200-132}{136}]3$

$= 165.5+[\dfrac{68}{136}]3 = 165.5+\dfrac{3}{2}$

$= 165.5+1.5$

Median

$= 167\,cm$

1178256_875340_ans_a24df465980943779fac55b10e4e27d3.JPG

Draw the or give the less than and more than of the date given below (Ex 14.03.02) and find the median
Class Interval Frequency
0-10 5
10-20 8
20-30 20
30-40 15
40-50 7
50-60 5
Total 60

Class Interval	Frequency	Less Than Cummulative Frequency	More Than Cummulative Frequency
$0-10$	$5$	$5$	$60$
$10-20$	$8$	$5+8=13$	$60-5=55$
$20-30$	$20$	$13+20=33$	$55-8=47$
$30-40$	$15$	$33+15=48$	$47-20=27$
$40-50$	$7$	$48+7=55$	$27-15=12$
$50-60$	$5$	$55+5=60$	$12-7=5$
$Total$	$60$

956143_1018579_ans_7086318dd9d84db8ba7111bf8adabb53.jpg

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is $50$ .
$x$ : $10$ $30$ $50$ $70$ $90$
$f$ : $17$ ${f}_{1}$ $32$ ${f}_{2}$ $19$ Total: $120$

Given frequency

$=120$

$\Rightarrow 17+f_{1}+32+f_{2}+19=120$

$\Rightarrow f_{1}+f_{2}=52$

Mean =

$\dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ mean = $\dfrac {(10\times 17)+(30\times f1)+(50 \times 32)+(72 \times f2)+(90 \times 19)}{120}$

$\Rightarrow$ mean =

$\dfrac {170+30f_{1}+1600+70f_{2}+1710}{120}=50$

$\Rightarrow 3f_{1}+7f_{2}=252$

$\Rightarrow 3f_{1}+7(52-f_{1})=252$

$\Rightarrow f_{1}=28$

$\Rightarrow f_{2}=24$

A consumer buys a certain quantity of a good at a price of $Rs. 10$ per unit. When price falls to $Rs. 8$ per unit, she buys $40$ per cent quantity. Calculate price elasticity of demand.

Suppose the consumer initially buys

$100$ units of the good.
New quantity

$= 100 + 40$ per cent of

$100 = 140$

$P = Rs. 10; P_{1} = Rs. 9; \triangle P = P_{1} - P = Rs. 8 - Rs. 10 = (-) Rs. 2$

$Q = 100\ units; Q_{1} = 140\ units; \triangle Q = Q_{1} - Q = (140 - 100) units = 40\ units$
Price elasticity of demand

$(E_{d}) = (-)\dfrac {P}{Q}\times \dfrac {\triangle Q}{\triangle P}$

$= (-) \dfrac {10}{100}\times \dfrac {40}{-2} = 2$
Percentage change in price

$= \dfrac {\triangle P}{P}\times 100$

$= \dfrac {8 - 10}{10}\times 100 =\dfrac {-2}{10}\times 100 = (-) 20\%$
Price elasticity of demand

$(E_{d}) = (-) \dfrac {\text {Percentage change in quantity demanded}}{\text {Percentage change in price}}$

$= (-) \dfrac {40\%}{(-)20\%} = 2$
Price elasticity of demand

$= 2$ .

Giving reason, comment on the shape of production possibilities curve based on the following schedule:
Good-X (Units) $0$ $1$ $2$ $3$ $4$
Good-Y (Units) $16$ $12$ $8$ $4$ $0$

Good-X (Units)	Good-Y (Units)	Marginal Rate of Transformation
$0$	$16$	-
$1$	$12$	$4Y : 1X$
$2$	$8$	$4Y : 1X$
$3$	$4$	$4Y : 1X$
$4$	$0$	$4Y : 1X$

Since marginal rate of transformation is constant, the production possibility curve is to be a straight line sloping downward.

The marginal utility from the consumption of commodity - X is given in following table:
Units of $X$ $1$ $2$ $3$ $4$ $5$
$MU_{x}$ $10$ $9$ $7$ $3$ $-5$
Calculate total utility for each unit of consumption of commodity - X.

Units of

$X$

$MU_{x}$

-5

$TU_{x}$

Total utility =

$TU_{x}$ is given above for each unit.

Write the frequency distribution table for the following data.
Marks Below $10$ Below $15$ Below $20$ Below $30$ Below $35$ Below $40$
Number of students $0$ $15$ $20$ $30$ $35$ $40$

$\Rightarrow$ The following table is frequency distribution table.

$Class\,Interval$	$Frequency$
$5-10$	$0$
$10-15$	$15$
$15-20$	$20$
$20-30$	$30$
$30-35$	$35$
$35-40$	$40$

Profits per shop $0-50$ $50-100$ $100-150$ $150-200$ $200-250$ $250-300$
No. of shops $12$ $18$ $27$ $20$ $17$ $6$
The monthly profits (in Rs.) of 100 shops are distributed as follows:

Draw the frequency polygon for it.

$Profit\,per\,shop$	$Mid-value$	$No.of\,shops$
Less than $0$	$0$	$0$
$0-50$	$25$	$12$
$50-100$	$75$	$18$
$100-150$	$125$	$27$
$150-200$	$175$	$20$
$200-250$	$225$	$17$
$250-300$	$275$	$6$
More than $300$	$0$	$0$

$\Rightarrow$ Now, we are going to draw graph of frequency of polygon between mid-value and number of shops.
1213255_1038792_ans_a7123c9cd18446f89005e44ab2149f98.png

1213255_1038792_ans_a7123c9cd18446f89005e44ab2149f98.png

A consumer buys $18$ units of a good at a price of $Rs. 9$ per unit. The price elasticity of demand for the good is $(-) 1$ . How many units the consumer will buy at a price of $Rs. 10$ per unit? Calculate.

Given,

$P = Rs. 9; P_{1} = Rs. 10; \triangle P = P_{1} - P = Rs. 10 - Rs. 9 = Rs. 1$

$Q = 18\ units; Q_{1} = ?$

$E_{d} = (-) 1$
Price elasticity of demand

$(E_{d}) = \dfrac {P}{Q} \times \dfrac {\triangle Q}{\triangle P}$

$-1 = \dfrac {9}{18}\times \dfrac {\triangle Q}{1}$

$\triangle Q = -2$
Now,

$\triangle Q = Q_{1} - Q$

$-2 = Q_{1} - 18$

$Q_{1} = 18 - 2 = 16$
Consumer will buy

$16$ units at a price of

$Rs. 10$ per unit.

For the frequency distribution given, find the class length of each class

$Class-interval$	$Frequency$	$Class\,length$
$0-9$	$10$	$9-0=9$
$10-24$	$20$	$24-10=14$
$25-49$	$30$	$49-25=24$
$50-74$	$20$	$74-50=24$
$75-100$	$10$	$100-75=25$

Production yield $(kg/ha)$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
Number of farms $10$ $15$ $30$ $32$ $8$ $5$
Construct a less than type ogive and hence find its median.

$\Rightarrow$ Now, we are going to form less than frequency distribution table :

$Production\,yield\,(kg/ha)$	$No.\,of\,farms$
Less than $20$	$10$
Less than $30$	$10+15=25$
Less than $40$	$25+30=55$
Less than $50$	$55+32=87$
Less than $60$	$87+8=95$
Less than $70$	$95+5=100$

$\Rightarrow$

$N=100$ then,

$\dfrac{N}{2}=\dfrac{100}{2}=50$

We are going to draw horizontal line from

$50$ on

$y-axis$ and draw vertical from where it touches curve.

The point at which line touches

$x-axis$ is our median.

$\therefore$

$Median=38$

1219324_1132867_ans_a3f635f371ad493999b116a4533c3fc4.jpeg

Construct a frequency distribution table with class interval of 5 for the following data:
2,4,8,11,14,16,20,28,25,40,30,48,5,22,29,13,22,17,17,7.

We have,

$2,4,8,11,14,16,20,28,25,40,30,48,5,22,29,13,22,17,17,7$

Take a look at the table below to understand the concept better:-
1166289_1155378_ans_da280463cf6149f4b145f9a7cc18b403.png

Which value decides the width of each rectangle in the histogram ?

1160317_1206067_ans_d236fcdcb9d04a68a0cfd1ab375a891f.jpg

Draw the frequency polygon of the following data:
Marks 0-10 10-20 20-30 30-40 40-50
Number of students 5 10 12 13 15

$Marks$	$Mid-value$ $(Marks)$	$Number\,of\,students$
Less than $0$	$0$	$0$
$0-10$	$5$	$5$
$10-20$	$15$	$10$
$20-30$	$25$	$12$
$30-40$	$35$	$13$
$40-50$	$45$	$15$
More than $50$	$0$	$0$

1216393_1126690_ans_3f7ed81be097480cab80648d05142642.png

Construct a histogram for the marks obtained by $600$ students in the VII class annual examinations.
Marks $360$ $400$ $440$ $480$ $520$ $560$
No. of students $100$ $125$ $140$ $95$ $80$ $60$

1178720_1074058_ans_f46e78384d5a4c278c74eebc4acd0e06.jpg

Given below are the ages of 45 people in a colony.
33 8 7 25 31 26 5 20 25 48 56
33 28 22 15 62 59 16 14 19 24 35
26 9 12 46 15 42 63 32 5 22 11
42 23 52 48 62 10 24 43 51 37 48
36

Construct group frequency distribution for the given data with 6 class intervals.

$\Rightarrow$ Maximum value

$=63$

$\Rightarrow$ Minimum value

$=5$

$\Rightarrow$ Range

$=63-5=58$

$\Rightarrow$ The number of class interval

$=6$

$\Rightarrow$ Length of class interval

$=\dfrac{58}{6}=9.6\approx 10$

$Class-Interval$	$Frequency$
$0-9$	$5$
$10-19$	$8$
$20-29$	$11$
$30-39$	$7$
$40-49$	$7$
$50-59$	$4$
$60-69$	$3$
$Total$	$45$

If the collection of data in Q.no.1 be grouped into the class intervals $40-45,45-50,50-55,55-60,60-65,65-70$ then answer the following.
1)What are the class limits and class boundaries of the class interval $60-65$ ?
2)What are the class size and the class mark of the class interval $45-50$ ?

1160337_1051717_ans_f170ab4e387b4bc18cd6f8a80467babe.jpg

Time alloted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.
Time (minutes) $60-80$ $80-100$ $100-120$ $120-140$ $140-160$
No. of students $14$ $20$ $24$ $22$ $16$

$\Rightarrow$ The above image is required histogram.

$\Rightarrow$

$x-axis$ contain Time (minutes) and

$y-axis$ contains number of students.

1216339_1117158_ans_2e125cc830a34242bfcda6b64da40cc7.jpeg

Find the class mark of class interval $50-60.$

$\Rightarrow$

$Class\,mark=\dfrac{Lower\,limit+Upper\,limit}{2}$

$\Rightarrow$ Given class interval is

$50-60$

$\Rightarrow$ Lower limit

$=50$ and upper limit

$=60$

$\therefore$

$Class\,mark=\dfrac{50+60}{2}$

$=\dfrac{110}{2}$

$=55$

$\therefore$ The class mark of class onterval

$50-60$ is

$55$ .

The newspapers Hindu (H), Times of India (T), Indian Express (I), and Deccan chronicle (D) are read by 24 members are as follows:

H, H, I, D, T, H, T, D, I, T, D, D, H, I, T, H, D, I, D, D, H, I, T, T

Represent the above data in the form of a frequency distribution table.

Newspaper	Frequency
Hindu	6
Times of India	6
Indian Express	5
Deccan Chronicle	7

Frequency distribution table:

Hindu newspaper was read by 6 members
Hence, frequency of Hindu

$=6$

Indian Express newspaper was read by 5 members

Frequency of Indian Express

$=5$

Deccan Chronicle was read by 7 members

Frequency

$=7$

Times of India was read by 6 members

Frequency of Times of India

$=6$

If the class intervals of a frequency distribution be $16-25, 26-35, 36-45, 46-55$ , then determine
(a) the class limits and class boundaries of the class $36-45$
(b) the class size and the class mark of the class interval $26-35$
(c) the class intervals when changed into overlapping class interval

(a)

Lower class limit of the interval 36 - 45 = 36 and

Upper class limit of the interval 36 - 45 = 45

(b)

Class Size of the interval 26 - 35 = 10 and

Class Mark of the interval 26 - 35 = 30 (Class Mark is the average of the limits of the class interval)

(c)

The class intervals when changed into overlapping class intervals are

15.5 - 25.5

25.5 - 35.5

35.5 - 45.5

45.5 - 55.5

The following table gives the distribution of students of two sections according the marks obtained by them:
Section A Section B
Marks Frequency Marks Frequency
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50 3
9
17
12
9 0 - 10
10 - 20
20 - 30
30 - 40
40 - 50 5
19
15
10
1
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.

$Marks$	$Class-mark$	$Section\,A\,frequency$	$Section\,B\,frequency$
$0-10$	$5$	$3$	$5$
$10-20$	$15$	$9$	$19$
$20-30$	$25$	$17$	$15$
$30-40$	$35$	$12$	$10$
$40-50$	$45$	$9$	$1$

Since, in the graph section

$A$ is to the right of section

$B$ , more students of section

$A$ has highest marks.

1243363_1181338_ans_917da5eefc7d43c1a54ab435406b94ee.png

C.I 0-4 4-8 8-12 12-16 16-20 20-24
frequency 4 7 10 15 11 6
Construct frequency polygon for the following table

$C.I.$	$class-mark$	$Frequency$
Less than $0$	$0$	$0$
$0-4$	$2$	$4$
$4-8$	$6$	$7$
$8-12$	$10$	$10$
$12-16$	$14$	$15$
$16-20$	$18$	$11$
$20-24$	$22$	$6$
More than $24$	$0$	$0$

1222733_1175761_ans_658b963e65bc4a1eb3c7694c7780985c.png

Weekly wages of 250 worker in a factory are given in the following table. Construct the histogram and frequeney polygon on the same graph for the data given:
Weekly wage 500-550 550-600 600-650 650-700 700-750 750-800
No of worker 30 42 50 55 45 28

So here the frequency table is obtained as follows-

Weekly wages $\left( \text{in Rs.} \right)$	No. of workers
$500-550$	$30$
$550-600$	$42$
$600-650$	$50$
$650-700$	$55$
$700-750$	$45$
$750-800$	$28$

1179633_1156260_ans_d5e8e4f7e9e040258b24b9db6fc63cca.png

Draw a histogram for the following distribution table of marks scored by $75$ students of class $IV$ :
Marks obtained $0-10$ $10-20$ $20-40$ $40-45$ $45-60$ $60-70$ $70-85$
No. of students $4$ $8$ $20$ $10$ $12$ $6$ $15$

1123982_1198983_ans_aa2a3a5a4d5349b6aa8b0afc17e33386.jpg

draw a histogram to represent the following data of the earnings of workers :
Monthly earning Number of workers
80-120 4
120-160 7
160-200 13
200-240 8
240-280 5
280-320 2

Now that we have organized our data by classes, we are ready to draw our histogram.

$1.$ Draw a horizontal line. This will be where we denote our classes.

$2.$ Place evenly spaced marks along this line that correspond to the classes.

$3.$ Label the marks so that the scale is clear and give a name to the horizontal axis.

$4.$ Draw a vertical line just to the left of the lowest class.

Choose a scale for the vertical axis that will accommodate the class with the highest frequency.

Scale:

$1$ cm

$=1$ worker

$y-$ axis

$5.$ Label the marks so that the scale is clear and give a name to the vertical axis.

$6.$ Construct bars for each class. The height of each bar should correspond to the frequency of the class at the base of the bar. We can also use relative frequencies for the heights of our bars.

1480525_1203666_ans_1e7741900d85495bb08de7afcd39a0e3.PNG

Following is the choice of 20 students of class VI of a school:
Banana, Apple, Guava, Orange, Apple, Banana, Orange, Guava, Banana, Guava, Banana, Banana, Apple, Banana, Apple, Banana, Orange, Guava, Apple, Banana, Guava, Banana.
(i) Arrange the name of fruits in a table using tally marks.
(ii) Which fruit is liked by maximum number of students?
(iii) Which fruit is liked by minimum number of students?

(i) Arrangement has been done in the image attached.
(ii) Frequency is maximum for Banana. Hence, banana is most liked fruit.
(iii) Frequency is minimum for Orange. Hence, orange is least liked fruit.
1752017_1199377_ans_a96663ef95d34540851b29759a095c38.png

1752017_1199377_ans_a96663ef95d34540851b29759a095c38.png

If class mark is $25$ and class width is $10$ , then find the class.

${\textbf{Step -1: Assume the lower and upper limit.}}$

${\text{Let lower limit be L}}.$

$\therefore {\text{upper limit}} = L + 10{\text{ [class width}} = 10{\text{ given]}}$

${\textbf{Step -2: Find the lower limit using class mark formula.}}$

${\text{As we know that,}}$

${\text{Class Mark = (Lower Limit + Upper Limit)}} \div {\text{2}} \ldots \left( 1 \right)$

$\Rightarrow 25 = \left( {L + L + 10} \right) \div 2$ $\left[ {{\text{Given class mark: }}25} \right]$

$\Rightarrow 25 \times 2 = \left( {2L + 10} \right)$

$\Rightarrow 50 = 2L + 10$

$\Rightarrow 50 - 10 = 2L$

$\Rightarrow 40 = 2L$

$\Rightarrow L = \dfrac{{40}}{2}$

$\Rightarrow L = 20$

$\therefore {\text{Lower Limit of the class }} = {\text{ }}20$

${\text{Upper Limit of the class }} = 20 + 10$

${\text{Upper Limit of the class }} = 30$

${\text{So, the required class is 20 - 30}}{\text{.}}$

${\textbf{ Hence, the required class is 20 - 30.}}$

For a given frequency distribution, $Z + M = 34$ and $M + \bar{x} = 40$ . Then, find $M$ .

