Class 10 Maths - Statistics

For a certain frequency distribution, if median $= 156$ and mode $= 180$ , then obtain the value of mean

By empirical formula
mode

$=3\times$ median

$-2\times$ mean

$\Rightarrow 180=3\times156-2\times$ mean

$\Rightarrow 2\times$ mean

$=468-180$

$\Rightarrow$ mean

$=144$

Find the mean of the data whose median is $64$ and mode is $45$ .

Median

$=64$ , Mode

$=45$ , Mean

$=?$
Mode

$=3$ Median

$-2$ Mean

$45=3\times 64-2$ Mean

$2$ Mean

$=192-45$

$2$ Mean

$=147$
Mean

$=\cfrac{147}{2}$
Mean

$=73.5$

The demand of different shirt sizes, as obtained by a survey, is given. Calculate the Mode.
Size $38$ $39$ $40$ $41$ $42$ $43$ $44$
Number of persons(wearing it) $27$ $40$ $51$ $16$ $14$ $8$ $6$

In this table,

$51$ is the greatest frequency which corresponds to the size

$40$

Hence, mode is $40$ .

Find the mode for the following table.
Temperature in $^oC$ $29$ $32.4$ $34.6$ $36.9$ $38.7$ $40$
Number of days $7$ $2$ $6$ $4$ $8$ $3$

The score whose frequency occurs more is the mode of data

Temperature( $^{o}C$ )	29	32.4	34.6	36.9	38.7	40
Number of days	7	2	6	4	8	3

In this table, $8$ is the greatest frequency where $x=38.7$

Hence, mode is $38.7$

Find the mode for the following frequency table.
Wages(Rs.) $250$ $300$ $350$ $400$ $450$ $500$
Number of workers $10$ $15$ $16$ $12$ $11$ $13$

Wages(Rs.)	Number of workers
$250$	$10$
$300$	$15$
$350$	$16$
$400$	$12$
$450$	$11$
$500$	$13$

We observe from the table that the maximum frequency is

$16$ . The value of the variate(wage) corresponding to the maximum frequency

$16$ is

$350$ . This is the mode of the given data.

The Mean and Median of a Uni-modal grouped data are $36.5$ and $38.5$ respectively. Find the Mode of the data.

We know that, Mode

$=3\times$ median

$-\, 2\times$ mean

Mode

$=3(36.5)-2(38.5)$

Mode

$=109.5-77=32.5$

Four integers are added to a group of integers $3, 4, 5, 5$ and $8$ and the mean, median, and mode of the data increases by $1$ each. What is the greatest integer in the new group of integers?

Let the Mean

$=\dfrac{(3+4+5+5+8)}{5}=\dfrac{25}{5}=5$

Median

$=5$

Mode

$=5$ (

$5$ has frequency

$2$ )

Four integers are added to a group of integers data increases by

$1$ each therefore,

Mode

$=6$

In group of integers

$6$ must be

$3$ times,therefore Let fourth integer

$=x$

Group of integers

$=3,4,5,5,6,6,6,8$ and

$x$

Therefore new mean is

$6$

$\dfrac{(3+4+5+5+6+6+6+8+x)}{9}=6$

Implies that

$\dfrac{(43+x)}{9}=6$

Implies that

$43+x=54$

$x=54-43$

$x=11$ .

Observations of a data are $16, 72, 0, 55, 65, 55, 10,$ and $41$ . Chaitanya calculated the mode and median without taking the zero into consideration. Did Chaitanya do the right thing?

Arranging the data in ascending order

$0$ ,

$10$ ,

$16$ ,

$41$ ,

$55$ ,

$65$ ,

$72$

Here

$n=8$

Median

$=$ mean of

${\left( {\frac{n}{2}} \right)^{th}}$ and

${\left( {\frac{n}{2} + 1} \right)^{th}}$ observations

$=$ mean of

${\left( {\frac{8}{2}} \right)^{th}}$ and

${\left( {\frac{8}{2} + 1} \right)^{th}}$ observation

$=$ mean of

${\left( 4 \right)^{th}}$ and

${\left( 5 \right)^{th}}$ observations

$=$

$\frac{{ 41 + 55}}{2}$

$=$

$48$

Mode

$=$

$55$

Now, if zero is not taken into consideration

arrange data in ascending order

$0$ ,

$10$ ,

$16$ ,

$41$ ,

$55$ ,

$65$ ,

$72$

Here

$n=7$

Median

$= {\left( {\frac{{n + 1}}{2}} \right)^{th}}$

$= {\left( {\frac{{7 + 1}}{2}} \right)^{th}}$

$= {4^{th}}$

$=55$

Mode

$=55$

Therefore Mode is correct but Median is wrong

If the mean of the following distribution be $50$ , then find the values of ${ f }_{ 1 }$ and ${ f }_{ 2 }$ .

Class 0-20 20-40 40-60 60-80 80-100 Total

Frequency 17 28 ${ f }_{ 1 }$ 24 ${ f }_{ 2 }$ 120

For the given distribution firstly we have

$17+28+f_1+24+f_2=120$

or,

$f_1+f_2=51$ .........(1).

Now, middle value of the classes are

$10,30,50,70,90$ respectively.

Now, mean

$=\large{\frac{17\times 10+28\times 30+f_1\times 50+24\times 70+f_2\times 90}{120}}=\frac{17+84+5f_1+168+9f_2}{12}$ .

According to the problem we have

$\large{\frac{17+84+5f_1+168+9f_2}{12}}$

$=50$

or,

$5f_1+9f_2+269=600$

or,

$5f_1+9f_2=331$ .......(2).

Now, (2)-5x(1) gives

$4f_2=76$

or,

$f_2=19$ .

Using value of

$f_2$ in (1) we get,

$f_1=32$ .

Find the mode of the following data:
Marks Below 10 Below 20 Below 30 Below 40 Below 50
No. of students 8 20 45 58 70

In the given frequency distribution ,

$h = 10$

Class

$20 - 30$ has maximum frequency. i.e.,

$25$

Therefore,

$20 - 30$ is the model class in which

$l = 20$ and

$f_m = 25.$

Here,

$f_1 = 12$ and

$f_2 = 13$

$Mode = l + \left(\dfrac{f_m - f_1}{2f_m - f_1 - f_2}\right) \times h$

$= 20 + \left(\dfrac{25-12}{50-12-13} \right)\times 10$

$= 20 + \dfrac{130}{25}$

$= 20 + 5.2$

$= 25.2$ (Ans)

986049_1039136_ans_7c2fdf1ec1c8414e8c80a88173088c45.jpg

Answer the following :
The median of 230 observation isFind a and b.
Class 10-20 20-30 30-40 40-50 50-60 60-70 80-90
Frequency 12 30 a 65 b 25 18

First finding Cumulative Frequencies(cf) of all classes

cf of class

$10$ to

$20=12$

cf of class

$20$ to

$30=12+30=42$

cf of class

$30$ to

$40=42+a$

cf of class

$40$ to

$50=107+a$

cf of class

$50$ to

$60=107+a+b$

cf of class

$60$ to

$70=132+a+b$

cf of class

$70$ to

$80=150+a+b$

As total number of observations are,

$n=230$ so

$230=150+a+b$ , because cf of last interval is equal to

$n$ .

$a+b=80$ ...

$(1)$

Median

$=46$ that is between

$40$ and

$50$ , hence median class

$=40-50$

$f=$ frequency of median class

$=65$

$cf=$ cumulative frequency

$=42+a$ ... (cf of previous class)

$l=$ lower limit of median class

$=40$

$h=50-40=10$

Median

$=l+\dfrac{\dfrac{n}{2}-cf}{f}\times h$

$46=40+\dfrac{\dfrac{230}{2}-(42+a)}{65}\times 10$

$\dfrac{6}{10}=\dfrac{73+a}{65}$

$a=34$

From

$(1)$ :

$34+b=80$

$b=46$

Hence

$a=34$ and

$b=46$

Find the mode for the following frequency distribution.
Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$ $500-600$
Frequency $64$ $62$ $77$ $62$ $66$ $54$

Modal class is the class with maximum frequency. The modal class here is

$200-300$ .

$l$ is the lower limit of modal class, so

$l=200$

$h$ is the size of class interval, so

$h=300-200=100$

$f_1$ is the frequency of modal class, so

$f_1=77$

$f_0$ is the frequency of the class preceding the modal class, so

$f_0=62$

$f_2$ is the frequency of the class succeeding the modal class, so

$f_2=62$

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

$Mode=200+\dfrac{77-62}{2(72)-62-62}\times 100$

$Mode=200+\dfrac{15}{20}\times 100$

$Mode=200+75$

$Mode=275$

Hence the mode is

$275$ .

Find the mode of the following frequency distribution.
Age (years) 14 13 12 15 10
Numbers of students 4 8 7 2 9
Shirt size(cm) 38 36 40 42 44 37 34 41
Numbers of students 7 12
18 6 2 14 3 8

Mode = which occurs more / frequency is more.

(1) number of students

$= 9$ , Age

$=10$

$\therefore$ Mode

$=10$

(2) Number of students

$= 18$ , Shirt size

$=40cm$

$\therefore$ Mode

$=40$

Find the median of the following data.
$x$ $10$ $20$ $30$ $40$ $50$
$f$ $2$ $3$ $2$ $3$ $1$

Median is

$30$ as the given data is already in increasing order.

The distribution of sale of shirts sold in a month in a departmental store is as under. Calculate the model class of shirts sold.

Size (in cms) 80-85 85-90 90-95 95-100 100-105 105-110 110-115
No. of shirts sold 33 27 85 155 110 45 15

The number of shirts sold under the class

$95-100$ is more compared to other sizes.

Hence, the model class of shirts sold is

$95-100$ .

Calculate mode of the following data.
Marks $0-20$ $20-40$
$40-60$ $60-80$
$80-100$
Number of students $8$ $10$ $12$ $6$ $3$

Highest frequency

$12$ , Modal class

$=40-60$

Mode

$=I+\frac{f_1-f_0}{2f_1-f_0-f_2} \times h$

Where ,

$I=$ lower limit of the modal class

$=40$

$f_1 =$ frequency of modal class

$=12$

$f_2$ frequency of class after the modal class

$=6$

$f_0$ frequency of class before the modal class

$=10$

$h=$ class width

$=20$

putting the values in the equation

$Mode =50$

The mode of the following data isFind the missing frequency $\chi$ in it.
CLASS 0-10 10-20 20-30 30-40 40-50 50-60 60-70
FREQUENCY 8 10 $\chi$
16 12 6 7

$\Rightarrow$ Modal class

$=30-40$

So,

$f_0=x,\,f_1=16,\,f_2=12,\,l=30$ and

$h=10$

$\Rightarrow$

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

$\Rightarrow$

$36=30+\dfrac{16-x}{2\times 16-x-12}\times 10$

$\Rightarrow$

$6=\dfrac{16-x}{20-x}\times 10$

$\Rightarrow$

$120-6x=160-10x$

$\Rightarrow$

$4x=40$

$\therefore$

$x=10$

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is $24$ .
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
Frequency $5$ $25$ $?$ $18$ $7$

$CI$	$Frequency$	$c.f$
$0-10$	$5$	$5$
$10-20$	$25$	$30$
$20-30$	$x$	$30+x$
$30-40$	$18$	$48+x$
$40-50$	$7$	$55+x$
	$N=55+x$

Median

$=24$

Median class

$=20-30$

Here,

$l=20,\,h=10,\,f=x,\,cf=30,\,\dfrac{N}{2}=\dfrac{55+x}{2}$

$\Rightarrow$

$Median=l+\dfrac{\dfrac{N}{2}-cf}{f}\times h$

$\Rightarrow$

$24=20+\dfrac{\dfrac{55+x}{2}-30}{x}\times 10$

$\Rightarrow$

$4=\dfrac{\dfrac{55+x}{2}-30}{x}\times 10$

$\Rightarrow$

$4x=\left(\dfrac{55+x}{2}-30\right)\times 10$

$\Rightarrow$

$4x\times 2=(55+x-60)\times 10$

$\Rightarrow$

$8x=550+10x-600$

$\Rightarrow$

$10x-8x=600-550$

$\Rightarrow$

$2x=50$

$\therefore$

$x=25$

The sum of squares of deviations for $10$ observations taken from mean $50$ is $250$ . The co-efficient of variations is

We know that, Deviation $=\left| {{x}_{i}}-\overline{x} \right|$

But given that,

${{\left| {{x}_{1}}-\overline{x} \right|}^{2}}+{{\left| {{x}_{2}}-\overline{x} \right|}^{2}}+{{\left| {{x}_{3}}-\overline{x} \right|}^{2}}+.............+{{\left| {{x}_{10}}-\overline{x} \right|}^{2}}=250$

$\sum\limits_{i=1}^{10}{{{\left| {{x}_{i}}-\overline{x} \right|}^{2}}}=250$

$S.D.=\sqrt{\dfrac{1}{n}\sum\limits_{i=1}^{10}{{{\left| {{x}_{i}}-\overline{x} \right|}^{2}}}}$

$=\sqrt{\dfrac{1}{10}\times 250}=\sqrt{25}=5$

Coefficient of variation

$=\dfrac{S.D.}{x}\times 100$

$=\dfrac{5}{50}\times 100=10%$

Hence, this is the answer.

The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.
No. of bulbs produced (Thousands) $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$
No. of factories $12$ $35$ $20$ $15$ $8$ $7$ $8$

$No.\,of\,bulbs\,produced$	$No.\,of\,factories$ $f$	$c.f.$
$30-40$	$12$	$12$
$40-50$	$35$	$47$
$50-60$	$20$	$67$
$60-70$	$15$	$82$
$70-80$	$8$	$90$
$80-90$	$7$	$97$
$90-100$	$8$	$105$
	$\sum f=105$

$\Rightarrow$

$\dfrac{N}{2}=\dfrac{105}{2}=52.5$

$\therefore$ Median class

$=50-60$

So,

$l=50,\,c.f=47,\,f=20$ and

$h=10$

$\Rightarrow$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

$=50+\dfrac{52.5-47}{20}\times 10$

$=50+2.75$

$=52.75$ thousand bulbs

$=52750$ bulbs.

Hence, the median of the production is

$52750$ bulbs

Find the mean of the following data $30, 32, 24, 34,28, 30, 35, 33, 25$
(i) Show that the sum of the deviations of all the given observations from the mean is zero.
(ii) Find the median of the given data.

Arranging given numbers in ascending order,

$24,25,26,28,30,30,32,33,34,35$

$Mean(\overline{x})=\dfrac{24+25+26+28+30+30+32+33+34+35}{10}=\dfrac{297}{10}=29.7$

$(i)$

$\Rightarrow$

$(24-29.7)+(25-29.7)+(26-29.7)+(28-29.7)+(30-29.7)+(30-29.7)+(32-29.7)+(33-29.7)$

$+(34-29.7)+(35-29.7)$

$\Rightarrow$

$-5.7-4.7-3.7-1.7+0.3+0.3+2.3+3.3+4.3+5.3$

$\Rightarrow$

$0$

Hence, we proved that the sum of the deviations of all the given observations from the mean is zero.