1186006_1176813_ans_53af2da50ca8464290b29a422b2315bf.jpeg

In certain towns, literacy rate (in %) is given in the following table:
Literacy rate
(in %) 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
Number
of towns 5 8 10 x 15 8 y 6

If mode of the data is $53\%$ , find the missing frequencies, when if is given that sum of the frequencies is $70$ .

$\Rightarrow$

$Mode=53\%$

$\therefore$ Modal class

$=50-60$

Then,

$l=50,\,f_0=x,\,f_1=15,\,f_2=8$ and

$h=10$

$\Rightarrow$

$Mode=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times h$

$\Rightarrow$

$53=50+\dfrac{15-x}{2(15)-x-8}\times 10$

$\Rightarrow$

$3=\dfrac{150-10x}{22-x}$

$\Rightarrow$

$66-3x=150-10x$

$\Rightarrow$

$7x=84$

$\therefore$

$x=12$

$\Rightarrow$

$\sum f_i=70$ [ Given ]

$\therefore$

$5+8+10+12+15+8+y+6=70$

$\therefore$

$64+y=70$

$\therefore$

$y=6$

$\therefore$ Missing frequencies are

$12$ and

$6$ .

Class intervals in a grouped frequency distribution are given as $4 - 11,12 - 19,20 - 27$ $28 - 35,36 - 43 .$ Write the next two class intervals.

Given class interval :

$4-11$ ,

$12-19$ ,

$20-27$ ,

$28-35$ ,

$36-43$

The next two class intervals are

$44-51$ ,

$52-59$

length of each class interval is

$8$ .

Class marks are :

Class Interval Class marks

$4-11$

$7.5$

$12-19$

$15.5$

$20-27$

$23.5$

$28-35$

$31.5$

$36-43$

$39.5$

$44-51$

$47.5$

$52-59$

$55.5$

The following are the daily expeditures (in $Rs$ ) of $30$ student in a school.
Ependiture (in $Rs$ )
No. of students
$10-15$ $3$
$15-20$ $5$
$20-25$ $10$
$25-30$ $6$
$30-35$ $4$
$35-40$ $2$
Total $30$
What is the upper-limit of the class interval $30-35$ ?

Upper limit of class interval 30-35 is

$35$

Show the following data by a frequency polygon.
Electricity bill ( $Rs$ ) $0-200$ $200-400$ $400-600$ $600-800$ $800-1000$
Families $240$ $300$ $450$ $350$ $160$

Electricity Bill	x	F
0-200 200-400 400-600 600-800 800-1000	100 300 500 700 900	240 300 450 350 160

1344336_1242991_ans_29c1a158c98542ca8d55ea86d580b7ab.png

The median of the following data is $525$ . Finds the values of $x$ and $y$ if the total frequency is $100$ .
Class interval Frequency
$0-100$ $2$
$100-200$ $5$
$200-300$ $x$
$300-400$ $12$
$400-500$ $17$
$500-600$ $20$
$600-700$ $y$
$700-800$ $9$
$800-900$ $7$
$900-1000$ $4$

Class	frequency (f)	c.f
$0 - 100$	$2$	$2$
$100 - 200$	$5$	$7$
$200 - 300$	$x$	$7+x$
$300 - 400$	$12$	$19+x$
$400 - 500$	$17$	$36 + x$
$500 - 600$	$20$	$56 + x$
$600 - 700$	$y$	$56 + x + y$
$700 - 800$	$9$	$65 + x + y$
$800 - 900$	$7$	$72 + x + y$
$900 - 1000$	$4$	$76 + x+ y$
	$n = \sum f$ $= 76 + x+y$

Given,

medium

$=525, n = \sum f = 76 + x+y = 100$

Now,

$\dfrac{n}{2} = \dfrac{100}{2}$

$\Rightarrow \dfrac{n}{2} = 50$

$\therefore 500 - 600$ is median class

then,

$f = 20$ and

$c.f = 36 +x$

$l = 500,\, h = 100$

Now,

$n = 76 + x+ y = 100$

$\Rightarrow x+y = 24$ ....eq.(i)

and, Median

$= l + \left(\dfrac{\dfrac{n}{2} - c.f}{f}\right)\times h$

$\Rightarrow 525 = 500 + \left(\dfrac{50 - 36 -x}{20}\right)\times 100$

$\therefore x = 9$

and,

$y = 15$

$[ \because x + y = 24]$

The blood groups of 30 students of class X are: AB, O, O, A, B, O, A,O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, B, O, B, A, B, O.
i) Prepare a frequency distribution table from the data.
ii) Which is the most common and which is the rarest blood among these students ?

The blood groups of students are:-

AB,O,O,A,B,O,A,O,B,A,O,B,A,O,O,A,AB,O,A,A,O,O,B,O,B,A,B,O.

Most common blood group: O

Most rare blood group: AB

1193341_1239836_ans_b8a5eb545a244778905a639181ced45a.png

Three dice were thrown, the outcomes are noted and added (The results can be $3,4,5,6 , \dots , 18$ ). Four players played the game $10$ times each. Their scores were:
$5,7,9,8,6,4,6,12,15,18,17,12,10,3,13,7,11,17,10,5 ,$
$4,6,8,16,18,14,6,18,9,10,15,14,3,7,9,17,11,4,12,16$ .
Divide this data into class intervals of size $3$ and draw a histogram.

Now that we have organized our data by classes, we are ready to draw our histogram.

Players	Outcomes
$3-6$	$7$
$6-9$	$9$
$9-12$	$8$
$12-15$	$6$
$15-18$	$7$
$18-21$	$3$
Total	$40$

$1$ Draw a horizontal line. This will be where we denote our classes.

$2.$ Place evenly spaced marks along this line that correspond to the classes.

$3.$ Label the marks so that the scale is clear and give a name to the horizontal axis.

$4.$ Draw a vertical line just to the left of the lowest class.

Choose a scale for the vertical axis that will accommodate the class with the highest frequency.

Scale:

$1$ cm

$=1$ outcome

$y-$ axis

$5.$ Label the marks so that the scale is clear and give a name to the vertical axis.

$6.$ Construct bars for each class. The height of each bar should correspond to the frequency of the class at the base of the bar. We can also use relative frequencies for the heights of our bars.

1480744_1231014_ans_36f182b91dd14359a8f1ebb30407e68b.png

Draw a histogram to represent the following data :
Marks obtained
$0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
No. of student $4$ $7$ $12$ $20$ $9$ $2$

1196597_1223343_ans_01a6e2ce82b240daa1fa01a259c94270.png

Use the table given below to find:
(a) The actual class limits of the fourth class
(b) The class boundaries of the sixth class.
(c) The class mark of the third class.
(d) The upper and lower limits of the fifth class.
(e) The size of the third class.
Class interval $30-34$ $35-39$ $40-44$ $45-49$ $50-54$ $55-59$
Frequency $7$ $10$ $12$ $13$ $8$ $4$

1131868_1229799_ans_1cf367cd18f0474d90066398e8b4ea77.jpg

The following table gives the distribution of 45 students across the different levels of intelligent quotient. Draw the histogram for the data.

A histogram is a graphical display of data using bars of different heights. In a histogram, each bar groups numbers into ranges. Taller bars show that more data falls in that range. A histogram displays the shape and spread of continuous sample data.

The above graph shows histogram of given frequency table.

$x-$ axis− IQ Level

$y-$ axis− Number of students.

1480148_1231335_ans_7377edebdc01485393cc6890206f2f04.png

The following table shows the utilisation of electricity for difference appliances in a home. Represent the information using a pie chart.
Appliances Lamps Freeze Geiser Other
Utilisation of electricity $35\%$ $20\%$ $15\%$ $30\%$

The pie chart for the given information will be as shown above.

The entire area of the circle is divided into

$4$ parts, each part representing one appliance.

The area that a particular appliance gets depends upon the utilization of electricity.

For example, from the given information, we can see that lamps utilize

$35\%$ electricity. Hence, the get

$35\%$ of the area of the circle. Similarly, all the areas can be represented.

That is how we get the pie chart shown above.

2100682_1258811_ans_86c8df9fde9942e2bf08d9df36b5b950.png

Represent the following data by histogram.

Price of Sugar per kg
18-20
20-22
22-24
24-26
26-28
No. of weeks
4
8
22
12
6

1475674_1353364_ans_cf1eed228dae4e70aaf4157377332690.png

Marks secured by $60$ students of a class in a test are given below.
Draw histogram for the given data.
Marks $0 - 10$ $10 - 20$ $20 - 30$ $30 - 40$ $40 - 50$ $50 - 60$
Number of students $7$ $10$ $14$ $20$ $6$ $3$

Histogram for the given data is as follows.
Marks are taken on

$X$ -axis.

Number of students are taken on

$Y$ - axis with scale

$=3$ units

2112546_1317478_ans_646913990e3e4a6c9a423d78783fd07d.png

The median of the data in the following table is $27$ . Find the missing frequencies $f_{1}$ and $f_{2}$

Class*frequency (Cf)

$5$

$5+f_1$

$25+f_1$

$39+f_1$

$39+f_1+f_2$

$47+f_1+f_2$

$47+f_1+f_2=68\Rightarrow f_1+f_2=21$ ...(1)

$m=L+\left ( \dfrac{\frac{n}{2}-Cf}{f} \right )\times h$

$27=20+\left ( \dfrac{34-(5+f_1)}{20} \right )\times 10$

$7=\left ( \dfrac{34-(5+f_1)}{2} \right )$

$7=17-\left ( \dfrac{5+f_1}{2} \right )$

$\dfrac{5+f_1}{2} =17-7=10$

$5+f_1=20$

$\therefore f_1=15$

From (1)

$f_2=21-f_1=21-15$

$\therefore f_2=6$

The performance of Tanik during the two terms in different subjects is as given in the table.
English Hindi Maths Science S.Studies
1st term 75 68 80 85 78
2nd term 72 75 85 90 82
Draw a histogram to represent the above data.

1482808_1348852_ans_8ac6880cef19425da7ac89afad539360.png

Prepare a continuous grouped frequency distribution from the following data:
Mid-point 5 15 25 35 43
Frequency 4 8 13 12 6
Also find the size of class intervals.

We know that , if

$m$ is the midpoint fo a class and

$h$ is the class size,

then lower limit of class interval

$= m - \frac{h}{2}$

Upper limit of class interval

$= m + \frac{h}{2}$

We have class sixe,

$15-5=10$

so the class interval formed from the midpoint

$5$ is

$= \left( {5 - \frac{{10}}{2}} \right) - \left( {5 + \frac{{10}}{2}} \right)\,\,i.e\,\,\,0 - 10$

Containing in the same manner, the continuous classes formed are;

1213743_1305578_ans_f1d09426d46240ac99061a47cb5b04c6.JPG

Convert the following frequency table into a continuous grouped frequency table:
Class interval Frequency
$75-79$ $8$
$80-84$ $10$
$85-89$ $6$
$90-94$ $5$

Given frequency table

Class interval		frequency
75 -79		8
80 -84		10
85 -89		6
90 -94		5

The lower limit of a class interval - Upper limit of the succeeding class interval

$=80-79=1$

$\therefore \dfrac{1}{2}=0.5$

convert the classes by subtracting

$0.5$ from lower limits and adding

$0.5$ from upper limits.

then the continuous grouped frequency table

class interval		frequency
$75-0.5$	$79+0.5$	$8$
$80-0.5$	$84+0.5$	$10$
$85-0.5$	$89 + 0.5$	$6$
$90-0.5$	$94 + 0.5$	$5$

$\Rightarrow$

Class	interval	frequency
$74.5$ $-79.5$		$8$
$79.5-84.5$		$10$
$84.5-89.5$		$6$
$89.5-94.5$		$5$

Draw a histogram to represent the following data:
Pocket money (in $Rs$ )
$150-200$ $200-250$ $250-300$ $300-350$ $350-400$
Number of students $10$ $5$ $7$ $4$ $3$

Now that we have organized our data by classes, we are ready to draw our histogram.

$1$ Draw a horizontal line. This will be where we denote our classes.

$2.$ Place evenly spaced marks along this line that correspond to the classes.

$3.$ Label the marks so that the scale is clear and give a name to the horizontal axis.

$4.$ Draw a vertical line just to the left of the lowest class.

Choose a scale for the vertical axis that will accommodate the class with the highest frequency.

Scale:

$1$ cm

$=1$ student

$y-$ axis

$5.$ Label the marks so that the scale is clear and give a name to the vertical axis.

$6.$ Construct bars for each class. The height of each bar should correspond to the frequency of the class at the base of the bar. We can also use relative frequencies for the heights of our bars.

1480533_1352144_ans_f2ab3637773246dd84279da3c6c7fbb8.png

A total of $20$ patients admitted to a hospital have blood sugar levels as given below :
$67, 69, 74, 73, 70, 70,71, 68, 73, 74, 73,75, 69, 72, 70, 70, 72, 70, 73, 74$ . Make a frequency table.

Let us make interval of size

$2,$

Class Interval	Frequency
$67-69$	$4$
$70-72$	$8$
$73-75$	$8$
Total	$20$

Calculate crude death rates (CDR) for district A:
Age group (in years) number of person (in thousand ) Number of death
$0-15$ $1$ $20$
$15-60$ $3$ $30$
$60$ and above $2$ $40$

Death rate of 0-15 =

$\dfrac{20\times100}{1000}=2$ %

Death rate of 15-60 =

$\dfrac{30\times100}{3000}=1$ %

Death rate of 60 and above =

$\dfrac{40\times100}{2000}=2$ %

Find the difference between the population of $1961$ and $1971$
Year $1931$ $1941$ $1951$ $1961$ $1971$
Population (in crores) $23$ $32$ $35$ $44$ $54$

Population in the year

$1961$ is

$44$

Population in the year

$1971$ is

$54$

The difference in population is

$54-44=10$ crores.

Without drawing a histogram construct a frequency polygon for the given frequency distribution.
Class interval 0-10 10-20 20-30 30-40 40-50
frequency 50 40 45 25 5

1486402_1372370_ans_63d5a32b0cd44392a206afd07c40e9fc.jpg

The following data gives the marks of 20 students in Mathematics (X) and statistics (Y), each out of 10, expressed as (x, y). Construct an ungrouped frequency distribution, considering single numbers as a class:
(2, 7), (3, 8), (4, 9), (2, 8), (5, 6), (5, 7), (4, 9), (3, 8), (4, 8), (2, 9), (3, 8), (4, 8), (5, 6), (4, 7),
(4, 7), (4, 6), (5, 6), (5, 7), (4, 6), (5, 5).

Given are the marks of

$20$ students in Mathematics

$(X)$ and Statistics

$(Y)$ out of

$10$ , in the form of

$(x,y)$ , as:

$(2,7), (3,8), (4,9), (2,8), (5,6), (5,7), (4,9), (3,8), (4,8), (2,9),$

$(3,8), (4,8), (5,6), (4,7), (4,7), (4,6), (5,6), (5,7), (4,6), (5,5)$

For constructing an ungrouped frequency distribution table, we will take the marks in one column and the corresponding number of students in another. Note that, here we have marks for two subjects. Hence, we will need two columns for number of students, one for Mathematics and one for Statistics.

Hence, the ungrouped frequency distribution table will be as shown below:

$\text{Marks}$	$\text{Number of Students (Mathematics)}$	$\text{Number of Students (Statistics)}$
$1$	$0$	$0$
$2$	$3$	$0$
$3$	$3$	$0$
$4$	$8$	$0$
$5$	$6$	$1$
$6$	$0$	$5$
$7$	$0$	$5$
$8$	$0$	$6$
$9$	$0$	$3$
$10$	$0$	$0$

Marks secured by $60$ students of a class in a test are given below. Draw histogram for the given data.

Marks 0-10 10-20 20-30 30-40 40-50 50-60

No. of
Students 7 10 14 20 6 3

$1.$ Draw a horizontal line to denote the class intervals of marks.

$2.$ Place evenly spaced marks along this line that correspond to the classes.

$3.$ Label the marks so that the scale is clear and give a name to the horizontal axis.

$4.$ Draw a vertical line to denote the number of students corresponding to class intervals of marks

$5.$ Label the marks along the vertical axis and give a name to the vertical axis.

$6.$ Construct bars for each class. The height of each bar should correspond to the frequency(number of students) of the class.

Hence our Histogram is ready.

1480527_1387379_ans_c621efdc2de44115afe65c9114f1c912.png

The number of rooms in 25 houses of a locality is as given below:
$4,3,2,6,4,3,2,1,5,3,2,3,4,3,5,1,6,1,3,4,2,3,4,3,5$
Construct a frequency distribution table for the above data.