$(ii)$

Number of observation

$=10$

$Median=\dfrac{1}{2}\left[\left(\dfrac{10}{2}\right)^{th}obs+\left(\dfrac{10}{2}+1\right)^{th}obs\right]$

$Median=\dfrac{1}{2}[5^{th}obs+6^{th}obs]$

$5^{th}=30$ and

$6^{th}obs=30$

$\therefore$

$Median=\dfrac{30+30}{2}=30$

Find the median of the following data:
Class interval $0 - 20$ $20 - 40$ $40 - 60$ $60 - 80$ $80 - 100$ $100 - 120$ $120 - 140$
Frequency $6$ $8$ $10$ $12$ $6$ $5$ $3$

$Class\,Interval$	$Frequency$	$c.f.$
$0-20$	$6$	$6$
$20-40$	$8$	$14$
$40-60$	$10$	$24$
$60-80$	$12$	$36$
$80-100$	$6$	$42$
$100-120$	$5$	$47$
$120-140$	$3$	$50$
	$N=50$

Here,

$N=50$ , then

$\dfrac{N}{2}=\dfrac{50}{2}=25$

$\therefore$ Median class

$=60-80$

We have,

$l=60,\,c.f=24,\,f=12$ and

$h=20$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

$=60+\dfrac{25-24}{12}\times 20$

$=60+1.66$

$=61.66$

Finding median graphically for the following grouped data ?
Class Interval Frequency
0-5 03
5-10 06
10-15 11
15-20 08
20-25 06

$\Rightarrow$ We are going to form more than cumulative frequency table:

$Class-Interval$	$c.f.(more\,than)$
More than $0$	$34$
More than $5$	$31$
More than $10$	$25$
More than $15$	$14$
More than $20$	$6$

$\Rightarrow$ We are going to form less than cumulative frequency table:

$Class-Interval$	$c.f.(less\,than)$
Less than $5$	$3$
Less than $10$	$9$
Less than $15$	$20$
Less than $20$	$28$
Less than $25$	$34$

$\Rightarrow$ Median is where both less than and more then curves meet.

$\therefore$

$Median=12.5$

1220119_1152390_ans_f9c28e5193a14812a7228ab8fd89f431.jpeg

The following table given 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 modal class.

$\Rightarrow$ We know that the modal class is a class which has highest frequency.

$\Rightarrow$ In given table, class interval

$55-60$ has highest frequency

$20$ .

$\therefore$ Modal class is

$55-60$

If $mean = 60$ and $median = 50$ , then find mode using empirical relationship.

$\Rightarrow$ Mean

$=60$ and

$Median=50$

$\Rightarrow$

$2Median=2Mean+Mode$

$\Rightarrow$

$3(50)=2(60)+Mode$

$\Rightarrow$

$150=120+Mode$

$\Rightarrow$

$Mode=150-120$

$\therefore$

$Mode=30$

The weights (in kilograms) of ten puppies from the same litter are recorded below:
$1.23 \,\,\,\,1.42 \,\, \,\, 2.01 \,\, \,\, 1.56 \,\, \,\,1.66\,\,\,\,\, 0.92 \,\, \,\,1.38 \,\, \,\,1.44 \,\, \,\, 1.52 \,\, \,\,1.48$
Calculate: the median weight

Given entries are

$1.23, 1.42, 2.01, 1.56, 1.66,$

$0.92, 1.38, 1.44,1.52, 1.48$

Ascending order of the entries

$0.92, 1.23, 1.38, 1.42, 1.44, 1.48, 1.52, 1.56, 1.66, 2.01$

total no. of terms = 10 i.e., even

$\therefore$ median =

$\dfrac{\dfrac{n}{2}+(\dfrac{n}{2}+1)}{2}th$ term

$=\dfrac{(5^{th}+6^{th})}{2}$ term =

$\dfrac{1.44+1.48}{2}$

median = 1.46

Find the mode when median is 12 and mean is 16 of a data.

$Mode=3Median-2Mean$

$Mode = 3\times 12 -2\times 16=36-32=4$ .

Find the mode
Marks 1-5 6-10 11-16 16-20 21-25
No. of Students 7 10 6 32 24

$Marks$	$No.\,of\,students$
$0.5-5.5$	$7$
$5.5-10.5$	$10$
$10.5-16..5$	$6$
$16.5-20.5$	$32$
$20.5-25.5$	$24$

Here, class

$16.5-20.5$ contains maximum number of students.

$\therefore$ Modal class

$=16.5-20.5$

We have,

$l=16.5,f_0=6,f_1=32,f_2=24$ and

$h=4$

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

$=16.5+\dfrac{32-6}{2\times 32-6-24}\times 4$

$=16.5+3.05$

$=19.55$

Find the mode of the following distribution
Marks in Mathematics $45$ $60$ $80$ $90$ $95$ $100$
Number of students $5$ $15$ $6$ $4$ $6$ $4$

$\Rightarrow$ The mode is simply the number which appears most often.

$\Rightarrow$ In given frequency table we can see,

$60$ marks got by maximum number of students i.e.

$15$ .

$\therefore$

$Mode=60$

Find the mode of the following distribution
Size of shoes $6$ $7$ $8$ $9$ $10$
Number of student $10$ $15$ $9$ $4$ $2$

$\Rightarrow$ The mode is simply the number which appears most often.

$\Rightarrow$ Here in given frequency table,

$7$ size of shoes wear maximum number of students.

$\therefore$

$Mode=7$

The following table shows the difference of the ages $20$ couples. Find the mean by direct method.
Difference in ages (in years) 0-2 2-4 4-6 6-8 8-10 10-12
No. of couples 1 2 8 5 3 1

$Difference\,in\,ages$	$Class-Mark$ $x_i$	$Frequency$ $f_i$	$f_ix_i$
$0-2$	$1$	$1$	$1$
$2-4$	$3$	$2$	$6$
$4-6$	$5$	$8$	$40$
$6-8$	$7$	$5$	$35$
$8-10$	$9$	$3$	$27$
$10-12$	$11$	$1$	$11$
		$\sum f_i=20$	$\sum f_ix_i=120$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{120}{20}=6$

The following table shows the ages of the patients admitted in a hospital during a year:
Age(in years ) 5-15 15-25 25-35 35-45 45-55 55-65
Number of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measure of central tendency.

$Age$	$No.\,of\,patients$ $f_i$	$Class-mark$ $x_i$	$f_ix_i$
$5-15$	$6$	$10$	$60$
$15-25$	$11$	$20$	$220$
$25-35$	$21$	$30$	$630$
$35-45$	$23$	$40$	$920$
$45-55$	$14$	$50$	$700$
$55-65$	$5$	$60$	$300$
	$\sum f_i=80$		$\sum f_ix_i=2830$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{2830}{80}=35.37$

$\Rightarrow$ Modal class

$=35-45$

$l=35,h=10,f_0=21,f_1=23,f_2=14$

$\Rightarrow$

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

$=35+\dfrac{23-21}{2\times 23-21-14}\times 10$

$=35+1.81$

$=36.81$

The mode of the data shows that maximum number of patients is in the age group of

$36.81$ , while average age of all the patients is

$35.37.$

Find the median :
Weight No of students
$40-45$ $2$
$45-50$ $3$
$50-55$ $8$
$55-60$ $6$
$60-65$ $6$
$65-70$ $3$
$70-75$ $2$

$Weight$	$No.\,of\,students$	$c.f.$
$40-45$	$2$	$2$
$45-50$	$3$	$5$
$50-55$	$8$	$13$
$55-60$	$6$	$19$
$60-65$	$6$	$25$
$65-70$	$3$	$28$
$70-75$	$2$	$30$

$N=30$ and

$\dfrac{N}{2}=\dfrac{30}{2}=15$

$\therefore$ Median class

$=55-60$

$l=55,\,c.f=13,f=6$ and

$h=5$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

$=55+\dfrac{15-13}{6}\times 5$

$=55+1.66$

$=56.66$

Find the mode of
Class $0-5$ $5-10$ $10-15$ $15-20$ $20-25$
Frequency $2$ $3$ $10$ $1$ $4$

Here, we can see class

$10-15$ has maximum number of frequency.

$\therefore$ Modal class

$=10-15$

Then we have,

$l=10,f_0=3,f_1=10,f_2=1$ and

$h=5$

Since,

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

$=10+\dfrac{10-3}{2\times 10-3-1}\times 5$

$=10+2.18$

$=12.18$

$\therefore$

$Mode=12.18$

Hence, this is the answer.

The following table shows the number of workers in a factory and their daily wages. Find the median of the daily wages.
Daily wages(Rupees) $100-110$ $110-120$ $120-130$ $130-140$ $140-150$ $150-160$
No. of workers $37$ $38$ $40$ $33$ $28$ $24$

$\dfrac{N}{2}=100$ , corresponds to

$120-130$ class

Median

$=L+\left(\dfrac{\dfrac{N}{2}-F}{fm}\right)\times h$

where

$F=75$

$fm=40$

$L=120$

$h=10$

$=120+\left(\left(\dfrac{100-75}{40}\right)\times 10\right)$

$=120+\dfrac{25}{4}$

$=126.25$ .

1192516_1260775_ans_5d17cc6831104d449967e22d00459587.jpg

Find the mode of following data:

CI 5-15 15-25 25-35 35-45 45-55 55-65
f 6 11 21 23 14 5

The class 35-45 has maximum frequency. Therefore, this is the modal class.

$l=35, h=`10, f=23, f_{1}=21, f_{2}=14$

Therefore,

Mode =

$l+\dfrac{f-f_{1}}{2f-f_{1}-f_{2}}\times h$

$=35+\dfrac{23-21}{46-21-14}\times 10$

$=35+\dfrac{2}{11}\times 10$

$=35+\dfrac{20}{11}=36.81$

Find the mode of the following distribution:
Height(in cm) 30-40 40-50 50-60 60-70 70-80 80-90
No. of plants 4 3 8 11 6 2

Height No. of parents

$30-40$

$4$

$40-50$

$3$

$50-60$

$8$

$60-70$

$11$

$\rightarrow$ Modal class as it has highest frequency

$70-80$

$6$

$80-90$

$2$

$l_1=60$

$f_o=8$

$f_1=11$

$f_2=6$

$c=10$

Mode

$=l_1+\left(\dfrac{f_1-f_o}{2f_1-f_o-f_2}\times c\right)$

$=60+\dfrac{30}{3}$

$=63.7$ .

1192807_1291223_ans_a4f57f4b09844e57a43b56de200e677e.jpg

Find the mode of the following distribution

Height (in inches) $60$ $61$ $62$ $63$ $64$ $65$ $66$
Number of children $0$ $10$ $6$ $4$ $18$ $10$ $2$

The mode is the number that appears most often in the given data set.

From the given data, we can see that the maximum number of children have a height

$64$ inch.

$\therefore$

$Mode=64$

Find the median marks for the following.
Marks $25-35$ $35-45$ $45-55$ $55-65$ $65-75$ $75-85$
No. of students $7$ $31$ $30$ $18$ $12$ $2$

1227139_1227254_ans_b96e67dc8d314015be9ffb253e3e0838.jpg

Find the median of the following data:
monthly consumption no. of consumers
Below 85 4
Below 105 9
Below 125 22
Below 145 42
Below 165 56
Below 185 64

From the given data

Width of class is

$w=105-85=20$

total frequency is

$N=64$

Median class frequency is

$f=\dfrac{N}{2}=\dfrac{64}{2}=32$

So median class is

$125-145$

Cumulative frequency

$C=22$

Lower Boundary of median class

$L=125$

Median is

$L+\dfrac{\dfrac{N}{2}-C}{f}\times w=125+\dfrac{32-22}{42-22}\times 20=135$

Find the mode of the following frequency distribution.
Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 8 10 10 16 12 6 7

Mode is the one which has the highest frequency

Here the class

$30-40$ has the highest frequency

$16$

Modal class is

$30-40$

Lower boundary of modal class is

$l=30$

Height of each class is

$h=10-0=10$

modal frequency is

$f_m=16$

Frequency of class preceding to Modal class is

$f_1=10$

Frequency of class succeeding to Modal class is

$f_2=12$

Mode

$=l+h\bigg(\dfrac{f_m-f_1}{2 f_m-f_1-f_2}\bigg)=30+10\times \dfrac{16-10}{32-10-12}=30+6=36$

Write the formula to find the median of a grouped data and explain the alphabet in it.

For grouped data-

$\text{Median} = l + \cfrac{\cfrac{N}{2} - {f}_{0}}{{f}_{1}} \times h$

Whereas

$l =$ Lower limit of edian class

$N =$ Total frequency

${f}_{0} =$ Frequancy of class preceding the median class

${f}_{1} =$ Frequency of median class

$h =$ Width of median class

Find the median of the following data
$3,1,5,6,3,4,5$

Given Data is

$3,1,5,6,3,4,5$

Ascending order of the data is

$1,3,3,4,5,5,6$

The middle element in this order is

$4$

So the medial of the data is

$4$

A survey conducted on $20$ household in a locality by a group of statement resulted in the following frequency table for the number of family
Members in a house hold.
$\text {Family} \,\text {size}$ $1-3$ $3-5$ $5-7$ $7-9$ $9-11$
$\text {Number}\, \,of\, \text {families}$ $7$ $8$ $2$ $2$ $1$
Find the mode of the data.

Family size	Number of families
$1-3$	$7 = {f}_{0}$
$3-5$	$8 = {f}_{1}$
$5-7$	$2 = {f}_{2}$
$7-9$	$2$
$9-11$	$1$
	$\sum{f} = 20$

Mode

$= l + \cfrac{{f}_{1} - {f}_{0}}{2 {f}_{1} - {f}_{0} - {f}_{2}} \times h$

Whereas,

$l =$ Lower limit of modal class

$= 3$

$h =$ Class interval

$= 5 - 3 = 2$

${f}_{0} =$ frequency of class preceeding the modal class

$= 7$

${f}_{1} =$ frequency of modal class

$= 8$

${f}_{2} =$ frequancy of class succeding the modal class

$= 2$

Therefore,

Mode

$= 3 + \cfrac{8 - 7}{\left( 2 \times 8 \right) - 7 - 2} \times 2 \\ = 3 + \cfrac{1}{16 - 9} \times 2 \\ = 3 + \cfrac{2}{7} = 3.286$

Hence the mode of the given data is

$3.286$ .

If the mode of a distribution is $8$ and mean is also $8$ , then find median.

Mode

$=8$ ; Mean

$=8$ ; Median =?
Relation among mean, median and mode is

$3$ median = mode

$+2$ mean

$3\times$ median

$=8+2\times 8$
Median

$=\dfrac {8+16}{3}=\dfrac {24}{3}=8$

If median = 137 units and mean = 137.05 units, then find the mode.

We know,

Mode = 3(Median) - 2(Mean)

= 3(137) - 2(137.05)

= 411-274.10 = 136.90

Hence, mode is 136.90 units.

Find the mean of the following frequency distribution:
Class 0-20 20-40 40-60 60-80 80-100
Frequency 15 18 21 29 17

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	15	10	150
20-40	18	30	540
40-60	21	50	1050
60-80	29	70	2030
80-100	17	90	1530
Total	$\sum { f_{ i } } =100$		$\sum { { f }_{ i }{ x }_{ i } } =5300$

Now,

$\overline { x } =\cfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { f_{ i } } } =\cfrac { 5300 }{ 100 } =53$

The following data gives the information on the observed lifetimes (in hours) of $225$ electrical components:
Lifetimes (in hours) $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$
Frequency $10$ $35$ $52$ $61$ $38$ $29$
Determine the modal lifetimes of the components.

Here, the maximum class frequency is

$61$ and the class corresponding to this frequency is

$60-80$ . So, the modal class is

$60-80$ .
Here,

$I=60, h=20, f_{1}=61, f_{0}=52, f_{2}=38$

$\therefore$ Mode

$=l+\dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\times h=60+\dfrac{61-52}{2\times 61-52-38}\times 20=60+\dfrac{9}{122-90}\times 20$

$=60+\dfrac{9}{3}\times 20=60+\dfrac{45}{8}=60+5.625=65.625$
Hence, modal lifetimes of the components is

$65.625$ hours.