Given that the number of rooms in

$25$ houses of a locality is as given below:

$4, 3, 2, 6, 4, 3, 2, 1, 5, 3, 2, 3, 4, 3, 5, 1, 6, 1, 3, 4, 2, 3, 4, 3, 5$

Hence, the frequency distribution table for the given data will be as shown below:

$\text{Number of Rooms }(x)$	$\text{Number of houses (frequency)}$
$1$	$3$
$2$	$4$
$3$	$8$
$4$	$5$
$5$	$3$
$6$	$2$

$\text{Total } =25$

Among the group of students, $10\%$ scored marks below $20,\ 20\%$ scored marks between $20$ and $40,\ 35\%$ scored marks between $40$ and $60,\ 20\%$ scored marks between $60$ and $80.$ The remaining $30$ students scored marks between $80$ and $100$ . Prepare frequency distribution table from the above information.

1486418_1366303_ans_467bb271032b425f9441537d159cc7ba.jpg

In the table given below, Find the mean.
(i)
Class mark Frequency
5 3
15 9
25 15
35 13

$f_i=3,9,15,13\\\sum f_i=3+9+15+13=40\\x_i=5,15,25,35\\\sum x_if_i=15+135+375+455=980\\\bar x=\dfrac {980}{40}=24.5$

Find the class marks of classes $30-35, 35-40$ and $40-45$ in the data below:
Marks Number of student
$30-35$ $5$
$35-40$ $8$
$40-45$ $15$
$45-50$ $20$
$50-55$ $5$
$55-60$ $4$
$60-65$ $4$
$65-70$ $2$
$70-75$ $3$

Marks	Number of student
30-35	5
35-40	8
40-45	15
45-50	20
50-55	5
55-60	4
65-70	2
70-75	3

Class mark =

$\dfrac{upper limit + lower limit}{2}$

for 30-35 =

$\dfrac{30+35}{2}=32.5$

for 35-40 =

$\dfrac{35+40}{2}=37.5$

for 40-45=

$\dfrac{40+45}{2}=42.5$

The following table gives the life times of 400 neon lamps :
$\left(i\right)$ Represent the given information with the help of histogram.
Life time (in hours) Number of lamps
$300-400$ $14$
$400-500$ $56$
$500-600$ $60$
$600-700$ $86$
$700-800$ $74$
$800-900$ $62$
$900-1000$ $48$
$\left(ii\right)$ How many lamps have a life time more than $700$ hours?

REF.Image

Numbers of neon lamps having their life time more than 700

= No. of neon lamps having their life time as 700-1000

= No. of neon lamps having their life time (700-800) + = No. of neon lamps having their life time 800-900+ = No. of neon lamps having their life time 900-1000 hours

=74+62+48=184

1240662_1505541_ans_3be63a7be122416a8c176583c8cae736.png

How many student got mark between 45 to 60?
Marks Number of student
$30-35$ $5$
$35-40$ $8$
$40-45$ $15$
$45-50$ $20$
$50-55$ $5$
$55-60$ $4$
$60-65$ $4$
$65-70$ $2$
$70-75$ $3$

1207783_1506372_ans_944e4732f0cf407dbca5962c2aae4706.jpg

For the class $21$ to $30$ , write lower and upper class limit.

Given class is

$21$ to

$30.$

therefore the lower limit is

$21$

the upper limit is

$30$ .

The top speeds of 30 different land animals have been organised into a frequency table given below:
Maximum speed (in km/h) 10-20 20-30 30-40 40-50 50-60 60-70
Number of animals 3 5 10 8 0 2
Draw a histogram for the given data.

1689067_1444780_ans_8e763280c7ff422898a9e54fc97e7e06.png

The mean of the following distribution is 50
${ x }_{ i }$ 10 30 50 70 90
${ f }_{ i }$ 17 5a+3 32 7a-11 19
Find the value 'a'

Mean

$=\dfrac{\sum x_if_i}{\sum x_i}$

$\Rightarrow \dfrac{10(17)+30(5a+3)+50(32)+70(7a-11)+90(19)}{10+30+50+70+90}=50$

$\Rightarrow \dfrac{2800+640a}{250}=50$

$\Rightarrow 2800+640a=12500$

$640a=9700$

$a=15.15625$ .

1205675_1451348_ans_c6da6a162fa648429982b7b00067c769.jpg

In English exam, the average of Class $A$ was found to be $x$ marks. After deducting a computational error, the average marks of $100$ candidates got reduced from $74$ to $54$ . The average thus came down by $25$ marks. The total numbers of candidates who took the English exam were?

Let

$x$ be the total numbers of candidates who took the English exam.

According to the question,

$\Rightarrow$

$(74-54)\times 100=25\times x$

$\Rightarrow$

$20\times 100=25x$

$\Rightarrow$

$2000=25x$

$\Rightarrow$

$x=\dfrac{2000}{25}$

$\Rightarrow$

$x=80$

Age(in years)
Number of teachers
20-24 10
25-29 28
30-34 32
35-39 48
40-44 50
45-49 35
50-54 12
Draw a histogram to represent the following grouped frequency distribution.

The given table is in inclusive form, So, we first convert it into an exclusive form, as given below.

Ages (in years)	Number of teachers
$19.5-24.5$ $24.5-29.5$ $29.5-34.5$ $34.5-39.5$ $39.5-44.5$ $44.5-49.5$ $49.5-54.5$	$10$ $28$ $32$ $48$ $50$ $35$ $12$

A histogram for this table is shown in the figure given below:

1818465_1489107_ans_fa1b8731f01b407c87ac029348a83e49.png

Find the class - mark of the class $130- 150$ .

Class mark = average of 130 and 150

$=\dfrac{130+150}{2}$

$=\dfrac{280}{2}=140$

1380929_1525847_ans_3a3a38d144a743268a23fcd68023a6ac.png

For class interval $20-25$ , write the lower class limit and the upper class limit.

Given class interval

$20 - 25$

$\therefore$ lower class limit

$\rightarrow 20$

upper class limit

$\rightarrow 25$

What are primary data?

$\bullet$ Primary data is data originated for the first time by the researcher through direct efforts and experience, specifically for the purpose of addressing his research problem. Also known as the first hand or raw data.

$\bullet$ Primary data collection is quite expensive, as the research is conducted by the organisation or agency itself, which requires resources like investment and Manpower. The data collection is under direct control and supervision of the investigator.

The median of the following data isFind the missing frequencies a and b, if the total frequency is 70.
Class 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
Frequency 12 a 12 15 b 6 6 4

Class Interval	Frequency	$cf$
$0-5$	$12$	$12$
$5-10$	$a$	$12+a$
$10-15$	$12$	$24+a$
$15-20$	$15$	$39+a$
$20-25$	$b$	$39+a+b$
$25-30$	$6$	$45+a+b$
$30-35$	$6$	$51+a+b$
$35-40$	$4$	$55+a+b=70$

Here:

$N=70$

$N/2=35$
Median

$=16$ , Median class

$=15-20$
So,

$I=15,\ h=5,\ f=15$ and

$xf=xf$ of preceding class i.e.

$24+a$
We know,
Median

$=1+\left\{h\times \dfrac {\left(\dfrac {N}{2}-cf \right)}{f}\right\}$
Substitute all the value in the above formula, we get

$16=15+\left\{5\times (35-24-a)/ 15\right\}$

$\Rightarrow 16=15+(11-a)/3$
or

$a=8$
Now,

$70=55+a+b$

$70=55+8+b$

$\Rightarrow b=7$
so, missing frequincies are

$8$ and

$7$ .

Find the class-mark of the class $35-40$ .

Given that class is

$35 - 40$

Here

Upper class limit is

$40$

lower class limit is

$35$

Class mark it is also called class mid point

$\therefore$ class mark

$= \dfrac{upper \ limit + lower \ limit}{2}$

$= \dfrac{40 + 35}{2}$

$= \dfrac{75}{2}$

Class mark

$= 37.5$

In a survey of 60 families of a colony, the number of members in each family was recorded and the data has been represented by the bar graph shown above:
How many couples have no child?

The family/couple having no child will have only

$2$ members.

This is represented by the yellow bar in the given graph. The height of the bar is

$5$ .

Hence, there are

$5$ couples who do not have children.

The mean of the following is $52$ and the frequency interval $30 - 40$ is $'f$ . Find the value of .
Class interval 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80
Frequency 5 3 f 7 2 6 13

Mean

$= \dfrac{∑ fx}{∑ f} = \dfrac{1940 + 35f}{36 + f}$ …… (i)

But, given mean

$= 52$ …. (ii)

From (i) and (ii), we have

$\dfrac{1940 + 35f}{36 + f} = 52$

$1940 + 35f = 1872 + 52f$

$17f = 68$

Thus,

$f = 4$

1796724_1521206_ans_f1a3524ab894408d92771743e70632d1.PNG

Construct a histogram for the following data:
Monthly School fee (in Rs.) $30-60$ $60-90$ $90-120$ $120-150$ $150-180$ $180-210$ $210-240$
No.of Schools $5$ $12$ $14$ $18$ $10$ $9$ $4$

The class limits are represented along the

$x-$ axis and the frequencies along the

$y-$ axis on a suitable scale.

Taking class intervals as bases and corresponding frequencies as heights of the rectangles, the histogram of the given data can be obtained as shown in the figure above.

1480287_1586715_ans_57cb8044750e438db79620c035579150.png

What are secondary data? Which of the two - the primary or the secondary data - is more reliable and why?

Secondary data refers to data that is collected by someone other than the user. Common sources of secondary data for social science include censuses, information collected by government departments, organizational records and data that was originally collected for other research purpose.

Primary data is more reliable than secondary data because primary data are collected by doing original research and not through secondary sources that may subject to some errors or discrepancy and may even contain ut-dated information. Secondary data are less reliable than primary data.

Following table gives the distribution of students of section A and B of a class according to the marks obtained by them.
Section A Section B
Marks Frequency Marks Frequency
10-15 5 0-15 3
15-30 12 15-30 16
30-45 20 30-45 25
45-60 30 45-60 27
60-75 35 60-75 40
75-90 15 75-90 10
Represent the marks of students of both the section on the same graph by two frequency polygons. What do you observe?

1870708_1521863_ans_aecec7c52bd848ef8183cd9134b63995.jpg

The final marks in mathematics of 30 students are as follows:
$53,61,48,60,78,68,55,100,67,90$
$75,88,77,37,84,58,60,48,62,56$
$44,58,52,64,98,59,70,39,50,60$
Now answer the following:
How many have scored 75 or more?

Given that final marks in mathematics of 30 students are as follows

$53,61,48,60,78,68,55,100,67,90$

$75, 88,77,37,84,58,60,48,62,56$

$44, 58,52, 64,98,59,70,39,50,60$

First of all arrange the marks as groups ascending order means lowest number to highest number

group	ascending order
30 to 39 40 to 49 50 to 59 60 to 69 70 to 79 80 to 89 90 to 100	37, 39 44, 48, 48 50, 52, 53, 55, 56, 58, 58, 59 60, 60, 60, 61, 62, 64, 67, 68 70, 75, 77, 78 84, 88 90, 98, 100

We observe that

highest score

$=100$

lowest score

$= 37$

$\rightarrow$ Number of students scored 75 or more

$= 8$

Explain the meaning of the following terms: class limits

Class limits

These are corresponding to a class interval, the class limits may be defined as the minimum value and the maximum value the class interval may contain.

The minimum value is known as the lower-class limit

The maximum value is known as the upper-class limit
For example,

Class internal	frequency
$10-14$ Upper limit = 14, lower limit = 10 $15-19$ Upper limit = 19, lower limit = 15	$8$ $3$

The weight of new born babies (in kg) in a hospital on a particular day are as follows:
$2.3,2.2,2.1,2.7,2.6,3.0,2.5,2.9,2.8,3.1,2.5,2.8,2.7,2.9,2.4$
Determine the highest weight. (in kg)

Given that the weight of new born babies (in kg) in a hospital on a particular day are as follows:

$2.3, 2.2, 2.1, 2.7,2.6, 3.0, 2.5, 2.9, 2.8, 3.1 , 2.5, 2.8 , 2.7 , 2.9 , 2.4$

first of all arrange data in ascending order

$2.1, 2.2, 2.3, 2.4, 2.5, 2.5, 2.6, 2.7,2.7, 2.8,2.8 ,2.9, 2.9, 3.0, 3.1$

Here observe that highest weight of new born baby is

$3.1$ kg.

Prepare a frequency table of the following scores obtained by 50 students in a test:
42 51 21 42 37 37 42 49 38 52
7 33 17 44 39 7 14 27 39 42
42 62 37 39 67 51 53 53 59 41
29 38 27 31 54 19 53 51 22 61
42 39 59 47 33 34 16 37 57 43

Marks	No. of Students	Marks	No. of Students	Marks	No. of Students
7 14 16 17 19 21 22 27 29 31	2 1 1 1 1 1 1 2 1 1	33 34 37 38 39 41 42 43 44 47	2 1 4 2 4 1 6 1 1 1	49 51 52 53 54 57 59 61 62 67	1 3 1 3 1 1 2 1 1 1

Explain the meaning of the following terms: class-interval

Class interval refers to the numerical width of any class in a particular distribution.

Mathematically it is defined as the difference between the upper class limit and the lower class limit

Class Interval

$=$ Upper-Class limit

$–$ Lower class limit

In statistics, the data is arranged into different classes and the width of such class is called a class interval. Class intervals are generally equal in width but this might not be the case always.

For example,

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$2$	$4$	$0$	$1$	$2$

class interval

$= upper \ limit - lower \ limit$

$= 10 - 0$

$= 10$

The final marks in mathematics of 30 students are as follows:
$53,61,48,60,78,68,55,100,67,90$
$75,88,77,37,84,58,60,48,62,56$
$44,58,52,64,98,59,70,39,50,60$
Now answer the following:
If 40 is the pass mark how many have failed?

Given that final marks in mathematics of 30 students are as follows

$53,61,48,60,78,68,55,100,67,90$

$75, 88,77,37,84,58,60,48,62,56$

$44, 58,52, 64,98,59,70,39,50,60$

first of all arrange the marks as groups ascending order means lowest number to highest number

group	ascending order
30 to 39 40 to 49 50 to 59 60 to 69 70 to 79 80 to 89 90 to 100	37, 39 44, 48, 48 50, 52, 53, 55, 56, 58, 58, 59 60, 60, 60, 61, 62, 64, 67, 68 70, 75, 77, 78 84, 88 90, 98, 100

We observe that

highest score

$=100$

lowest score

$= 37$

$\rightarrow$ If 40 is the pass mark then No of the failed students

$= 2$

The final marks in mathematics of 30 students are as follows:
$53,61,48,60,78,68,55,100,67,90$
$75,88,77,37,84,58,60,48,62,56$
$44,58,52,64,98,59,70,39,50,60$
Now answer the following:
What is the lowest score?

Given

Final marks in mathematics of

$30$ students are as follows

$53,61,48,60,78,68,55,100,67,90$

$75,88,77,37,84,58,60,48,62,56$

$44,58,52,64,98,59,70,39,50,60$

the observations arrange in ascending order

$37,39,44,48,48,50,52,53,55,56,58,58,59,60,60,60,$

$61,62,64,67,68,70,75,77,78,84,88,90,98,100$

the lowest score (marks)

$= 37$

The weight of new born babies (in kg) in a hospital on a particular day are as follows:
$2.3,2.2,2.1,2.7,2.6,3.0,2.5,2.9,2.8,3.1,2.5,2.8,2.7,2.9,2.4$
Determine the lowest weight.

Given that the weight of new born babies (in kg) in a hospital on a particular day are as follows:

$2.3, 2.2, 2.1, 2.7,2.6, 3.0, 2.5, 2.9, 2.8, 3.1 , 2.5, 2.8 , 2.7 , 2.9 , 2.4$

first of all arrange data in ascending order

$2.1, 2.2, 2.3, 2.4, 2.5, 2.5, 2.6, 2.7,2.7, 2.8, 2.9, 2.9, 3.0, 3.1$

Here observe that

highest weight of new born baby is

$3.1 \ kg$ .

lowest weight of new born baby is

$2.1 \ kg$ .

Explain the meaning of the following terms: frequency

Frequency :

The frequency (or absolute frequency) of an event is the number of times the event occurred in an experiment or study.

These frequencies are often graphically represented in histogram

Ex.:-

Age (in years)	No of students
$7$ $8$ $9$ $10$	$12$ $14$ $17$ $8$

The weight of new born babies (in kg) in a hospital on a particular day are as follows:
$2.3,2.2,2.1,2.7,2.6,3.0,2.5,2.9,2.8,3.1,2.5,2.8,2.7,2.9,2.4$
How many babies were born on that day?

Given weights of new born babies (in kg) in a hospital on a particular day are as follows:

$2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1 , 2.5, 2.8 , 2.7 , 2.9 , 2.4$

first of all arrange data in ascending order

$2.1, 2.2, 2.3, 2.4, 2.5, 2.5, 2.6, 2.7,2.7, 2.8,2.8 2.9, 2.9, 3.0, 3.1$

No of babies were born on that day is

$'15'$

$\therefore$ No of new born babies weight below

$2.5 kg$ is

$4$

$\rightarrow$ No of new born babies weight more then

$2.8 kg$ is

$4$

and

No of new born babies weight

$2.8 kg$ is

$2$ .

Explain the meaning of the following terms: class-mark

Classmark: it is also called the midpoint of the class.

$\rightarrow$ The class midpoint (or classmark) is a specific point in the center of the class in a frequency distribution table and an average of the upper and lower class limits.

Ex: the class mark of

$180 - 199$

Average of the class limits

$= \dfrac{180 + 199}{2}$

$= \dfrac{379}{2}$

$= 189.5$

$\therefore$ class mark

$= 189.5$

The weight of new born babies (in kg) in a hospital on a particular day are as follows:
$2.3,2.2,2.1,2.7,2.6,3.0,2.5,2.9,2.8,3.1,2.5,2.8,2.7,2.9,2.4$
How many babies weigh below $2.5 kg$ ?

Given weights of new born babies (in kg) in a hospital on a particular day are as follows:

$2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1 , 2.5, 2.8 , 2.7 , 2.9 , 2.4$

first of all arrange data in ascending order

$2.1, 2.2, 2.3, 2.4, 2.5, 2.5, 2.6, 2.7,2.7, 2.8,2.8 2.9, 2.9, 3.0, 3.1$

No of babies were born on that day is

$'15'$

$\therefore$ No of new born babies weight below

$2.5\ kg$ is

$4$

$\rightarrow$ No of new born babies weight more then

$2.8\ kg$ is

$4$

and

No of new born babies weight

$2.8\ kg$ is

$2$ .

In s survey of $60$ families of a colony, the number of members in each family was recorded and the data has been represented by the bar graph given below:
Which type of family is the most common?

Since, the bar shows the families having four members has the highest height. Hence the families having two children are the most common.

What is the total number of persons living in the town in the age-groups $10-15$ and $60-65$ ?

Data that can be interpreted from the bar graph :

The number of persons in the age group $10-15$ is $=1400$ .

The number of persons in the age group $20-25$ is $=1200$ .

The number of persons in the age group $30-35$ is $=1100$ .

The number of persons in the age group $40-45$ is $=1000$ .

The number of persons in the age group $50-55$ is $=900$ .

The number of persons in the age group $60-65$ is $=800$ .

The number of persons in the age group $70-75$ is $=300$ .

Then, the total number of persons living in the town in the age-groups

$10-15$ and

$60-65$

$=$ number of persons living in the town in the age-group

$(10-15)$

$+$ number of persons living in the town in the age-group

$(60-65)$

$=1400+800=2200$ .

Hence, the total number of persons living in the town in the age-groups

$10-15$ and

$60-65$ is

$2,200$ .

Read the bar graph and answer the following question:
What is the information given by the bar graph?

The given bar graph represents the number of commercial banks in India during some years,

$1969, 1976, 1977, 1978, 1979,1980,1981,1982$ and

$1983$ .

Three coins were tossed 30 times. each time the number of heads occuring was noted down as follow:
$0,1,2,2,1,2,3,1,3,0$
$1,3,1,1,2,2,0,1,2,1$
$3,0,0,1,1,2,3,2,2,0$
Prepare a frequency distribution table for the data given above.

Frequency distribution table

Number of heads	Frequency
0 1 2 3	6 10 9 5

The weight of new born babies (in kg) in a hospital on a particular day are as follows:
$2.3,2.2,2.1,2.7,2.6,3.0,2.5,2.9,2.8,3.1,2.5,2.8,2.7,2.9,2.4$
How many babies weight more than 2.8 kg?

Given weights of new born babies (in kg) in a hospital on a particular day are as follows:

$2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1 , 2.5, 2.8 , 2.7 , 2.9 , 2.4$

first of all arrange data in ascending order

$2.1, 2.2, 2.3, 2.4, 2.5, 2.5, 2.6, 2.7,2.7, 2.8,2.8 2.9, 2.9, 3.0, 3.1$

$\therefore$ No of new born babies weight below

$2.5 kg$ is

$4$

$\rightarrow$ No of new born babies weight more then

$2.8 kg$ is

$4$

and

No of new born babies weight

$2.8 kg$ is

$2$ .

The weight of new born babies (in kg) in a hospital on a particular day are as follows:
$2.3,2.2,2.1,2.7,2.6,3.0,2.5,2.9,2.8,3.1,2.5,2.8,2.7,2.9,2.4$
How many babies weight 2.8 kg?

Given weights of new born babies (in kg) in a hospital on a particular day are as follows

$2.3, 2.2, 2.1, 2.7,2.6,3.0,2.5,2.9, 2.8, 3.1, 2.5, 2.8, 2.7, 2.9, 2.4$

First all arrange the data in an ascending order

$2.1, 2.2, 2.3,2.4,2.5,2.5,2.6,2.7,2.7,2.8,2.8,2.9,2.9,3.0,3.1$

$\therefore$ No of new born babies weight below

$2.5 kg$ is

$4$

$\rightarrow$ No of new born babies weight more than

$2.8 kg$ is

$4$ .

and No. of new born babies weight

$2.8 kg$ is

$2$ .

Study the bar graph representing the number of persons in various age groups in a town shown. Observe the bar graph and answer the following questions:
What is the total population of the town?

From the graph,

The number of persons in the age group $10-15$ is $=1400$ .

The number of persons in the age group $20-25$ is $=1200$ .

The number of persons in the age group $30-35$ is $=1100$ .

The number of persons in the age group $40-45$ is $=1000$ .

The number of persons in the age group $50-55$ is $=900$ .

The number of persons in the age group $60-65$ is $=800$ .

The number of persons in the age group $70-75$ is $=300$ .

Therefore, the total population in the town

$=1400+1200+1100+1000+900+800+300=6700$ .

Hence, the total population in the town is

$6700$ .

Read the above bar graph and answer the following question:
Which state has total production of rice and wheat as its maximum?

Consider a chart by using data from the bar graph:

$States$	$Rice\, production$	$Wheat \,production$	$Total\, production$
$U.P$	$8$	$16$	$24$
$W.P$	$10$	$2$	$12$
$M.P$	$5$	$5$	$10$
$Maharashtra$	$4$	$2$	$6$
$Haryana$	$3$	$6$	$9$

Clearly,

$U.P$ has the total production of rice and wheat as its maximum, i.e.

$24$ .

How many persons are more in the age-group $10-15$ than in the age group $30-35$ ?

From the graph,

The number of persons in the age group $10-15$ is $=1400$ .

The number of persons in the age group $20-25$ is $=1200$ .

The number of persons in the age group $30-35$ is $=1100$ .

The number of persons in the age group $40-45$ is $=1000$ .

The number of persons in the age group $50-55$ is $=900$ .

The number of persons in the age group $60-65$ is $=800$ .

The number of persons in the age group $70-75$ is $=300$ .

Now,

the number of persons in the age-group

$(10-15)$

$-$ the number of persons in the age-group

$(30-35)$ is

$=1400-1100=300$ .

Therefore, the number of persons more in the age-group

$(10-15)$ than in the age group

$(30-35)$ is

$300$ .

The number of runs scored by a cricket player in 25 innings are as follows:
$26,35,94,48,82,105,53,0,39,42,71,0,64,15,34,67,0,42,124,84,54,48,139,$
$64,47.$
How many times did the player not score a run?

Given that the number of runs scored by a cricket player in

$25$ innings are as follows

$26 , 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47$

Here observe that, no of times the player not a score a run

i.e

$0$ runs =

$3$ times

Following data gives the number of children in 41 families:
$1,2,6,5,1,5,1,3,2,6,2,3,4,2,0,0,4,4,3,2,2,0,0,1,2,2,4,3,2,1,0,5,1,2,4,3,4,1,6,2,2.$
Represent it in the form of a frequency distribution.

Given that

Data gives the number of children in 41 families

$1, 2, 6,5,1,5,1,3,2,6,2,3,4,2,0,0,4,4,3,2,2,0,0,1,2,2,4,3,2,1,0,4,1,2,4,3,4,1,6,2,2$

Frequency no of children's	Distribution no of families
$0$ $1$ $2$ $3$ $4$ $5$ $6$	$5$ $7$ $12$ $5$ $6$ $3$ $3$
	$41$

Here no of number in families is

$41$

Find the mean of first five natural numbers.

$\textbf{Step 1: Find the mean.}$

$\text{First five natural numbers are 1, 2, 3, 4 and 5.}$

$\text{To find the mean of 1, 2, 3, 4 and 5.}$

$\text{We know, mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}$

$=\dfrac{1+2+3+4+5}{5}$

$=\dfrac{15}{5}$

$=3$

$\textbf{Hence, the required mean is 3.}$

Draw a histogram for the daily earnings of 30 drug stores in the following table:
Daily earnings (in Rs.): 450-500 500-550 550-600 600-650 650-700
Number of Stores: 16 10 7 3 1

1482817_1671068_ans_1d88e18595d942508a9cb90a3f729536.png

Thirty children were asked about the number of hours they watched T.V. programmes in the previous week. The results were found as follows:
1 6 2 3 5 12 5 8 4 8
10 3 4 12 2 8 15 1 17 6
3 2 8 5 9 6 8 7 14 12
How many children watched television for 15 or more hours a week?

Thirty children were asked about the number of hours they watched T.V. programmes in the previous week.

The results were found as follows:

1 6 2 3 5 12 5 8 4 8

10 3 4 12 2 8 15 1 17 6

3 2 8 5 9 6 8 7 14 12

Here observe that children watched television for

$15$ or more hours a week is

$15$ &

$17$

$\therefore$ Number of children watched television for

$15$ for more hours a week is

$2.$

Construct a histogram for the following data:
Monthly School fee (in Rs): 30-60 60-90 90-120 120-150 150-180 180-210 210-240
No of Schools 5 12 14 18 10 9 4

1482416_1670988_ans_a9c711e96d79406f8ecbd4a06a87a64a.png

The following is the distribution of weights (in kg) of $52$ persons :
Weight (in kg) Persons
$30-40$ 10
$40-50$ 15
$50-60$ 17
$60-70$
6
$70-80$ 4
Find the class marks of the classes $40-50$ , $50-60$ .

We know that,

$\text{Class mark} = \dfrac{\text{lower limit + upper limit}}{2}$

Class mark of

$40-50$ is

Lower limit

$= 40$

Upper limit

$= 50$

Class mark

$= \dfrac{40+50}{2}=45$

Class mark of

$50-60$ is

Lower limit

$= 50$

Upper limit

$= 60$

Class mark

$= \dfrac{50+60}{2}=55$

Read the following bar graph and answer the following question:
Which state is the largest producer of rice?

Consider a chart by using data from the bar graph:

$States$	$Rice\, production$	$Wheat \,production$	$Total\, production$
$U.P$	$8$	$16$	$24$
$W.B$	$10$	$2$	$12$
$M.P$	$5$	$5$	$10$
$Maharashtra$	$4$	$2$	$6$
$Haryana$	$3$	$6$	$9$

The largest producer of rice will have the longest bar.

Here,

$W.B$ , i.e. West Bengal has the longest bar for rice.

Therefore, the largest producer of rice is

$W.B.$ .

Read the above bar graph and answer the following question:
Which state is the largest producer of wheat?

Consider a chart by using data from the bar graph:

$States$	$Rice\, production$	$Wheat \,production$	$Total\, production$
$U.P$	$8$	$16$	$24$
$W.B$	$10$	$2$	$12$
$M.P$	$5$	$5$	$10$
$Maharashtra$	$4$	$2$	$6$
$Haryana$	$3$	$6$	$9$

The largest producer of wheat will have the longest bar.

Here, $U.P$ , i.e. Uttar Pradesh has the longest bar for wheat.

Therefore, the largest producer of wheat is $U.P$ .

The time taken, in seconds, to solve a problem by each 25 pupils is as follows:
$16,20,26,27,28,30,33,37,38,40,42,43,46,46,46,48,49,50,53,54,58,59,60,62,70$
Draw a histogram to represent the frequency distribution.

Frequency distribution for the data is as follows:

Seconds	Frequency
$10-20$	$1$
$20-30$	$5$
$30-40$	$4$
$40-50$	$8$
$50-60$	$5$
$60-70$	$2$

1482771_1671031_ans_3fbf51c465d94d49a1c6348442a0e09e.png

The following is the distribution of total household expenditure (in Rs.) of manual worker in a city:
Expenditure (in Rs.): 100-150 150-200 200-250 250-300 300-350 350-400 400-450 450-500
No. of manual workers: 25 40 33 28 30 22 16 8
Draw a histogram and a frequency polygon representing the above data.

1671757_1671050_ans_3ee84a47943242008eebda1bec17aea6.jpg

The distribution of heights (in cm) of 96 children is given below. Construct a histogram and a frequency polygon on the same axes.
Height (in cm): 124
to
128 128
to
132 132
to
136 136
to
140 140
to
144 144
to
148 148
to
152 152
to
156 156
to
160 160
to
164
No. of Children: 5 8 17 24 16 12 6 4 3 1

To represent the given data by a histogram, we first draw horizontal and vertical axes.

Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.

The given data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights.

To draw the frequency polygon of the given data using histogram, obtain the mid-points of the upper horizontal side of each rectangle and then join these mid-points of the adjacent rectangles of the histogram by line segments.

Obtain the mid-points of two class-intervals of 0 frequencies, i.e. on the horizontal axis, one adjacent to the first, on its left and one adjacent to the last, on its right. These class-intervals are known as imagined class-intervals.

Complete the polygon by joining the mid-points of first and last class-intervals to the mid-points of imagined class-intervals adjacent to them. Let us take one vertical division is equal to

$4$ .

The heights of the different rectangles are as following

$1.$ The height of the rectangle corresponding to the class-interval

$124-128$ is

$\dfrac{5}{4}=1.25$ big divisions.

$2.$ The height of the rectangle corresponding to the class-interval

$128-132$ is

$\dfrac{8}{4}=2$ big divisions.

$3.$ The height of the rectangle corresponding to the class-interval

$132-136$ is

$\dfrac{17}{4}=4.25$ big divisions.

$4.$ The height of the rectangle corresponding to the class-interval

$136-140$ is

$\dfrac{24}{4}=6$ big divisions.

$5.$ The height of the rectangle corresponding to the class-interval

$140-144$ is

$\dfrac{16}{4}=4$ big divisions.

$6.$ The height of the rectangle corresponding to the class-interval

$144-148$ is

$\dfrac{12}{4}=3$ big divisions.

$7.$ The height of the rectangle corresponding to the class-interval

$1484-152$ is

$\dfrac{6}{4}=1.5$ big divisions.

$8.$ The height of the rectangle corresponding to the class-interval

$152-156$ is

$\dfrac{4}{4}=1$ big divisions.

$9.$ The height of the rectangle corresponding to the class-interval

$156-160$ is

$\dfrac{3}{4}=0.75$ big divisions.

$10.$ The height of the rectangle corresponding to the class-interval

$160-164$ is

$\dfrac{1}{4}=0.25$ big divisions.

The histogram and frequency polygon of the given data is the following:

1482764_1671020_ans_891789b326cb44d4a893cde3573dcae6.PNG

Find the mean of $x,x+2,x+4,x+6,x+8$ .

Given tha the observations are

$x,x+2,x+4,x+6,x+8$

We know mean =

$\dfrac{\text{sum of observations}}{\text{no' of observations}}$

$=\dfrac{x+x+2+x+4+x+6+x+8}{5}$

$=\dfrac{5x+20}{5}=\dfrac{5(x+4)}{5}$

$=x+4$

mean is

$x+4$

Find the mean of first 10 even natural numbers.

First

$10$ even natural numbers

i.e

$2,4,6,8,10,12,14,16,18,20$

No.'s of even natural numbers

$(n)=10$

$\therefore$ we know mean

$=\dfrac{\text{sum of first even natural no's}}{\text{no's of even natural no's}}$

$=\dfrac{2+4+6+8+10+12+14+16+18+20}{10}$

$=\dfrac{110}{10}$

$=11$

$\therefore$ mean of first

$10$ even natural no.'s is

$11$

The investment (in ten crores of rupees) of Life Insurance Corporation of India in different sectors are given below :
Sectors Investment
(in ten crores of rupees )
Central Government Securities 45
State Government Securities 11
Securities guaranteed by the Government
23
Private Sectors 18
Socially oriented sectors (Plan) 46
Socially oriented sectors (Non-Plan) 11
Represent the above data with the help of a bar graph.

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes.

Let us consider that the horizontal and vertical axes represent the sectors and the investment in ten Crores of rupees respectively. We have to draw

$6$ bars of different lengths given in the table.

At first, we mark

$6$ points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the investments of the Life Insurance Corporation of India.

The short forms used in the graph are:

CGS: Central Government Securities
SGS: State Government Securities
SGG: Securities Guaranteed by Government
PS: Private Sectors
SOS(P): Socially Oriented Sectors (Plan)
SOS(NP): Socially Oriented Sectors (Non-Plan)

1482629_1671258_ans_de8f34a988f64aa3bb36ff4223501669.png

Define the following terms :
Class interval

It is refers to the numerical width of any class in a particular distribution

mathematically defined as

Difference between the upper class limit and the lower class limit

$\to$ The data is arranged into different classes and the width of such class is called class interval.

Example

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequency	2	4	0	1	2

Class internal

$=$ upper limit

$-$ lower limit

$= 10 - 0$

$=10$

The mean of 200 items wasLater on, it was discovered that the two items were misread as 92 and 8 instead of 192 andFind the correct mean.