Weight (in kg)	$40-45$	$45-50$	$50-55$	$55-60$	$60-65$	$65-70$	$70-75$
Number of students	$2$	$3$	$8$	$6$	$6$	$3$	$2$

Write the modal class for the following frequency distribution:
Class interval $10-20$
$20-30$ $30-40$ $40-50$ $50-60$ $60-70$
Frequency $33$ $38$ $65$ $52$ $19$ $48$

Maximum frequency i.e.,

$65$ corresponds to the class

$30-40$

$\therefore$ Modal class is

$30-40$ .

A data constitutes of heights (in cm) of $50$ children. What do you understand by mode for the data?

We know that,

Mode is the data that occur most frequently in mention set.

In the given data, the mode will represent the height that most number of the student have.

Find the mean of the following frequency distribution:
Class 25-29 30-34 35-39 40-44 45-49 550-54 555-59
Frequency 14 22 16 6 5 3 4

Find the median class of the following distribution:
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
Frequency $4$ $4$ $8$ $10$ $12$ $8$ $4$

First we find the cumulative frequency

Classes	Frequency	Cumulative Frequency
$0-10$	$4$	$4$
$10-20$	$4$	$8$
$20-30$	$8$	$16$
$30-40$	$10$	$26$
$40-50$	$12$	$38$
$50-60$	$8$	$46$
$60-70$	$4$	$50$
Total	$50$

Here,

$\dfrac{n}{2}=\dfrac{50}{2}=25$

$\therefore$ Median class

$=30-40$ .

Find the mean of the following frequency distribution:
Class interval 50-60 60-70 70-80 80-90 90-100
Frequency 8 6 12 11 13

Class	Frequency $f_{ i }$	Class Mark ${ x }_{ i }$ ${ x }_{ i }=\frac {\text{ Upper class limit+ Lower class limit} }{ 2 }$	${ x }_{ i }{ f }_{ i }$
50-60	8	55	440
60-70	6	65	390
70-80	12	75	900
80-90	11	85	935
90-100	13	95	1235
Total	$\sum { f_{ i } } =50$		$\sum { { f }_{ i }{ x }_{ i } } =3900$

Now,

$\overline { x } =\cfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { f_{ i } } } =\cfrac { 3900 }{ 50 } =78$

Find the mean of the following frequency distribution:
Class interval 15-25 25-35 35-45 45-55 55-65
Frequency 60 35 22 18 15

Class	Frequency $(f_{ i })$	Class Mark $(x_{ i })$ ${ x }_{ i }=\frac {\text{ Upper class limit+ Lower class limit} }{ 2 }$	${ x }_{ i }{ f }_{ i }$
15-25	60	20	1200
25-35	35	30	1050
35-45	22	40	880
45-55	18	50	900
55-65	15	60	900
Total	$\sum { f_{ i } } =150$		$\sum { { f }_{ i }{ x }_{ i } } =4930$

Now,

$\overline { x } =\cfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { f_{ i } } } =\cfrac { 4930 }{ 150 } =32.87$

Find the mean of the following data:
Classes 0-100 100-200 200-300 300-400 400-500
Frequency 6 9 15 12 8

Find the mean of the following data:
Classes 0-10 10-20 20-30 30-40 40-50
Frequency 3 5 9 5 3

Class	Frequency $({ f }_{ i })$	Class Mark $({ x }_{ i })$ ${ x }_{ i }=\frac { Upper\quad class\quad limit+\quad Lower\quad class\quad limit }{ 2 }$	${ x }_{ i }{ f }_{ i }$
0-10	3	5	15
10-20	5	15	75
20-30	9	25	225
30-40	5	35	175
40-50	3	45	135
Total	$\sum { { f }_{ i } }=25$		$\sum { { f }_{ i }{ x }_{ i } }=625$

Now,

$\overline { x } =\cfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\cfrac { 625 }{ 25 } =25$

A hockey player has scored the following number of goals in $9$ matches.$$5, 4, 0, 2, 2, 4, 4, 3, 3$$.Find the mean, median, and mode of the data.

We have,
The mean of $35$ observations is $20$ .
Let x be the sum of $35$ observations
$\Rightarrow mean =\dfrac{\text{sum of observations number}}{{number of observations}}$
$\Rightarrow 20 =\dfrac{x}{35}$
$\Rightarrow x = 700$

Let y be the sum of the first $18$ observations.
Mean of first $18$ observations is $15$ .
$\dfrac{y}{18} = 15$
$\Rightarrow y = 270$

And z be the sum of last $18$ observations
Mean of the last $18$ observations is $25$ .
$\dfrac{z}{18} = 25$
$\Rightarrow z = 450$

let the $18th$ term be a.
$18th$ term is common in first $18$ and last $18$ observations
$x + a = y + z$
$700 + a = 270 + 450$
$a = 720 - 700$
$a = 20$
Hence, the $18th$ observation is $20$ .

The arithmetic mean of the following frequency distribution is 25.Determine the value of p:
Class
0-10
10-20
20-30
30-40
40-50
Frequency
7
8
p
15
4

Class Interval	$f_i$	Class Mark $x_i$	$x_iF_i$
0-10	7	5	35
10-20	8	15	120
20-30	P	25	25p
30-40	15	35	525
40-50	4	45	180
	$\Sigma f_i=34+p$		$\Sigma x_i f_i=860+25p$

$For\quad grouped\quad data\quad when\quad Class\quad Interval\quad \quad is\quad given,\quad we\quad find\quad the\quad Class\quad Mark.\\ Class\quad Mark\quad =\quad mid\quad point\quad of\quad an\quad interval\quad =\quad \frac { \left( Lower\quad limit\quad +\quad Upper\quad limit \right) }{ 2 } \\ Class\quad Mark\quad is\quad taken\quad as\quad { x }_{ i }\quad for\quad each\quad class\quad interval.\quad Find\quad { x }_{ i }{ f }_{ i }.\quad \\ From\quad the\quad table,\quad we\quad have\quad \Sigma { x }_{ i }{ f }_{ i }=860+25p\quad \& \quad \quad \Sigma { f }_{ i }=34+p\\ Given\quad Mean\quad =\quad 25.25\\ Mean\quad =\quad \frac { \Sigma { x }_{ i }{ f }_{ i } }{ \Sigma { f }_{ i } } \quad \\ \Rightarrow \quad 25.25\quad =\quad \frac { 860\quad +\quad 25p }{ 34+p } \\ \Rightarrow \quad 860\quad +\quad 25p\quad =\quad \left( 34+p \right) \quad \times \quad 25.25\\ \Rightarrow \quad 860\quad +\quad 25p\quad =\quad 858.5\quad +\quad 25.25p\\ \Rightarrow \quad 25.25p\quad -\quad 25p\quad =\quad 860\quad -\quad 858.5\\ \Rightarrow \quad 0.25p\quad =\quad 1.5\\ \Rightarrow \quad p\quad =\quad \frac { 1.5 }{ 0.25 } \\ \Rightarrow \quad p\quad =\quad \frac { 15\quad \times \quad 100 }{ 25\quad \times \quad 10 } \\ \Rightarrow \quad p\quad =\quad \frac { 15\quad \times \quad 4 }{ 10 } \\ \Rightarrow \quad p\quad =\quad \frac { 60 }{ 10 } \\ \Rightarrow \quad p\quad =\quad 6$

Class Interval
0-6
6-12
12-18
18-24
24-30
Frequency
6
8
10
9
7
If mean of above distribution is $15.45$ , If true then enter $1$ and if false then enter $0$