Since, mean of

$200$ items was

$50$ and number of items =

$200$

So, mean

$=50$

Also, we know mean =

$\dfrac{\text{sum of items}}{\text{number of items}}$

$\therefore 50=\dfrac{\text{sum of items}}{200}$

Sum of items =

$200\times 50$

$=10000$

Later on, it was discovered that the two items were misread as

$92$ and

$8$ instead of

$192$ and

$88$ respectively

Then misread instead Correct item

92 192 192-92=100

8 88 88-8=80

$\therefore$ Correct sum of items =

$10000+180$

$=10180$

$\therefore$ Correct mean =

$\dfrac{\text{sum of items}}{\text{number of items}}$

$=\dfrac{10180}{200}$

$=50.9$

Find the mean of first five multiples of 3.

First five multiplies of

$3$

i.e

$3,6,9,12,15$

No' of multiplies is

$5$

We know,

Mean =

$\dfrac{\text{sum of multiples of first five of 3}}{\text{numbers of multiples}}$

$=\dfrac{3+6+9+12+15}{5}$

$=\dfrac{45}{5}$

$=9$

$\therefore$ mean of first five multiplies of

$3$ is

$9$

In a hypothetical sample of $20$ people the amounts of money with them were found to be as follows:
$114, 108, 100, 98, 101, 109, 117, 119, 126, 131, 136, 143, 156, 169, 182, 195, 207, 219, 235, 118$ .
Draw the histogram of the frequency distribution (taking one of the class intervals as $50-100$ ).

1876087_1673899_ans_eccf570dfc9949dea597985f7d36018d.jpg

Number of workshops organized by a school in different areas during the last five years are as follows :
Years No.of workshops
$1995-1996$ $25$
$1996-1997$ $30$
$1997-1998$ $42$
$1998-1999$ $50$
$1999-2000$ $65$
Draw a histogram representing the above data.

1482626_1673894_ans_ed9acd7d194e4f20a545dfc797288337.png

The weights of new born babies (in kg) in a hospital on a particular day are as follows:
$2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1, 2.5, 2.8, 2.7, 2.9, 2.4$
How many babies weigh more than $2.8 kg$ ?

In Descending order: The lowest weight will come last and highest will come first.

$3.1, 3.0,2.9,2.9, 2.8, 2.8, 2.7, 2.7,2.6, 2.5,2.5, 2.4,2.3, 2.2,2.1$

$2.9,2.9,3.0,3.1$ are the

$4$ observations more than

$2.8$ .

$4$ babies weigh more than

$2.8$ kg.

Prepare a frequency table of the following scores obtained by $50$ students in a test:
$42, 51, 21, 42, 37, 37, 42, 49, 38, 52, 7, 33, 17, 44, 39, 7, 14, 27, 39, 42, 42, 62, 37, 39, 67, 51, 53, 53,$
$59, 41, 29, 38, 27, 31, 54, 19, 53, 51, 22, 61, 42, 39, 59, 47, 33, 34, 16, 37, 57, 43$

Marks	Tally Marks	Number of students
$7$	\|\|	$2$
$14$	\|	$1$
$16$	\|	$1$
$17$	\|	$1$
$19$	\|	$1$
$21$	\|	$1$
$22$	\|	$1$
$27$	\|\|	$2$
$29$	\|	$1$
$31$	\|	$1$
$33$	\|\|	$2$
$34$	\|	$1$
$37$	\|\|\|\|	$4$
$38$	\|\|	$2$
$39$	\|\|\|\|	$4$
$41$	\|	$1$
$42$	\|\|\|\| \|\|	$6$
$43$	\|	$1$
$44$	\|	$1$
$47$	\|	$1$
$49$	\|	$1$
$51$	\|\|\|	$3$
$52$	\|	$1$
$53$	\|\|\|	$3$
$54$	\|	$1$
$57$	\|	$1$
$59$	\|\|	$2$
$61$	\|	$1$
$62$	\|	$1$
$67$	\|	$1$

Study the bar graph representing the number of persons in various age groups in a town shown in figure. Observe the bar graph and answer the following questions:
What is the percentage of the youngest age-group persons over those in the oldest age group?

Observing the bar graph,

Number of persons in the youngest age group

$=1400$

Number of persons in the oldest age group

$=300$

$\therefore$ Percentage of the youngest age-group persons over those in the oldest age group

$=\dfrac{1400}{300}\times 100$

$=\dfrac{1400}3$

$=466.67\ \%$

In a study of number of accidents per day, the observations for 30 days were obtained as follows:
$6$ $3$ $5$ $6$ $4$ $3$ $2$ $5$ $4$ $2$
$4$ $2$ $1$ $2$ $2$ $0$ $5$ $4$ $6$ $1$
$6$ $0$ $5$ $3$ $6$ $1$ $5$ $5$ $2$ $6$
Prepare a frequency distribution table.

Given data

$6$

$3$

$5$

$6$

$4$

$3$

$2$

$5$

$4$

$2$

$4$

$2$

$1$

$2$

$0$

$5$

$4$

$6$

$1$

$6$

$0$

$5$

$3$

$6$

$1$

$5$

$2$

$6$

Number of accidents:	$0$	$1$	$2$	$3$	$4$	$5$	$6$
Number of days:	$2$	$3$	$6$	$3$	$4$	$6$	$6$

Following data gives the number of children in 40 families:
$1,2,6,5,1,5,1,3,2,6,2,3,4,2,0,0,4,4,3,2$
$2,0,0,1,2,2,4,3,2,1,0,5,1,2,4,3,4,1,6,2$
Represent it in the form of a frequency distribution.

Given data of the number of children in

$40$ families:

$1,2,6,5,1,5,1,3,2,6,2,3,4,2,0,0,4,4,3,2$

$2,0,0,1,2,2,4,3,2,1,0,5,1,2,4,3,4,1,6,2$

$No.\ of\ Children$	$Frequency$
0	5
1	7
2	11
3	5
4	6
5	3
6	3
Total	$40$

The population of Delhi State in different census years is as given below:
Census year 1961 1971 1981 1991 2001
Population in Lakhs 30 55 70 110 150
Represent the above information with the help of a bar graph.

Given data

The population of Delhi state in different census years is

Census year

$1961$

$1971$

$1981$

$1991$

$2001$

Population(in lakhs)

$30$

$55$

$70$

$110$

$150$

1481327_1675358_ans_7e068fd65ce14525b40e53b30c465f66.png

Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of $28$ laborers working in a factory, taking one of the class intervals as $210-230$ ( $230$ not included):
$220, 268, 258, 242, 210, 268, 272, 242, 311, 290,$
$300, 320, 319, 304, 302, 318, 306, 292, 254,278,$
$210, 240, 280, 316, 306, 215, 256, 236$ .

Given that,

The monthly wages (in rupees) of

$28$ laborers working on a factory are:

$220, 268, 258, 242, 210, 268, 272, 242, 311, 290,$

$300, 320, 319, 304, 302, 318, 306, 292, 254,278,$

$210, 240, 280, 316, 306, 215, 256, 236$

We need to construct a continuous frequency table to represent the given data, taking one of the class as

$210-230 (230\ not\ included)$

Hence, the frequency table will be as shown below:

Class interval	Frequency
210 - 230	4
230 - 250	4
250 - 270	5
270 - 290	3
290 - 310	7
310 - 330	5
Total	28

The marks obtained by $40$ students of class VIII in an examination are given below:
$16,17,18, 3, 7, 23, 18,13,10, 21, 7, 1, 13, 21,13, 15, 19, 24, 16,$
$3, 23, 5, 12, 18, 8, 12, 6, 8, 16, 5, 3, 5, 0, 7, 9, 12 , 20, 10, 2, 23$ .
Divide the data into five groups, namely $0-5$ , $5-10$ , $10-15$ , $15-20$ and $20-25$ and prepare a grouped frequency table.

Given that

The marks obtained by

$40$ students of class

$VIII$ in an examination are given below

$16,17,18, 3, 7, 23, 18,13,10, 21, 7, 1, 13, 21,13, 15, 19, 24, 16,$

$3, 23, 5, 12, 18, 8, 12, 6, 8, 16, 5, 3, 5, 0, 7, 9, 12 , 20, 10, 2, 23$ .

Divide the data into five groups

Prepare a grouped frequency table

Class group	frequency
0 - 5	6
5 - 10	10
10 - 15	8
15 - 20	9
20 - 25	7
Total	40

A die was thrown 25 times and following scores were obtained:
1 5 2 4 3
6 1 4 2 5
1 6 2 6 3
5 4 1 3 2
3 6 1 5 2
Prepare a frequency table of the scores.

Given Data:

1 5 2 4 3

6 1 4 2 5

1 6 2 6 3

5 4 1 3 2

3 6 1 5 2

Score :	1	2	3	4	5	6
Number of times:	5	5	4	3	4	4

The following bar graph shows the number of houses (out of $100$ ) in a town using different types of fuels for cooking. Read the bar graph and answer the following questions:
How many houses are using coal as fuel?

Observing the bar graph,

The number of houses using coal

$=10 \times 5=50$

Construct a frequency table with class-intervals $0-5$ ( $5$ not included) of the following marks obtained by a group of $30$ students in an examination:
$0, 5, 7, 10, 12, 15, 20, 22, 25, 27, 8, 11, 17, 3, 6, 9, 17, 19, 21, 29, 31, 35, 37, 40, 42, 45, 49, 4, 50, 16$ .

Given that marks obtained by a group of

$30$ students in an examination

$0, 5, 7, 10, 12, 15, 20, 22, 25, 27, 8, 11, 17, 3, 6, 9, 17, 19, 21, 29, 31, 35, 37, 40, 42, 45, 49, 4, 50, 16$ .

We can prepare frequency table with class - intervals (0-5) 5 not included

Class intervals	Frequency
0 - 5	3
5 - 10	5
10 - 15	3
15 - 20	5
20 - 25	3
25 - 30	3
30 - 35	1
35 - 40	2
40 - 45	2
45 - 50	2
50 - 55	1
Total	30

Construct a frequency distribution table for the following marks obtained by 25 students in a history test in class VI of a school:
$9,17,12,20,9,18,25,17,19,9,12,9,12,18,17,19,20,25,9,12,17,19,19,20,9$
The number of students who obtained $20\ marks =$ ____.

Given that

the marks obtained by

$25$ students in a history test in class VI of a school

$9, 17, 12, 20, 9, 18, 25, 17, 19, 9, 12, 9, 12, 18, 17, 19, 20, 25, 9, 12, 17, 19, 19, 20, 9$

Frequency Distribution table:

Marks No. of students obtained that marks

$9$

$6$

$12$

$4$

$17$

$4$

$18$

$2$

$19$

$4$

$20$

$3$

$25$

$2$

Thus, the number of students who obtained

$20\ marks = 3$

The following data gives the amount of loans (in crores of rupees) disbursed by a bank during some years:
Year $1992$ $1993$ $1994$ $1995$ $1996$
Loan(in crores of rupees) $28$ $33$ $55$ $55$ $80$
With the help of the bar graph, indicate the year in which amount of loan is not increased over that of the preceding year.

Given data is the amount of loans(in crores of rupees) disturbed by a bank-during some years.

Years

$1992$

$1993$

$1994$

$1995$

$1996$

Loan(in crores of rupees)

$28$

$33$

$55$

$80$

It is seen from the bar graph that the heights of the bars in the year

$1994$ and

$1995$ are the same.

Hence, the amount of loan is not increased in the year

$1995$ over the preceding year

$1994$ .

Therefore,

$1995$ is the required answer.

1480884_1675537_ans_65f680e8a0214d98b52386d28720cd8a.png

The bar graph as shown in above figure represents the heights (in cm) of $50$ students of Class $XI$ of a particular school. Study the graph and answer the following questions:
How many students are in the range of shortest height of the class?

The given bar graph represents the height (in cm) of $50$ students of class $XI$ of a particular school.

We can prepare frequency table from given bar graph as follows:

Heights(in cm) No. of students

$140-144$ $7$

$145-149$ $11$

$150-154$ $17$

$155-159$ $9$

$160-164$ $6$

$165$ -

Total number of students $=50$ .

Here observe that:

In class, the range of minimum/shortest height is

$140-144$ .

And the number of students in this range is

$7$ .

Therefore, there are

$7$ students in the range of shortest height of the class.

Read the following bar graph (Fig.) and answer the following question:What information is given by the bar graph?

Industrial production of cement in different years in India is given by the bar graph.

Read the following bar graph and answer the following questions:
What information is given by the bar graph?

Production of rice and wheat in different states is the information given by the bar graph.

The numbers of children in 10 families of a locality are: 2, 4, 3, 4, 2, 3, 5, 1, 1,Find the mean number of children per family.

Given data:

The numbers of children in

$10$ families of a locality are

$2, 4, 3, 4, 2, 3, 5, 1, 1, 5$

We know mean

$=\dfrac{sum \ of \ observations}{number \ of \ observations}$

$=\dfrac{2+4+3+4+2+3+5+1+1+5}{10}$

$=\dfrac{30}{10}$

$=3$

$\therefore$ Mean number of children per family

$=3$ .

Two hundred students of class VI and VII were asked to name their favourite colours so as to decide upon what should be the colour of their school house. The results are shown in the following table.
Colour: Red Green Blue Yellow Orange
Number of students: 43 19 55 49 34
Represent the given data on a bar graph.
Which is the most prrferred colour and which is least?

Since, number of students are from 19 to 55.

We make a bar graph by taking the following scale,

$1$ unit

$=10$ students

In the bar graph,

The longest bar is the blue one.

Therefore, Blue is the most preferred color.

The shortest bar is the green one.

Therefore, Green is the least preferred color.

1694307_1677160_ans_47f1a62f857142ef9fd7cdc055686013.jpg

The following data gives the amount of manure (in thousand tonnes) manufactured by a company during some years:
Year $1992$ $1993$ $1994$ $1995$ $1996$ $1997$
Manure (in thousand tonnes) $15$ $35$ $45$ $30$ $40$ $20$
Indicate with the help of the bar graph, the year in which the amount of manure manufactured by the company was maximum.

Given data gives the amount of manure(in thousand tonnes) manufactured by a company during some years:

Year

$1992$

$1993$

$1994$

$1995$

$1996$

$1997$

Manure(in thousand tonnes)

$15$

$35$

$45$

$30$

$40$

$20$

Clearly, in the year

$1994$ the amount of manure manufactured by the company was maximum, i.e.

$45$ .

1480933_1675582_ans_494011bc61014de287dc263edd77705a.png

What information is given by the bar graph?

Production of rice and wheat in different states is the information given by the bar graph.

Draw a histogram for the frequency distribution of the following data:
Class interval

$20-25$

$25-30$
$30-35$

$35-40$

$40-45$

$45-50$

Frequency
$320$

$780$

$160$

$540$

$260$

$100$

1689121_1775140_ans_80e2d8600a134232ac23b3fed1d41efd.jpg

The mean of marks scored by 100 students was found to beLater on it was discovered that a score of 53 was misread asFind the correct mean.

$\textbf{Step 1: Find the initial sum of observations}$

$\text{Initial Mean}=40$

$\text{Number of Observations}=100$

$\text{Mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}$

$40=\dfrac{\text{Sum of observations}}{100}$

$\text{Sum of observations}=100\times 40=4000$

$\textbf{Step 2: Find the corrected mean}$

$\text{A score of 53 was misread as 83}$

$\text{New sum of observations}=4000-(83-53)$

$=3970$

$\text{New Mean}=\dfrac{\text{New Sum of Observations}}{\text{Number of observations}}$

$\text{New Mean}=\dfrac{3970}{100}=39.7$

$\textbf{Therefore, the correct mean is 39.7}$

Find the mean of all factors of 10.

Factors of

$10$ are

$1, 2, 5, 10$ because,

$10=1\times 10$

$=2\times 5$

No. of factors

$(n)=4$

Mean

$=\dfrac{sum \ of \ factors}{no. \ of \ factors}$

$\therefore$ Mean

$=\dfrac{1+2+5+10}{4}$

$=\dfrac{18}{4}$

$=4.5$ .

If the mean of $5$ observations $x, x+2, x+4, x+6$ and $x+8$ is $11$ , find the value of $x$ .

We know that,

Mean

$=\dfrac {Sum\ of\ the\ given\ observation}{Total\ number\ of\ observation}$

Sum of the given observations

$=x+(x+2)+(x+4)+(x+6)+(x+8)$

$=x+x+2+x+4+x+6+x+8$

$=5x+20$

Total number of observations

$=5$

Therefore, Mean

$=\dfrac{(5x+20)}{5}$

Also, Mean

$=11$

(given)

$\Rightarrow 11=\dfrac{(5x+20)}{5}$

$\Rightarrow 55=5x+20$

$x=7$

The percentage of marks obtained by students of a class in mathematics are: 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3,Find their mean.

Given that the percentage of marks obtained by students of a class in mathematics are

$64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1$

No. of students

$=12$

We know mean

$=\dfrac{sum \ of \ observations(marks \ of \ student)}{No. \ of \ observations(students)}$

$=\dfrac{64+36+47+23+0+19+81+93+72+35+3+1}{12}$

$=\dfrac{474}{12}$

Mean

$=39.5$ .

Following are the weights (in kg) of 10 new born babies in a hospital on a particular day: 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.Find the mean $\overline {X}.$ (in kg)

Given that the weights(in kg) of

$10$ new born babies in a hospital on a particular day i.e., no. of new born babies

$(n)=10$

$3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6$

We know that

Mean(

$\bar{x}$ )

$=\dfrac{sum \ of \ observations}{number \ of \ observations}$

$\bar{x}=\dfrac{3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.6}{10}$

$=\dfrac{40.0}{10}$

$=\dfrac{40}{10}$

$\bar{x}=4$

$\therefore$ Mean weight of new born babies

$(\bar{x})=4$ kg.

If the mean of the following frequency distribution is $24$ , find the value of $p$ .
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
Frequency $3$ $4$ $p$ $3$ $2$

Class Interval	Frequency $f_i$	Mid-Value $x_i$	$f_i\ x_i$
$0-10$	$3$	$5$	$15$
$10-20$	$4$	$15$	$60$
$20-30$	$p$	$25$	$25p$
$30-40$	$3$	$35$	$105$
$40-50$	$2$	$45$	$90$
	$\sum f_i=12+p$		$\sum f_ix_i=270+25p$

Mean

$=\dfrac {\displaystyle \sum f_ix_i}{\displaystyle \sum f_i}$
Mean

$=24$ (given)

$\Rightarrow 24=\dfrac{(270+25p)}{ (12+p)}$

$\Rightarrow 24(12+p)=270+25p$
or

$p=18$

Attempt the following question.
Represent the following data by histogram.
Price of Sugar
(per kg in ) Number of weeks
$18 - 20$
$20 - 22$
$22-24$
$24-26$
$26-28$
$28-30$ $4$
$8$
$22$
$12$
$6$
$8$

Price of sugar (per kg in)	Number of weeks
$18-20$	$4$
$20-22$	$8$
$22-24$	$22$
$24-26$	$12$
$26-28$	$6$
$28-30$	$8$

1502200_1701109_ans_8b4a90a0aab74fefb696fa5c8aa66b66.png

The ages (in years) of $360$ patients treated in a hospital on a particular day are given below:
Age (in years )

$10-20$

$20-30$

$30-40$

$40- 50$

$50 -60$

$60-70$

Number of
patients $90$

$40$

$60$

$20$

$120$

$30$

Draw a histogram for the given data.

Consider the class intervals that is age in years on the X – axis and number of patients on the Y- axis and draw rectangles which provides the required histogram. ( Take

$10$ small divisions=

$10$ patients)
1617681_1775134_ans_ee2d329ab3a74a739d19a572c83d6c05.png

Look at the bar graph given.
In which subject is the student best?

The bar represents the marks obtained by a student in Mathematics is the highest or the height of the bar is high in Mathematics.

Hence, the student is best in Mathematics subject.

The number of children in $25$ families of a colony are given below:
$2,0,2,4,2,1,3,3,1,0,2,3,4,3,1,1,1,2,2,3,2,4,1,2,2.$
Represent the above data in the form of a frequency distribution table.

1623084_1778420_ans_fff4bf5f0e8b4e6bbcdf2f03865b472a.png

Look at the bar graph given.
In which subject is the student poorest?

The bar represents the marks obtained by a student in Science is the lowest. Hence, the student is poorest in Science subject.

In s survey of $60$ families of a colony, the number of members in each family was recorded and the data has been represented by the bar graph given below:
What information does the bar graph give?

The given bar graph gives the information about the number of members is

$60$ families of a colony.

Look at the bar graph given.
What information does the bar graph give?

The give bar graph shows the marks obtained by a student in four different subjects.

English

$=40$ marks

Hindi

$=55$ marks

Mathematics

$=70$ marks

Science

$=25$ marks

Fill in the blanks:
The number of times a particular observation occurs is called its..........

Frequency
According to the definition, the number of times a particular observation occurs is called its frequency.

Look at the bar graph given.
In which subjects did the student get more than $40$ marks?

Looking at the bar graph we can figure out that he scored

$55$ marks in Hindi and

$70$ marks in Mathematics.

Hence, the student scored more than

$40$ marks in Hindi and Mathematics.

The scale of shoes of various sizes at a shop on a particular day is given below:
$6,9,8,5,5,4,9,8,5,6,9,9,7,8,9,7,6,9,8,6,7,5,8,9,4,5,8,7.$
Represent the above data in the form of a frequency distribution table.

Observation (Shoe Size)	Frequency (number of shoes)
4	2
5	5
6	4
7	4
8	6
9	7

If the mean of the following distribution is 7.5, find the missing frequency "f".

Given mean

$=7.5$

$\therefore \text{Mean}=\dfrac{\Sigma fx}{\Sigma f}$

$7.5=\dfrac{563+f}{74+7f}$

On cross multiplication, we get:

$7.5\times(74+7f)=(563+7f)$

$7.5f-7f=563-555$

$0.5f=8$

$\therefore f=8/0.5=16$
Hence the missing value of

$f$ is

$16$ .

1712885_1784026_ans_68a3a834651d4b2b8dee64d5d29a3336.jpg

Find the class mark of the class $10-25$ .

Class mark

$=\dfrac {Upper\ limit+Lower\ limit}{2}=\dfrac {10+25}{2}=\dfrac {35}{2}17.5$

The following distribution shows the daily pocket allowance for children of a locality. The mean pocket allowance is Rs. $18$ . Find the value of f.
Daily pocket allowance in Rs.
11-13
13-15
15-17
17-19
19-21
21-23
23-25
No. of children
3
6
9
13
f
5
4

Class mark,

$x_i$ = (upper class limit + lower class limit)/2

Daily pocket allowance in Rs.	No. of children $f_i$	Class mark $x_i$	$f_ix_i$
11-13	3	12	36
13-15	6	14	84
15-17	9	16	144
17-19	13	18	234
19-21	f	20	20f
21-23	5	22	110
23-25	4	24	96
Total	$\Sigma f_i = 40+f$		$\Sigma f_ix_i = 704+20f$

Given mean $= 18$

$Mean = \Sigma f_ix_i/ \Sigma f_i$

$18 = (704+20f)/( 40+f)$

$18\times(40+f) = 704+20f$

$720 +18f = 704+20f$

$720-704 = 20f-18f$

$16 = 2f$

$f = 16/2 = 8$

Hence the value of f is $8$ .

The following bar graph shows the number of houses (out of $100$ ) in a town using different types of fuels for cooking. Read the bar graph and answer the following questions:
Which fuel is used in a maximum number of houses?

From the height of the bars representing a particular fuel, we can conclude that the height of the bar representing LPG is the largest which is in

$35$ houses out of

$100$ (without giving consideration to scale). Hence, LPG is used in the maximum number of houses.

Fill in the blanks
In a bar graph, ________ can be drawn horizontally or vertically.

1899488_1788736_ans_197f172369a44de5bbef224b17545df9.jpg

The following bar graph represents the data for different sizes of shoes worn by the students in a school. Read the graph and answer the following questions.
Which shoe size is worn by the maximum number of students?

From the graph, it can be observed that the bar against the 7 size shoe has maximum length.

$\therefore, 7$ number sized shoe is worn by maximum number of students.

The following bar graph represents the data for different sizes of shoes worn by the students in a school. Read the graph and answer the following questions.
What is the number of students wearing shoe size $6$ ?

Observing the bar graph,

The number of students wearing shoe size

$6=300$ students.

The following graph gives information about
the number of railway tickets sold for different cities
on a railway ticket counter between $6.00$ am to $10.00$ am. Read the bar graph and answer the following
questions
How many tickets were sold in all?

Number of all ticket sold

$= 80 + 50 + 100 + 20 + 40$

$= 290$

The following bar graph shows the number of houses (out of $100$ ) in a town using different types of fuels for cooking. Read the bar graph and answer the following questions:
Suppose that the total number of houses in the town is $1$ lakh. From the above graph estimate, the number of houses using electricity.

Observing the graph, we get,

Number houses using electricity out of

$100=5$

$\therefore$ Number houses using electricity out of

$1$ lakh will be

$=\dfrac{5}{100}\times100000=5000$ houses

The following bar graph represents the data for different sizes of shoes worn by the students in a school. Read the graph and answer the following questions.
What are the different sizes of the shoes worn by the students?

There are

$5$ types (

$4, 5, 6, 7$ and

$8$ ) of shoes worn by
the students.

The following graph gives information about the number of railway tickets sold for different cities on a railway ticket counter between $6.00$ am to $10.00$ am. Read the bar graph and answer the following questions
For which city were the maximum numbers of tickets sold?

Observing the bar graph,

The bar for Delhi is the longest.

$\therefore$ Maximum number of tickets are sold for Delhi.

The following table represents the income of a Gram Panchayat from different sources in a particular year:
Sources Income (in Rs)
Income from local taxes
Funds received from government
Donations
Income from other resources 75000
150000
25000
50000
Draw a bar graph to represent the above information

Take Income on the vertical axis and Sources on the horizontal axis.
1646284_1789441_ans_3d072c0e2ac7405ebdd3112e6f8688de.png

The following graph gives information about the number of railway tickets sold for different cities on a railway ticket counter between $6.00$ am to $10.00$ am. Read the bar graph and answer the following questions
Fill in the blanks:
The number of tickets sold for Delhi and Jaipur together exceeds the total number of tickets sold for Patna and Chennai by _______.

From the bar graph,

Number of tickets sold for different cities are,

Patna

$=80$

Jaipur

$=50$

Delhi

$=100$

Chennai

$=20$ and

Guwahati

$=40$

$\therefore$ The number of tickets sold for Delhi and Jaipur together

$=100+50=150$

The number of tickets sold for Patna and Chennai together

$=80+20=100$

Now, Required difference

$=150-100=50$

Hence, the number of tickets sold for Delhi and Jaipur together exceeds the total number of tickets sold for Patna and Chennai by

$50$

The bar graph given below represents the approximate length (in kilometers) of some National
Highways in India. Study the bar graph and answer
the following questions:
Which National Highway (N.H.) is the longest among
the above?

N.H

$2$ is the longest National Highway

The following graph gives information about the number of railway tickets sold for different cities on a railway ticket counter between $6.00$ am to $10.00$ am. Read the bar graph and answer the following questions
For which city were the minimum numbers of tickets sold?

Observing the bar graph,

The bar forChennai is the shortest.

∴ Minimum number of tickets are sold for Chennai.

Read the bar graph given below and answer the
following question:
In which year is the number of students maximum, and what was the number of students in that year?

From the bar graph, we can say, in

$2005-06$ the number of student is maximum.

And the number of students in this year is

$350$

Read the bar graph given below and answer the
following questions:
What information is given by the bar graph?

Increment and decrement of students per year.

The bar graph shown above represents the circulation of newspapers in different languages in a town. Study the bar graph and answer the following question:
What is the circulation of the English newspaper?

Given that,

The bar graph shows the circulation of newspapers in different languages in a town.

To find out,

The circulation of English newspaper.

Looking at the bar graph, carefully, we will see that the length of the horizontal bars represent the number of newspapers.

The length of the bar representing English newspapers is

$1000$ .

Hence, the number of circulation of English newspaper is

$1000$ .

The bar graph given below represents the approximate length (in kilometers) of some National Highways in India. Study the bar graph and answer the following questions:
Which National Highway is the shortest among the
above?

N.H

$9$ is

$900$ km in distance.

The bar graph given below represents the approximate length (in kilometers) of some National Highways in India. Study the bar graph and answer the following questions:
Which National Highway is the shortest among the
above?

N.H

$10$ is the shortest National Highway.

The bar graph given below represents the
circulation of newspapers in different languages in a
town. Study the bar graph and answer the following
questions:
By how much is the circulation of newspapers in Hindi
more than the newspaper in Bengali?

$1400 – 600 = 800$ the circulation of newspapers in
Hindi more than the newspaper in Bengali.

The bar graph given below represents the
circulation of newspapers in different languages in a
town. Study the bar graph and answer the following
questions:
Name the two languages in which the circulation of the
newspaper is the same.

In Bengali and Marathi languages in which the
circulation of the newspaper is the same.

Read the bar graph given below and answer the
following questions:
In which year is the increase in the number of students
maximum as compared to the previous year?

In year

$2004-05$ maximum increment occurs as
compare to

$2003-04$

Read the bar graph given below and answer the
following questions:
In which year did the number of students decrease as
compared to the previous year?

In year

$2003-04$ the number of students decreases as
compare to

$2002-03$ .

Read the bar graph given below and answer the
following questions:
In which year is the number of students twice as that of $2001-02$ ?

In year

$2004-05$ year is the number of students twice
as that of

$2001-02$ .

Prepare a bar graph of the data.
Crop Area of land (in million hectors)
Rice
Wheat 50
30
Pulses
Sugarcan
Cotton 20
25
15

Take Area of land (in million hectares) on the vertical axis and
Crops on the horizontal axis.
1646501_1789511_ans_388449a875144219b6d5e8dc78f34e0d.png

1646501_1789511_ans_388449a875144219b6d5e8dc78f34e0d.png

Prepare a bar graph of the data.

Number of two-wheelers	Number of families
0 1 2 3 4	2 28 14 4 1

Take Number of families on the vertical axis and Number of
two-wheelers on the horizontal axis.
1646471_1789481_ans_23aa35d931a04f1284eb11a5c6aead16.png

1646471_1789481_ans_23aa35d931a04f1284eb11a5c6aead16.png

Number of mobile phone users in various age
groups in a city are listed below:
Age group (in years) Number of mobile users
1 - 20
21 - 40
41 - 50
61 - 80 25000
40000
35000
10000
Draw a bar graph to represent the above information.

Take Number of the mobile user on the vertical axis and Age group
in the year on the horizontal axis.
1646271_1789433_ans_ea42472bb37441f18f0f6ef320d8156f.png

1646271_1789433_ans_ea42472bb37441f18f0f6ef320d8156f.png

The following table gives the number of vehicles passing through a toll gate, every hour from $8.00$ am. to $1.00$ pm:
Time Interval 8.00 to 9.00 9.00 to 10.00 10.00 to 11.00 11.00 to 12.00 12.00 to 1.00
Number of vehicles 250 450 300 250 150
Draw a bar graph representing the above data.
Solution:

Take the Number of vehicles used on the vertical axis and Time the interval on the horizontal axis.
1646275_1789437_ans_ae609135db154c66a0a0f6cc86405d8f.png

1646275_1789437_ans_ae609135db154c66a0a0f6cc86405d8f.png

In a botanical garden, the numbers of different types of plants are found as follows:
Type of the plant Number of plants
Herb
Shrub
Creeper
Climber
Tree 50
60
20
45
95
Draw a bar graph to represent the above information and
answer the following questions:

Take Number of Plants on the vertical axis and Type of Plant
on the horizontal axis.
1646424_1789455_ans_2a9ee58d82214b19acd378f35b87cf11.png

The following table gives the data on the number of schools (stage-wise) of a country in the year $2002$ .
Stage Number of schools (in thousands)
Primary
Upper primary
Secondary
Higher Secondary 80
55
30
20
Draw a bar graph to represent the above data:

Take Number of School on the vertical axis and Stages on
horizontal axis.
1646389_1789445_ans_7c070f7fec054cbcaa7b128f4e8537d7.png

Prepare a bar graph of the data given in the question.
Surname Number of people
Khan $3\times 100 + 50 = 350$
Patel
Rao
Roy
Saikia
Singh $5\times 100 = 500$
$4\times 100 = 400$
$4\times 100 = 400$
$2\times 100 = 200$
$3\times 100 = 300$

Take Number of People on the vertical axis and
Surname on the horizontal axis.
1646440_1789462_ans_f522093f1e514baa8bf704a358293098.png

The number of people owning book less than $40$ is ______

The correct fill up is

$22$ .

The number of people owning books less than

$40=8+14=22.$

The representation of data with bars of uniform width is called ______.

We know,

A bar graph is a graphical representation of data using bars of equal width drawn vertically or horizontally with equal spacing between them and different heights.

Hence, the representation of data with bars of uniform width is called Bar graph.

The marks in a subject for 12 students are as follows:
$31, 37, 35, 38, 42, 23, 17, 18, 35, 25, 35, 29$
For the given data, find the
(a) Range (b) Mean (c) Median (d) Mode

Rearranging the data into ascending order, we get

$17,18,23,25,29,31,35,35,35,37,38,42$

Here,

(a) Range

$=$ greatest observation

$-$ lowest observation

$=42-17=25$

(b) Mean

$=$ Sum of observations

$\div$ number of observations

$=\dfrac{17+18+23+25+29+31+35+35+35+37+38+42}{12}$

$=\dfrac{365}{12}= 30.41$

Median

$=\dfrac{31+35}{2}= \dfrac{66}{2}$

$=33$

(d) The mode is

$35$ because it occurs most frequently i.e.,

$3$ times.

Fill in the blanks(Refer fig)
The total number of people owning books more than $60$ is _____

The correct fill up is

$8.$

The number of people owning books more than

$60=6+2=8$ .

When the number of observations is large , the observation are usually organised in groups of equal width called __________

The correct fill up is class intervals.

The number of times a particular observation occurs in a given data is called its _____

The number of times a particular observation occurs in a given data is called it's frequency.

The above histogram shows the number of literate females in the age group of 10 to 40 years in town.
(a) Write the classes assuming all the classes are of equal width.
(b) What is class's width?
(c) In which age group are literate females the least?
(d) In which age group is the number of literate females the highest?