Calculate the missing frequency $f$ from the following distribution, it is being given that the median of the distribution is $24$ .
Class
0-10
10-20
20-30
30-40
40-50
Frequency
5
25
f
18
7

Frequency distribution table :

$ci$	$f_i$	$cf$
$0-10$	$5$	$5$
$10-20$	$25$	$5+25=30$
$20-30$	$f$	$30+f$
$30-40$	$18$	$48+f$
$40-50$	$7$	$55+f$

Median

$=l +\dfrac {\bigg(\dfrac{N}{2}-cf\bigg)}{ f_m}\times c$

Here,

Class Interval

$c=10$

$N=\Sigma f_i=55+f$

Median

$=24$

$\Rightarrow$ Median class

$=20-30$

$\Rightarrow$ Frequency of median class

$f_m=f$

Lower boundary of median class

$l=20$

previous cumulative frequency of median class

$cf=30$

$\therefore24$

$=20 +\dfrac {\bigg(\dfrac{55+f}{2}-30\bigg)}{ f}\times 10$

$\therefore24$

$=20 +\dfrac {(f-5)}{ 2f}\times 10$

$\therefore24$

$=\dfrac{40f + 10f-50}{2f}$

$\therefore 50f-48f=50$

$\therefore f =25$

Find the median for the following frequency distribution:
Class Interval
0-10
10-20
20-30
30-40
40-50
50-60
Frequency
6
9
14
2
19
10

$Class-Interval$	$Frequency$	$Cumlative\,frequency$
$0-10$	$6$	$6$
$10-20$	$9$	$15$
$20-30$	$14$	$29$
$30-40$	$2$	$31$
$40-50$	$19$	$50$
$50-60$	$10$	$60$
	$N=60$

Here,

$N=60$ , then

$\dfrac{N}{2}=\dfrac{60}{2}=30$

$\therefore$ Median class

$=30-40$

So,

$l=30,\,c.f=29,\,f=3$ and

$h=10$

$\Rightarrow$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

$=30+\dfrac{30-29}{2}\times 10$

$=30+5$

$\therefore$

$Median=35$

Find the median for the following frequency distribution:

Class Interval
25-35
35-45
45-55
55-65
65-75
Frequency
20
25
5
7
4
Give answer upto the nearest integer value.

$For\quad continuous\quad data,\quad median=\quad L\quad +\quad \left( \frac { \frac { N }{ 2 } -C }{ f } \right) \quad \times \quad H\\ Where\quad N\quad =\quad Total\quad cumulative\quad frequency\\ L\quad =\quad lower\quad limit\quad of\quad the\quad median\quad class\quad Interval$

$\quad ;\quad Median\quad Class\quad =\quad the\quad class\quad interval\quad having\quad cumulative\quad freq\quad \ge \quad \frac { N }{ 2 } \quad lies\\ C\quad =\quad Cumulative\quad frequency\quad of\quad the\quad class\quad interval\quad just\quad before\quad the\quad median\quad class\quad interval\\ f\quad =\quad frequency\quad of\quad the\quad median\quad class\quad interval\quad ;\quad H\quad =\quad width\quad of\quad the\quad median\quad class$

$\quad interval// Here\quad N=61\\ \frac { N }{ 2 } =30.5\quad ,\quad this\quad cumulative\quad freq\quad lies\quad within\quad interval\quad \left( 35-45 \right) ;\quad L=35\\ C=20\quad ;f=25;\quad H=10\\ Hence\quad median=\quad 35+\left( \frac { \frac { 61 }{ 2 } -20 }{ 25 } \times 10 \right) =\quad 35+\left( \frac { 30.5-20 }{ 25 } \times 10 \right)$

$=\quad 35+\left( \frac { 10.5 }{ 25 } \times 10 \right) =\quad 35+\frac { 105 }{ 25 } =\quad 35+4.2=39.2$

The.following table gives the distribution of IQ of 100 students. Find the median IQ.
IQ
Frequency
75-84
8
85-94
11
95-104
26
105-114
31
115-124
18
125-134
4
135-144
2

$\quad having\quad cumulative\quad freq\quad \ge \quad \frac { N }{ 2 } \quad lies\\ C\quad =\quad cumulative\quad frequency\quad of\quad the\quad class\quad interval\quad just\quad before\quad the\quad median\quad class\quad interval\\ f\quad =\quad frequency\quad of\quad the\quad median\quad class\quad interval\quad ;\quad H\quad =\quad width\quad of\quad the\quad median\quad class$

$\quad interval\\$

$Here\quad N=100\\ \frac { N }{ 2 } =50\quad ,\quad this\quad cumulative\quad freq\quad lies\quad within\quad interval\quad \left( 35-40 \right) ;\\ L=105\quad ;C=45\quad ;f=31;\quad H=5\\ Hence\quad median=\quad 105+\left( \frac { \frac { 100 }{ 2 } -45 }{ 31 } \times 10 \right) =\quad 105+\left( \frac { 50-45 }{ 31 } \times 10 \right) =\quad 105+\left( \frac { 5 }{ 31 } \times 10 \right) =\quad 105+\frac { 50 }{ 31 }$

$=\quad 105+1.61=\quad 106.6$

Find the median from the following data:
Class groups
Frequency
110-120
6
120-130
25
130-140
48
140-150
72
150-160
116
160-170
60
170-180
38
180-190
22
190-200
3
mean is 153.8
If true then enter $1$ and if false then enter $0$

$\quad interval\quad having\quad cumulative\quad freq\quad \ge \quad \frac { N }{ 2 } \quad lies\\ C\quad =\quad cumulative\quad frequency\quad of\quad the\quad class\quad$

$interval\quad just\quad before\quad the\quad median\quad class\quad interval\\ f\quad =\quad frequency\quad of\quad the\quad median\quad class\quad interval\quad ;\quad H\quad =\quad width\quad of\quad the\quad median\quad class\quad$

$interval\\$

$Here\quad N=390\\ \frac { N }{ 2 } =195\quad ,\quad this\quad cumulative\quad freq\quad lies\quad within\quad interval\quad \left( 150-160 \right) ;\\ L=150\quad ;C=151\quad ;f=116;\quad H=10\\ Hence\quad median=\quad 150+\left( \frac { \frac { 390 }{ 2 } -151 }{ 116 } \times 10 \right) =\quad 150+\left( \frac { 195-151 }{ 116 } \times 10 \right) =\quad 150+\left( \frac { 44 }{ 116 } \times 10 \right)$

$=150+\left( \frac { 22 }{ 58 } \times 10 \right) =\quad 150+\left( \frac { 11 }{ 29 } \times 10 \right) =\quad 150+\left( \frac { 110 }{ 29 } \right) =\quad 150+3.79=150+3.8=\quad 153.8$

The following table gives the marks obtained by 50 students in a class test :
Marks
No. of students
11-15
2
16-20
3
21-25
6
26-30
7
31-35
14
36-40
12
41-45
4
46-50
2
Find the median

$ci$	$f_i$	$cf$
$10.5-15.5$	$2$	$2$
$15.5-20.5$	$3$	$2+3=5$
$20.5-25.5$	$6$	$5+6=11$
$25.5-30.5$	$7$	$11+7=18$
$30.5-35.5$	$14$	$18+14=32$
$35.5-40.5$	$12$	$32+12=44$
$40.5-45.5$	$4$	$44+4=48$
$45.5-50.5$	$2$	$48+2=50$

Median

$=l +\dfrac {\bigg(\dfrac{N}{2}-cf \bigg)}{ f_m}\times c$

Here,

Class Interval

$c=5$

$N=\Sigma f_i=50$

$\dfrac {N}{2}=25$

$\Rightarrow$ Median class

$=30.5-35.5$

$\Rightarrow$ Frequency of median class

$f_m=14$

Lower boundary of median class

$l=30.5$

previous cumulative frequency of median class

$cf=18$

$\therefore$ Median

$=30.5+\dfrac {\bigg(\dfrac{50}{2}-18\bigg)}{ 14}\times 5$

$\therefore$ Median

$=30.5+\dfrac {(7)}{14}\times 5$

$\therefore$ Median

$=33$

Find the value of, the missing frequency, If the mean of the following distribution is 67.
Class Interval
Frequency
25-35
10
35-45
6
45-55
4
55-65
f
65-75
4
75-85
12
85-95
26
Mean is 23.71
If true then enter $1$ and if false then enter $0$