(a)

As per the given information, age group is

$10-40$ years.

We can see that there are

$6$ columns in the histograms.

So, it can be divided in

$6$ classes of equal class intervals with class size

$=\cfrac{40-10}{6}=5$ , starting from

$10$

Hence, the classes are

$10-15, \ 15-20, \ 20-25, \ 25-30,\ 30-35,\ 35-40$

(b)

We know that, Class width = upper limit - Lower limit =

$15 -10$

$\therefore\$ Class width =

$5$

(c)

As per the given histogram, class

$10 -15$ consist of the least number of literate females.

(d)

Looking at the histogram, we can say that class

$15-20$ consists of the highest number of literate females.

In a histogram, ________ are drawn with width equal to a class interval without leaving any gap in between.

In a histogram, bars are drawn with width equal to a class interval without leaving any gap in between.

The class size of the interval $80 -85$ is ______

Class size of the interval

$80-85$ is

$=$ Upper class limit

$-$ lower class limit

$= 85-80$

$=5$

Given below are heights of $15$ boys of a class measured in cm:
$128,144,146,143,136,142,138,129,140,152,144,140,150,142,154$
Find
(a) The height of the tallest boy.
(b) The height of the shortest boy.
(c) The range of the given data.
(d) The median height of the boys.

The given observations arranged in ascending order are:

$128,129,136,138,140,140,142,142,143,144,144,146,150,152,154$

Hence

(a) The height of the tallest boy is

$154$ cm.

(b) The height of the shortest boy is

$128$ cm.

$=(154-128)cm= 26$ cm

(d) Median:

Here,

$N=15$ (odd)

So, Median

$M=\left(\dfrac{N+1}{2}\right) ^{th}\text{observation}$

$\Rightarrow M=\left(\dfrac{15+1}{2}\right) ^{th}\text{observation}$

$\Rightarrow M=8^{th} \text{observation}$

$\Rightarrow M=142$

Therefore, median height

$=142$ cm

The top speeds to $30$ different land animals have been organized into a frequency table. Draw a histogram for the given data.
Class interval Frequency
$10 -20$ $5$
$20 -30$ $5$
$30 - 40$ $10$
$40 -50$ $8$
$50 -60$ $0$
$60 -70$ $2$

The histogram for the given data is shown above where the class interval is represented on

$x-axis$ and frequency is equal to the height of the bars on

$y-axis.$

1689463_1793548_ans_1b5cb01945544ce38dc268feb062814c.png

Given below is a frequency distribution table. read it and answer the questions that follow.
Class interval Frequency
$10-20$ $5$
$20-30$ $10$
$30- 40$ $4$
$40 -50$ $15$
$50 -60 \quad \quad \quad 12$
(a) What is the lower limit of the second class interval ?
(b) What is the upper limit of the last class interval ?
(c) What is the frequency of the third class ?
(d) Which interval has a frequency of $10$ ?
(e) Which interval has the lowest frequency ?
(f) What is the class size ?

From observing the given frequency table:

(a) Lower limit of second class interval

$( 20 -30)$ is

$20$

(b) The upper limit of the last class interval

$( 50 -60)$ is

$60 .$

$( 30 -40)$ is

$4.$

(d) The interval

$( 20-30)$ has frequency of

$10.$

(e) The interval

$( 30-40)$ has the lowest frequency i,e,

$4$

(f)

$\text{Class size = upper class limit - lower class limit.}$ , for a class.

Let us take first class i.e.

$10-20$ from the given classes.

Therefore

$20 - 10 = 10$ is the class size.

Look at the histograms below answer the questions that follow
(a) How many students have height more than or equal to $135\ cm$ but less than $150\ cm$ ?
(b) Which class interval has the least number of students ?
(c) What is the class size ?
(d) How many students have height less than $140\ cm$ ?

(a)

There are three columns

$(135-140),(140-145),(145-150)$ who represent height more than or equal to

$135\ cm$ but less than

$150\ cm$

So, No. of students that have height more than or equal to

$135\ cm$ but less than

$150\ cm = 14 +18 +10 = 42$

(b)

From histogram least number of students in any class interval are

$4$

The least number of students are in the class interval

$150 -155$ .

(c)

$\text{Class size = Upper limit - Lower limit}$

$= 130-125= 5$ is the class size.

(d)

There are three columns

$(125-130),(130-135),(135-140)$ who represent height less than

$140\ cm$

Students that have height less than

$140\ cm = 6 + 8 + 14 = 28$

The graph shows the maximum temperatures recorded for two consecutive weeks of a town. Study the graph and answer the questions that follow.
What were the temperatures for both the weeks on Thursday?

The temperature for the first week on Thursday was 37°C and the temperature for the second week on the same day was 34°C

Expenditure on Education of a country during a five year period (2002-2006), in crores of rupees, is given below:

Elementary Education 240
Secondary Education 120
University Education 190
Teacher's Training 20
Social Education 10
Other Education programmes 115
Culture Programmes 25
Technical Education 125
Represent the information above by a bar graph.

1904646_1795088_ans_666b8df3af304fa0a075f41278562f7f.PNG

Observe the given bar graph carefully and answer the question that follow.
(a) What information does the bar graph depict?
(b) How many motor bikes were produced in the first three months?
(c) Calculate the increase in production in May over the production in January.
(d) In which month the production was minimum and what was it?
(e) Calculate the average (mean) production of bikes in 6 months.

(a) The bar graph shows the information about the production of motorbikes that are manufactured by XYZ automobiles Ltd. from January to June.

(b) The production of the motor-bikes in the first three months =600+800+700=2100

Production in May=900

$\therefore$ the increment in the production in May over January =

$900-600=300$

(d) The minimum production was in January and it was 600.

(e) Average production=

$\dfrac{600+800+700+1100+900+900}{6}$

$\dfrac{5000}{6}=833.33$

Observe the given data:
Day of the
Week Mon Tues Wed Thurs Fri Sat
Numbers of Mobile Phone Sets Sold
50 45 30 55 27 60
(a) Draw a bar graph to represent the above given information.
(b) On which day of the week was the sales maximum?
(c) Find the total sales during the week.
(d) Find the ratio of the minimum sale to the maximum sale.
(e) Calculate the average sale during the week.
(f) On how many days of the week was the sale above the average sales?

(a) see image

(b) The sales were maximum on Saturday.

$45+50+30+55+27+60=267$

(d) Minimum sale=

$27$

Maximum sale =

$60$

The required ratio=

$\dfrac{27}{60}= \dfrac{9}{20}$

i.e, 9:20

(e) Average sale =

$\dfrac{Total\, sales\, during\, the\, week}{total\, number\, of\, days}$

Average sale=

$\dfrac{267}{6}=44.5$

(f) 4 days in a week has the above value of sale than the average sale that is Monday, Tuesday, Thursday, and Saturday.

1904671_1795906_ans_e1b1fb56524447b8a2174bced165f397.PNG

Study the bar graph given below and answer the questions that follow:
(a) What information is depicted from the bar graph?
(b) In which subject is the student very good?
(c) Calculate the average marks of the student.
(d) If 75 and above marks denote a distinction, then name the subjects in which the student got distinction.
(e) Calculate the percentage of marks the student got out of 500.

(a) The bar graph depicts the information about marks obtained by a student in 5 different subjects.

(b) 82, is the highest marks that are obtained by students in mathematics.

Therefore, we can say that students are good in Maths.

$\dfrac{64+75+82+71+49}{5}=68.2$

(d) The student got 82 marks in math and 75 marks in Hindi.

Therefore, the students got a distinction in Maths and Hindi.

(e) Total marks secured by the student

$64+75+82+71+49=341$

Hence, percentage marks=

$\dfrac{341}{500}= 68.2\%$

The following table gives the frequencies of most commonly used letters $a,e,i,o,r,t,u$ from a page of a book:
Letters a e i o r t u
Frequency 75 125 80 70 80 95 75
Represent the information above by a graph.

1904644_1795091_ans_8035b77b2e8943c6a630b8b907392e16.PNG

Head	Food	Education	Clothing	House rent	Others	Saving
Expenditure (rs)	$4000$	$2500$	$1000$	$3500$	$2500$	$1500$

Marks (out of 90)	Number of candidates
More than or equal to 80	4
More than or equal to 70	6
More than or equal to 60	11
More than or equal to 50	17
More than or equal to 40	23
More than or equal to 30	27
More than or equal to 20	30
More than or equal to 10	32
More than or equal to 0	34

Marks	Below 20	Below 40	Below 60	Below 80	Below 100
Number of students	17	22	29	37	50

Marks	Number of students
Below 10	10
Below 20	50
Below 30	130
Below 40	270
Below 50	440
Below 60	570
Below 70	670
Below 80	740
Below 90	780
Below 100	800

Weight (in $kg$ )	$50-55$	$55-60$	$60-65$	$65-70$	$70-75$	$75-80$
Number of patients	$7$	$4$	$4$	$9$	$2$	$3$

Investment (Thousand Rupees)	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$
No. of families	$30$	$50$	$60$	$55$	$15$

Use of electric (unit)	$50-70$	$70-90$	$90-110$	$110-130$	$130-150$	$150-170$
No. of families	$150$	$400$	$460$	$540$	$600$	$350$

Height of student ( $\text{cm}$ )	$135 - 140$	$140 - 145$	$145 - 150$	$150 - 155$
No. of students	$4$	$12$	$16$	$8$

TIme (minutes)	$60 - 80$	$80 - 100$	$100 - 120$	$120 - 140$	$140 - 160$
No.of students	$14$	$20$	$24$	$22$	$16$

Classes	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	Total
Frequency	17	$f_{1}$	32	$f_{2}$	19	$N = \sum f_{i} = 120$

Class-interval	Frequency
0-10	5
10-20	$x$
20-30	20
30-40	14
40-50	$y$
50-60	8
Total	68

Class Interval	$0-20$	$20-40$	$40-60$	$60-80$	$80 -100$
Frequency	$5$	$6$	$12$	$4$	$3$

Class interval	$5 - 9$	$10 - 14$	$15 - 19$	$20 - 24$
Frequency	$3$	$5$	$4$	$2$

Class Interval	$3-6$	$6-12$	$12-13$	$13-14$	$14-15$
Frequency	$150$	$420$	$100$	$110$	$50$

Marks	Number of Students
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50	25 $x$ 30 2 $x$ 65

21	20	55	39	48	46	36	54	42	30
29	42	32	40	34	31	35	37	52	44
54	36	41	33	23	39	28	44	45	38
39	45	37	33	51	53	52	46	43	47
51	26	52	48	25	34	37	33	36	27

Mass in (gms)	$10-19$	$20-24$	$25-34$	$35-49$	$50-54$
No. of objects	$6$	$4$	$12$	$18$	$8$

Marks	Frequency
20-30 30-40 40-50 50-60 60-70 70-80 80-90	p 15 25 20 q 8 10

Ependiture (in $Rs$ )	No. of students
$10-15$	$3$
$15-20$	$5$
$20-25$	$10$
$25-30$	$6$
$30-35$	$4$
$35-40$	$2$
Total	$30$

Class	$5-14$	$14-23$	$23-32$	$32-41$	$41-50$	$50-59$	$59-68$
Frequency	$5$	$11$	$a$	$53$	$b$	$16$	$10$

Statistics - Class 9 Maths - Extra Questions

The expenditure of a family on different heads in a month is given below:HeadFoodEducationClothingHouse rentOthersSavingExpenditure (rs)400040002500250010001000350035002500250015001500Draw a bar graph to represent the data above.

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.Answer the following question:What information does the above bar graph give ?

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.Answer the following question:On which day were the maximum number of shirts sold ? How many shirts were sold on that day ?

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.Answer the following question:What is the scale chosen on the horizontal line representing number of shirts ?

Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class:MarksBelow 20Below 40Below 60Below 80Below 100Number of students1722293750Form the frequency distribution table for the data.

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.Answer the following question:How many shirts were sold on Thursday?

Fill in the blanks:The number of times a particular observation occurs in a data is called ....... of the observation.

The following table shows the cumulative frequency distribution of marks of 800 students in an examination:MarksNumber of studentsBelow 1010Below 2050Below 30130Below 40270Below 50440Below 60570Below 70670Below 80740Below 90780Below 100800Construct a frequency distribution table for the data above.

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.Answer the following question:On which day were the minimum number of shirts sold?

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:Name the subject(s) in which she got 8080 or more marks.

Given below a graph:Read the bar graph carefully and answer the following question:What is the information given by the bar graph?

Fill in the blanks:In a bar graph, the bars are of uniform...........

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:What is the scale of this bar graph ?

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:In which subject she obtained minimum marks ?

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:In which subject Gurmeet obtained maximum marks ?

The weight of 2929 patients in a hospital were recorded as follows:Weight (in kgkg)50−5550-5555−6055-6060−6560-6565−7065-7070−7570-7575−8075-80Number of patients774444992233Draw a histogram to represent this data visually.

Given below a graph:Read the bar graph carefully and answer the following question:On which item the expenditure is minimum?

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:Answer the following question:Numbers of workers whose weakly wages are more than or equal to Rs.400Rs. 400 but less than Rs.450Rs. 450

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:Answer the following question:How many workers get more than Rs.425Rs. 425?

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.Answer the following:For how many hours did the maximum number of students watch T.VT.V?

Given below a graph:Read the bar graph carefully and answer the following question:State whether true or false:Expenditure on education is twice the expenditure on clothing.

In a study diabetic patients, the following data was obtained:Age (in years)10−2010-2020−3020-3030−4030-4040−5040-5050−6050-6060−7060-7070−8070-80Number of patients3388303036362727151566Represents the above data by a histogram.

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:Answer the following question:What is the class having a minimum frequency?

Given below a graph:Read the bar graph carefully and answer the following question:On which item the expenditure is maximum?

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:Answer the following question:How many workers get less than Rs.475Rs. 475?

The following observations have been arranged in ascending order. If the median the data is 47.547.5, find the value of xx.17,21,23,29,39,40,x,50,51,54,59,67,91,9317,21,23,29,39,40,x,50,51,54,59,67,91,93

In a Science test given to a group of students, the marks scored by them (out of 100) are:41,39,52,48,54,62,46,52,40,96,42,40,98,60,5241,39,52,48,54,62,46,52,40,96,42,40,98,60,52Find the mean and median of this data.

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.Answer the following:For how many students spent more than 22 hours but less than 44 in watching T.VT.V?

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.Answer the following:For how many students spent more than 55 hours in watching T.VT.V?

The number of literate females in the group of 1010 to 4040 years in a town is shown in the histogram alongside.Answer the following question:What is the class size?

Rearrange the following raw data in descending order:5.3, 5.2, 5.1, 5.7, 5.6, 6.0, 5.5, 5.9, 5.8, 6.1, 5.5, 5.8, 5.7, 5.95.3, 5.2, 5.1, 5.7, 5.6, 6.0, 5.5, 5.9, 5.8, 6.1, 5.5, 5.8, 5.7, 5.9 and 5.45.4. The write the:range of values

The number of literate females in the group of 1010 to 4040 years in a town is shown in the histogram alongside.Answer the following question:In which age group is the number of literate females the highest?

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.Answer the following:How many students watched T.VT.V for less than 44 hourse?

The number of literate females in the group of 1010 to 4040 years in a town is shown in the histogram alongside.Answer the following question:Write the classes assuming all the classes are of equal width.

The number of literate females in the group of 1010 to 4040 years in a town is shown in the histogram alongside.Answer the following question:In which age are the literate females the least?

The mean of 18,\ 28,\ x,\ 32,\ 1418,\ 28,\ x,\ 32,\ 14 and 3636 is 2323. Find the value of xx Sum of data

The bar graph drawn below shoes the number of tickets sold during a fair by 66 students A, B, C, D, EA, B, C, D, E and FFUsing the Bar graph, answer the following question:Who sold the maximum number of tickets?

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.Use the graph to answer the following:On which day were the maximum number of patients examined?

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.Use the graph to answer the following:On which day were the least number of patients examined?

The above bar graph shows the number of children, in various classes, in a school in Delhi.Using the given bar graph, find:How many more children are there in Class 5 compared to Class 6?

The bar graph drawn below shoes the number of tickets sold during a fair by 66 students A, B, C, D, EA, B, C, D, E and FFUsing the Bar graph, answer the following question:How many tickets were sold by D, ED, E and FF taken together?

The bar graph drawn below shoes the number of tickets sold during a fair by 66 students A, B, C, D, EA, B, C, D, E and FFUsing the Bar graph, answer the following question:How many tickets were sold by A, BA, B and CC taken together?

The following bar graph shows the number of children, in various classes, in a school in Delhi.Using the given bar graph, find:the average number of children in a class

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.Use the graph to answer the following:On which days were equal number of patients examined?

The bar graph drawn below shoes the number of tickets sold during a fair by 66 students A, B, C, D, EA, B, C, D, E and FFUsing the Bar graph, answer the following question:Who sold the least number of tickets?

The mean of the number 6, y, 7, x6, y, 7, x and 1414 is 8.8. Express yy in terms of xx.