$For\quad grouped\quad data\quad when\quad Class\quad Interval\quad \quad is\quad given,\quad we\quad find\quad the\quad Class\quad Mark.\\ Class\quad Mark\quad =\quad mid\quad point\quad of\quad an\quad interval\quad =\quad \dfrac { \left( Lower\quad limit\quad +\quad Upper\quad limit \right) }{ 2 } \\ Class\quad Mark\quad is\quad taken\quad as\quad { x }_{ i }\quad for\quad each\quad class\quad interval.\quad Find\quad { x }_{ i }{ f }_{ i }.\quad \\ From\quad the\quad table,\quad we\quad have\quad \Sigma { x }_{ i }{ f }_{ i }=4320+60f\quad \& \quad \quad \Sigma { f }_{ i }=62+f\\ Given\quad Mean\quad =\quad 67\\ Mean\quad =\quad \frac { \Sigma { x }_{ i }{ f }_{ i } }{ \Sigma { f }_{ i } } \quad \\ \Rightarrow \quad 67\quad =\quad \frac { 4320+60f }{ 62+f } \\ \Rightarrow \quad \left( 62+f \right) \quad \times \quad 67\quad =\quad 4320\quad +\quad 60f\\ \Rightarrow \quad 4154\quad +\quad 67f\quad =\quad 4320\quad +\quad 60f\\ \Rightarrow \quad 67f\quad -60f\quad =\quad 4320\quad -\quad 4154\\ \Rightarrow \quad 7f\quad =\quad 166\\ \Rightarrow \quad f\quad =\quad \frac { 166 }{ 7 } \\ \Rightarrow \quad f\quad =\quad 23.71$

Calculate the mode for the following frequency distribution.
Class
Frequency
0-10
5
10-20
8
20-30
7
30-40
12
40-50
28
50-60
20
60-70
10
70-80
10
mode is 4.67
If true then enter $1$ and if false then enter $0$

Range(Xi)	Freq(Fi)	Cum Freq
0 to 10	5	5
10 to 20	8	13
20 to 30	7	20
30 to 40	12	32
40 to 50	28	60	Modal range
50 to 60	20	80
60 to 70	10	90
70 to 80	10	100
Total	100

$Mode=\quad l\quad +\frac { \left( { f }_{ m }-{ f }_{ m-1 } \right) }{ \left( { f }_{ m }-{ f }_{ m-1 } \right) +\left( { f }_{ m }-{ f }_{ m+1 } \right) } \times \quad w\\ l\quad is\quad the\quad lower\quad class\quad boundary\quad of\quad the\quad modal\quad class\quad interval;\quad { f }_{ m-1 }\quad is\quad the\quad frequency\quad of\quad the\quad group\quad before\quad the\quad modal\quad class\quad interval.\\ { f }_{ m }\quad is\quad the\quad frequency\quad of\quad the\quad modal\quad class\quad interval;{ f }_{ m+1 }\quad is\quad the\quad frequency\quad of\quad the\quad group\quad after\quad the\quad modal\quad class\quad interval.\\ w\quad is\quad the\quad width\quad of\quad the\quad class\quad interval.\\ Here\quad modal\quad range\quad is\quad 40-50.\\ l=40,\quad { f }_{ m }=28,\quad { f }_{ m-1 }=12,{ f }_{ m+1 }=20,\quad w=10\\ Mode\quad =40+\frac { 28-12 }{ \left( 28-12 \right) +\left( 28-20 \right) } \times 10=40+\frac { 16\times 10 }{ 16+8 } =40+\frac { 160 }{ 24 } =40+\frac { 20 }{ 3 } =40+6.67=46.67$

Compute the mode of the following data:
Class Interval
Frequency
1-5
3
6-10
8
11-15
13
16-20
18
21-25
28
26-30
20
31-35
13
36-40
8
41-45
6
46-50
4
Mode $23.22$ .
If true then enter $1$ and if false then enter $0$

Class intervals	Freq(Fi)
1-5	3
6-10	8
11-15	13
16-20	18
21-25	28
26-30	20
31-35	13
36-40	8
41-45	6
46-50	4
Total	121

$Mode=\quad l\quad +\frac { \left( { f }_{ m }-{ f }_{ m-1 } \right) }{ \left( { f }_{ m }-{ f }_{ m-1 } \right) +\left( { f }_{ m }-{ f }_{ m+1 } \right) } \times \quad w\\ l\quad is\quad the\quad lower\quad class\quad boundary\quad of\quad the\quad modal\quad class\quad interval;\quad { f }_{ m-1 }\quad is\quad the\quad frequency\quad of\quad the\quad group\quad before\quad the\quad modal\quad class\quad interval.\\ { f }_{ m }\quad is\quad the\quad frequency\quad of\quad the\quad modal\quad class\quad interval;{ f }_{ m+1 }\quad is\quad the\quad frequency\quad of\quad the\quad group\quad after\quad the\quad modal\quad class\quad interval.\\ w\quad is\quad the\quad width\quad of\quad the\quad class\quad interval.\\ Here\quad modal\quad range\quad is\quad 21-25.\\ l=21,\quad { f }_{ m }=28,\quad { f }_{ m-1 }=18,{ f }_{ m+1 }=20,\quad w=4\\ Mode\quad =21+\frac { 28-18 }{ \left( 28-18 \right) +\left( 28-20 \right) } \times 4=21+\frac { 10\times 4 }{ 10+8 } =21+\frac { 40 }{ 18 } =21+2.22=23.22$

The following frequency distribution gives the monthly consumption of electricity of $68$ consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption (in units)
Number of consumers
65-85
4
85-105
5
105-125
13
125-145
20
145-165
14
165-185
8
185-205
4

Monthly consumption (in units)	Number of consumers $f_i$	Cumulative frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	8	64
185-205	4	68
Total	$n=68$

$n=68$ gives

$\frac {n}{2}=34$
So, we have the median class (125-145)

$l=125, n=68, f=20, cf=22, h=20$

$Median=l+\left \{\frac {\frac {n}{2}-cf}{f}\right \}\times h$

$=125+\left \{\frac {34-22}{\cdot 20}\right \}\times 20=137 units$ .

(ii) Modal class is (25-145) having maximum frequency

$f_m=20, f)1-13, f_2=14, l=125$ and

$h=20$

$Mode=l+\left \{\frac {f_m-f_1}{2f_m-f_1-f_2}\right \}\times h$

$=125+\left \{\frac {20-13}{40-13-14}\right \}\times 20=125+\frac {7\times 20}{13}$

$=125+\frac {140}{13}=125+10.76=135.76 units$

(iii)

$n=68, a=135, h=20$ and

$\sum f_iu_i=7$

Monthly consumption (in units)	Number of consumers	Class mark $x_i$	$u_i=\frac {x_i-135}{20}$	$f_i\times u_i$
65-85	4	75	-3	-12
85-105	5	95	-2	-10
105-125	13	115	-1	-13
125-145	20	135=a	0	0
145-165	14	155	1	14
165-185	8	175	2	16
185-205	4	195	3	12
Total	n=68			7

$n=68, a=135, h=20$ and

$\sum f_iu_i=7$
By step-deviation method.

$Mean=a+h\times \frac {1}{n}\times \sum f_iu_i=135+20\times \frac {1}{68}\times 7$

$=135+\frac {35}{17}=135+2.05=137.05 units$ .