If the mean of 6, 4, 7,6, 4, 7, aa and 1010 is 88. Find the value of aa

Find the actual lower and upper class limits and also the class marks of the classes:1.1-2.0,2.1-3.01.1-2.0,2.1-3.0 and 3.1-4.03.1-4.0

Following data shows the weekly wages (in ) of 1010 workers in a factory.3500, 4250, 4000, 4250, 4000, 3750, 4750, 4000, 42503500, 4250, 4000, 4250, 4000, 3750, 4750, 4000, 4250 and 40004000.What is the range of wages (in )?

Fill in the blanks:The range of the data 7,13,6,25,18,20,16 is .......

Arrange the following data in descending order: 3.3, 3.2, 3.1, 3.7, 3.6, 4.0, 3.5, 3.9, 3.8, 4.1, 3.5, 3.8, 3.7, 3.93.3, 3.2, 3.1, 3.7, 3.6, 4.0, 3.5, 3.9, 3.8, 4.1, 3.5, 3.8, 3.7, 3.9 and 3.43.4.Determine the range.

Uses the table given to find:The actual class limits of the fourth class

Consider the following numbers:68, 76, 63, 75, 93, 83, 70, 115, 82, 105, 90, 103, 92, 52, 99, 73, 75, 63, 7768, 76, 63, 75, 93, 83, 70, 115, 82, 105, 90, 103, 92, 52, 99, 73, 75, 63, 77 and 7171.What is the range of these numbers?

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.Use the graph to answer the following:What is the total number of patients examined in the week?

Hundred students from a certain locality use different modes of travelling to school as given below. Draw a bar graph.BusCarRickshawBicycleWalk323216162424202088

Fill in the blanks in the following table:Class intervalFrequencyCumulative frequency25-3425-34____151535-4435-44___282845-5445-542121____55-6455-641616____65-7465-74____737375-8475-841212____

Uses the table given to find:The class boundary of the sixth class

In the following table, the investment made by 210210 families is shown. Present it in the form of a histogram.Investment(Thousand Rupees)10 - 1510 - 1515 - 2015 - 2020 - 2520 - 2525 - 3025 - 3030 - 3530 - 35No. of families30305050606055551515

Uses the table given to find:The class mark of the third class

Draw a histogram for the following frequency distribution.Use of electric(unit)50-7050-7070-9070-9090-11090-110110-130110-130130-150130-150150-170150-170No. of families150150400400460460540540600600350350

Draw a histogram of the following data.Height of student (\text{cm}\text{cm})135 - 140135 - 140140 - 145140 - 145145 - 150145 - 150150 - 155150 - 155No. of students441212161688

Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.TIme (minutes)60 - 8060 - 8080 - 10080 - 100100 - 120100 - 120120 - 140120 - 140140 - 160140 - 160No.of students14142020242422221616

Classify the following information as primary or secondary data. Information of attendance of every student collected by visiting every class in a school.

Classify the following information as primary or secondary data. In the village Nandpur, the information collected from every house regarding students not attending school.

Classify the following information as primary or secondary data. The information of the heights of students was gathered from school records and sent to the head office, as it was to be sent urgently.

Classify the following information as primary or secondary data: For a science project, information of trees gathered by visiting a forest.

If the mean of the following data is 20.220.2, then find the value of p.x_ix_i1015202530f_if_i68p106

If the median of the distribution given is 28.5, find the values of 'x' and 'y'Class IntervalFrequency0 - 10510 - 20x20 - 302030 - 401540 - 50y50 - 605Total60

Find the class mark of the class interval 30 - 40.

If the mean of the following distribution is 6, find the value of P. x_{i} x_{i} 24610P + 5 f_{i} f_{i} 32312

If the classmark is 1010 and class width is 66 then find the class.

Draw a histogram to represent the following frequency distribution:Class intervalFrequency20 - 2520 - 255525 - 3025 - 30101030 - 3530 - 35181835 - 4035 - 40141440 - 4540 - 451212

If median of the following distribution is 27, then find the values of x and y.Class-intervalFrequency0-10510-20 x x 20-302030-401440-50 y y 50-608Total68

If mean of following data is 21.5, then find K :Class0-1010-2020-3030-4040-50Frequency643K2

Draw a histogram for the following frequency distribution .

Class Interval0-20 0-20 20-40 20-40 40-60 40-60 60-8060-80 80 -10080 -100Frequency 556 6 12 12 4 4 33 Construct histogram for the following frequency table:

Draw a histogram for the following frequency distribution

Construct histogram for the following frequency table : Class Interval0-50-5 5-105-10 10-15 10-15 15-2015-20 20-2520-25 Frequency 18 1815 15 14 14 8 8 1010

The value of \pi \pi upto 5050 decimal places is given below.3.141592653589793238462643383279502884197169399375103.14159265358979323846264338327950288419716939937510(ii) What are the most and the least frequency occurring digits?

The expenditure of a family on different heads in a month is given below:

Head Food Education Clothing House rent Others Saving
Expenditure (rs) $4000$ $2500$ $1000$ $3500$ $2500$ $1500$

Draw a bar graph to represent the data above.

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
What information does the above bar graph give ?

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
On which day were the maximum number of shirts sold ? How many shirts were sold on that day ?

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
What is the scale chosen on the horizontal line representing number of shirts ?

Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class:
Marks
Below 20 Below 40 Below 60 Below 80 Below 100
Number of students 17 22 29 37 50
Form the frequency distribution table for the data.

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
How many shirts were sold on Thursday?

Fill in the blanks:
The number of times a particular observation occurs in a data is called ....... of the observation.

Observe the bar graph given above which is showing the sale of shirts in a readymade garment shop from Monday to Saturday.
Answer the following question:
On which day were the minimum number of shirts sold?

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:
Name the subject(s) in which she got $80$ or more marks.

Given below a graph:
Read the bar graph carefully and answer the following question:
What is the information given by the bar graph?

Fill in the blanks:
In a bar graph, the bars are of uniform...........

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:
What is the scale of this bar graph ?

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:
In which subject she obtained minimum marks ?

Observe the above bar graph, showing the marks scored by Gurmeet in the annual examination in different subjects: Answer the following question:
In which subject Gurmeet obtained maximum marks ?

The weight of $29$ patients in a hospital were recorded as follows:
Weight (in $kg$ )
$50-55$ $55-60$ $60-65$ $65-70$ $70-75$ $75-80$
Number of patients $7$ $4$ $4$ $9$ $2$ $3$
Draw a histogram to represent this data visually.

Given below a graph:
Read the bar graph carefully and answer the following question:
On which item the expenditure is minimum?

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:
Answer the following question:
Numbers of workers whose weakly wages are more than or equal to $Rs. 400$ but less than $Rs. 450$

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:
Answer the following question:
How many workers get more than $Rs. 425$ ?

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.
Answer the following:
For how many hours did the maximum number of students watch $T.V$ ?

Given below a graph:
Read the bar graph carefully and answer the following question:
State whether true or false:
Expenditure on education is twice the expenditure on clothing.

In a study diabetic patients, the following data was obtained:
Age (in years) $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
$70-80$
Number of patients $3$ $8$ $30$ $36$ $27$ $15$ $6$
Represents the above data by a histogram.

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:
Answer the following question:
What is the class having a minimum frequency?

Given below a graph:
Read the bar graph carefully and answer the following question:
On which item the expenditure is maximum?

The histogram showing the weekly wages (in Rs) of workers in a factory is given alongside:
Answer the following question:
How many workers get less than $Rs. 475$ ?

The following observations have been arranged in ascending order. If the median the data is $47.5$ , find the value of $x$ .
$17,21,23,29,39,40,x,50,51,54,59,67,91,93$

In a Science test given to a group of students, the marks scored by them (out of 100) are:
$41,39,52,48,54,62,46,52,40,96,42,40,98,60,52$
Find the mean and median of this data.

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.
Answer the following:
For how many students spent more than $2$ hours but less than $4$ in watching $T.V$ ?

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.
Answer the following:
For how many students spent more than $5$ hours in watching $T.V$ ?

The number of literate females in the group of $10$ to $40$ years in a town is shown in the histogram alongside.
Answer the following question:
What is the class size?

Rearrange the following raw data in descending order:
$5.3, 5.2, 5.1, 5.7, 5.6, 6.0, 5.5, 5.9, 5.8, 6.1, 5.5, 5.8, 5.7, 5.9$ and $5.4$ . The write the:
range of values

The number of literate females in the group of $10$ to $40$ years in a town is shown in the histogram alongside.
Answer the following question:
In which age group is the number of literate females the highest?

The number of hours for which students of a particular class watched television during holidays is shown in the histogram below.
Answer the following:
How many students watched $T.V$ for less than $4$ hourse?

The number of literate females in the group of $10$ to $40$ years in a town is shown in the histogram alongside.
Answer the following question:
Write the classes assuming all the classes are of equal width.

The number of literate females in the group of $10$ to $40$ years in a town is shown in the histogram alongside.
Answer the following question:
In which age are the literate females the least?

The mean of $18,\ 28,\ x,\ 32,\ 14$ and $36$ is $23$ . Find the value of $x$ Sum of data

The bar graph drawn below shoes the number of tickets sold during a fair by $6$ students $A, B, C, D, E$ and $F$
Using the Bar graph, answer the following question:
Who sold the maximum number of tickets?

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
On which day were the maximum number of patients examined?

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
On which day were the least number of patients examined?

The above bar graph shows the number of children, in various classes, in a school in Delhi.
Using the given bar graph, find:
How many more children are there in Class 5 compared to Class 6?

The bar graph drawn below shoes the number of tickets sold during a fair by $6$ students $A, B, C, D, E$ and $F$
Using the Bar graph, answer the following question:
How many tickets were sold by $D, E$ and $F$ taken together?

The bar graph drawn below shoes the number of tickets sold during a fair by $6$ students $A, B, C, D, E$ and $F$
Using the Bar graph, answer the following question:
How many tickets were sold by $A, B$ and $C$ taken together?

The following bar graph shows the number of children, in various classes, in a school in Delhi.
Using the given bar graph, find:
the average number of children in a class

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
On which days were equal number of patients examined?

The bar graph drawn below shoes the number of tickets sold during a fair by $6$ students $A, B, C, D, E$ and $F$
Using the Bar graph, answer the following question:
Who sold the least number of tickets?

The mean of the number $6, y, 7, x$ and $14$ is $8.$ Express $y$ in terms of $x$ .

If the mean of $6, 4, 7,$ $a$ and $10$ is $8$ . Find the value of $a$

Find the actual lower and upper class limits and also the class marks of the classes:
$1.1-2.0,2.1-3.0$ and $3.1-4.0$

Following data shows the weekly wages (in ) of $10$ workers in a factory. $3500, 4250, 4000, 4250, 4000, 3750, 4750, 4000, 4250$ and $4000$ .What is the range of wages (in )?

Fill in the blanks:
The range of the data 7,13,6,25,18,20,16 is .......

Arrange the following data in descending order: $3.3, 3.2, 3.1, 3.7, 3.6, 4.0, 3.5, 3.9, 3.8, 4.1, 3.5, 3.8, 3.7, 3.9$ and $3.4$ .
Determine the range.

Uses the table given to find:
The actual class limits of the fourth class

Consider the following numbers:
$68, 76, 63, 75, 93, 83, 70, 115, 82, 105, 90, 103, 92, 52, 99, 73, 75, 63, 77$ and $71$ .
What is the range of these numbers?

The column graph, given above, shows the number of patients, examined by Dr. V.K. Bansal, on different days of a particular week.
Use the graph to answer the following:
What is the total number of patients examined in the week?

Hundred students from a certain locality use different modes of travelling to school as given below. Draw a bar graph.
Bus Car Rickshaw Bicycle Walk
$32$ $16$ $24$ $20$ $8$

Fill in the blanks in the following table:
Class interval Frequency Cumulative frequency
$25-34$ __ $15$
$35-44$ _ $28$
$45-54$ $21$
$55-64$ $16$

$65-74$ $73$
$75-84$ $12$

Uses the table given to find:
The class boundary of the sixth class

In the following table, the investment made by $210$ families is shown. Present it in the form of a histogram.
Investment
(Thousand Rupees) $10 - 15$ $15 - 20$ $20 - 25$ $25 - 30$ $30 - 35$
No. of families $30$ $50$ $60$ $55$ $15$

Uses the table given to find:
The class mark of the third class

Draw a histogram for the following frequency distribution.
Use of electric
(unit) $50-70$ $70-90$ $90-110$ $110-130$ $130-150$ $150-170$
No. of families $150$ $400$ $460$ $540$ $600$ $350$

Draw a histogram of the following data.
Height of student ( $\text{cm}$ ) $135 - 140$ $140 - 145$ $145 - 150$ $150 - 155$
No. of students $4$ $12$ $16$ $8$

Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.
TIme (minutes) $60 - 80$ $80 - 100$ $100 - 120$ $120 - 140$ $140 - 160$
No.of students $14$ $20$ $24$ $22$ $16$

Classify the following information as primary or secondary data:
For a science project, information of trees gathered by visiting a forest.

If the mean of the following data is $20.2$ , then find the value of p.
$x_i$ 10 15 20 25 30
$f_i$ 6 8 p 10 6

If the median of the distribution given is 28.5, find the values of 'x' and 'y'
Class Interval Frequency
0 - 10 5
10 - 20 x
20 - 30 20
30 - 40 15
40 - 50 y
50 - 60 5
Total 60

If the mean of the following distribution is 6, find the value of P.
$x_{i}$ 2 4 6 10 P + 5
$f_{i}$ 3 2 3 1 2

If the classmark is $10$ and class width is $6$ then find the class.

Draw a histogram to represent the following frequency distribution:
Class interval Frequency
$20 - 25$ $5$
$25 - 30$ $10$
$30 - 35$ $18$
$35 - 40$ $14$
$40 - 45$ $12$

If median of the following distribution is 27, then find the values of x and y.
Class-interval Frequency
0-10 5
10-20 $x$
20-30 20
30-40 14
40-50 $y$
50-60 8
Total 68

If mean of following data is 21.5, then find K :
Class 0-10 10-20 20-30 30-40 40-50
Frequency 6 4 3 K 2

Class Interval $0-20$ $20-40$ $40-60$ $60-80$ $80 -100$
Frequency $5$ $6$ $12$ $4$ $3$
Construct histogram for the following frequency table:

Construct histogram for the following frequency table :

Class Interval $0-5$ $5-10$ $10-15$ $15-20$ $20-25$
Frequency $18$ $15$ $14$ $8$ $10$

The value of $\pi$ upto $50$ decimal places is given below.
$3.14159265358979323846264338327950288419716939937510$
(ii) What are the most and the least frequency occurring digits?

Find the range ad coefficient of range of the following table -
x 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5
f 4 5 6 3 2 1 3 5

Variate (x) 5 10 15 20 25 30
Frequency (f) 2 6 4 1 5 2
Construct a frequency polygon for the following frequency distribution.

Class interval $5 - 9$ $10 - 14$ $15 - 19$ $20 - 24$
Frequency $3$ $5$ $4$ $2$
Construct histogram for the following frequency distribution.

Mid values 8 14 20 26 32
Frequency 10 15 25 9 6
Construct histogram from the following distribution.

Class Interval $3-6$ $6-12$ $12-13$ $13-14$ $14-15$
Frequency $150$ $420$ $100$ $110$ $50$
Construct histogram for the following frequency table:

The following frequency distribution table shows marks obtained by 180 students in Mathematics examination :
Marks Number of Students
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50 25
$x$
30
2 $x$
65
Find the value of $x$ .
Also draw a histogram representing the above information.

The mean of five numbers is $27$ . If one number is excluded, then the new mean is $25$ . Find the excluded number.

Find the missing frequencies in the following frequency distribution, if the mean of the distribution is $1.46$ .

21	20	55	39	48	46	36	54	42	30
29	42	32	40	34	31	35	37	52	44
54	36	41	33	23	39	28	44	45	38
39	45	37	33	51	53	52	46	43	47
51	26	52	48	25	34	37	33	36	27

Height (in cm)	102 - 108	112 - 118	122 - 128	132 - 138	142 - 148	152 - 158
No. of pupils	8	12	5	16	35	23

Profits per shop:	0-50	50-100	100-50	150-200	200-250	250-300
No. of shops:	12	18	27	20	17	6

Cost of living index:	$440-460$	$460-480$	$480-500$	$500-520$	$520-540$	$540-560$	$560-580$	$580-600$
No. of months:	$2$	$4$	$3$	$5$	$3$	$2$	$1$	$4$

Country	India	Nepal	China	Pakistan	U.K.	U.S.A.
Average age (in years)	55	52	60	50	70	75

Class interval	$600-640$	$640-680$	$680-720$	$720-760$	$760-800$	$800-840$
Frequency	$18$	$45$	$153$	$288$	$171$	$63$

Year	$1960-1961$	$1970-1971$	$1980-1981$	$1990-1991$	$2000-2001$
Route length (in thousand kilometres)	$56$	$60$	$61$	$74$	$98$

Year	1995-96	1996-97	1997-98	1998-99	1999-2000
Interest (in lakhs of rupees)	20	25	15	18	30

Class Interval	Frequency
150-153	7
154-157	7
158-161	15
162-165	10
166-169	5
170-173	6

Life time (hours)	Number of tubes	Life time (hours)	Number of tubes
$300-399$	$14$	$800-899$	$62$
$400-499$	$46$	$900-999$	$48$
$500-599$	$58$	$1000-1099$	$22$
$600-699$	$76$	$1100-1199$	$6$
$700-799$	$68$
			Total $400$