The following number of goals were scored by a team in a series of $10$ matches:
$2$ , $3$ , $4$ , $5$ , $0$ , $1$ , $3$ , $3$ , $4$ , $3$
Find the mean, median and mode of these scores.

Hence,the data has

$10$ values. We arrange these values in the ascending order

$0,1,2,3,3,3,3,4,4,5$

(i)

$\text{Mean} = \dfrac { 0+1+2+3+3+3+3+4+4+5 }{ 10 } =\dfrac { 28 }{ 10 }= 2.8$

(ii) Here, the two middle most values are

$5th$ and

$6th$ each equal to

$3$ . Hence,

$\text{Median}=\dfrac { 3+3 }{ 2 } =3$

(iii) The value

$3$ has maximum frequency. Hence,

$\text{Mode} = 3$

The following table shows the ages of the patients admitted in a hospital during a year.
Age (in years)
No. of patients
5-15
6
15-25
11
25-35
21
35-45
23
45-55
14
55-65
5
Find the mode and the mean of the data given above. Compare and interpret the measures of central tendency.

Since the maximum frequency is

$23$ and it corresponds to the class

$35-45$ .

$\therefore\$ Modal class

$=35-45$

We know that,

Mode

$=l + \displaystyle\cfrac{f-f_1}{2f - f_1 - f_2}\times h$

Where,

$l\rightarrow$ lower limit of the modal class

$f\rightarrow$ frequency of the modal class

$f_1\rightarrow$ frequency of the class preceding the modal class

$f_2\rightarrow$ frequency of the class succeeding the modal class

$h\rightarrow$ class width

Here,

$l=35, h=10, f=23, f_{1}=21, f_{2}=14$

$\therefore \$ Mode

$=35+\left (\dfrac {23-21}{2(23)-21-14}\right )\times 10$

$=35+\dfrac {10(2)}{11}\\$

$=35+\dfrac {20}{11}\\$

$=35+1.82\\$

$=36.82$ years (approximately)

Again, we have:

Age (in years)	Number of patients $(f_i)$	Mid-value $(x_i)$	$f_ix_i$
$5-15$	$6$	$10$	$60$
$15-25$	$11$	$20$	$220$
$25-35$	$21$	$30$	$630$
$35-45$	$23$	$40$	$920$
$45-55$	$14$	$50$	$700$
$55-65$	$5$	$60$	$300$
	$\sum f_i=80$		$\sum f_ix_i=2830$

We know that, Mean,

$\overline x=\dfrac {\sum f_ix_i}{\sum f_i}$

$\therefore \ \overline{x}=\dfrac {2830}{80}\\$

$=\dfrac {283}{8}\\$

$=35.38$ years

Hence, for the given data, mean

$=35.38$ years and mode

$=36.82$ years.

This shows that maximum number of patients admitted in the hospital are of the age

$36.82$ years, while on an average the age of a patient admitted to the hospital is

$35.38$ years.

If the mode of the following distribution is given by $M$ , then find the value of a nearest integer.
Daily Wages 31 - 36 37 - 42 43 - 48 49 - 54 55 - 60 61 - 66
No.of workers 6 12 20 15 9 4

Daily Wages	No.of workers	Daily wages	No.of workers	$\displaystyle f_{i}u_{i}$
31 - 36	6	30.5 - 36.5	6	-15
37 - 42	12	36.5 - 42.5	12	-12
43 - 48	20	42.5 - 48.5	20	-8
49 - 54	15	48.5 - 54.5	15	0
55 - 60	9	54.5 - 60.6	4	6
61 - 66	4	60.5 - 66.5	4	6
		$\displaystyle \sum f_{i}=40$		$\displaystyle \sum f_{i}u_{i}=-23$

Modal class frequency is 42.5 - 48.5

$\displaystyle l=42.5,f_{1}=20,f_{o}=12,f_{2}=15,h=6$

$\displaystyle Mode=42.5+\frac{20-12}{2(20)-12-15}\times 6 \\\: \: \: \: \therefore Mode=46.2$

The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital
Age (in Years) 5 - 14 15 - 24 25 - 34 35 - 44 45 - 54 55 - 64
No. of Cases 6 11 21 23 14 5
Find the nearest integer to the modal age.

Here class intervals are not in inclusive form. So, first convert the above frequency table in inclusive form and it is given by

Age(in years)	$4.5-14.5$	$14.5-24.5$	$24.5-34.5$	$34.5-44.5$	$44.5-54.5$	$54.5-64.5$
No. of classes	$6$	$11$	$21$	$23$	$14$	$5$

Class

$34.5 - 44.5$ , has maximum frequency.

$\implies 34.5-44.5$ is the modal calss

$\therefore \displaystyle l=34.5,h=10,f_{o}=21,f_{1}=23\: and\: f_{2}=14$
Mode

$\displaystyle = l+\frac{f_{1}-f_{o}}{2f_{1}-f_{o}-f_{2}}\times h$

$\implies$ Mode

$\displaystyle =34.5+\frac{23-21}{46-21-14}\times 10=34.5+\frac{2}{11}\times 10=36.31$

Hence, the answer is

$36$ .

The daily wages (in rupees) of $100$ workers in a factory are given below
Daily wages (in Rs.) $125$ $130$ $135$ $140$ $145$ $150$ $160$ $180$
No. of workers $6$ $20$ $24$ $28$ $15$ $4$ $2$ $1$
Find the median wages of a worker (in the nearest integer value).

Daily wages (in Rs.)	No. of workes	Cumulative frequency
125	6	6
130	20	26
135	24	50
140	28	78
145	15	93
150	4	97
160	2	99
180	1	100

$= 100$ (even)

Median

$\displaystyle =\frac{\left ( \frac{N}{2} \right )^{th}observation+\left ( \frac{N}{2}+1 \right )^{th}observation}{2}$

Median

$\displaystyle =\frac{50^{th}observation+51^{st}observation}{2}=\frac{135+140}{2}=137.50$

$\therefore$ Median wages of a workers in the factory is Rs.

$137.50$ .

Calculate the median for the following distribution class
Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60
Frequency 5 10 20 7 8 5

(i) First we find

$\displaystyle \left ( \frac{N}{2} \right )^{th}$ value i.e.

$\displaystyle \left ( \frac{55}{2} \right )^{th}=27.5^{th}$
which lies in 20 - 30

$\therefore$ 20 - 30 class in median class

Class	f	c.f.
0 - 10	5	5
10 - 20	10	15
20 - 30	20	35
30 - 40	7	42
40 - 50	8	50
50 - 60	5	55

Here

$\displaystyle l=20,\frac{N}{2}=27.5,C=15,f=20,h=10$

$\therefore$ median

$\displaystyle =20+\frac{27.5-15}{20}\times 10$ Median = 26.25

If the median of the following frequency distribution is $46$ . find the absolute difference of missing frequencies
Variable $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Total
Frequency $12$ $30$ $f_{1}$ $65$ $f_{2}$ $25$ $18$ $229$

Class Interval	Frequency	C.F.
$10-20$	$12$	$12$
$20-30$	$30$	$42$
$30-40$	$\displaystyle f_{1}$	$\displaystyle 42+f_{1}$
$40-50$	$65$	$\displaystyle 107+f_{1}$
$50-60$	$\displaystyle f_{2}$	$\displaystyle 107+f_{1}+f_{2}$
$60-70$	$25$	$\displaystyle 132+f_{1}+f_{2}$
$70-80$	$18$	$\displaystyle 150+f_{1}+f_{2}$

Let the frequency of the class

$30-40$ be

$\displaystyle f_{1}$ and that of the class

$50-60$ be

$\displaystyle f_{2}$ . The total frequency is

$229$ .

$12 + 30 +$

$\displaystyle f_{1}$

$+ 65 +$

$\displaystyle f_{2}$

$+25 + 18 = 229$

$\displaystyle \Rightarrow f_{1}+f_{2}=79$
It is given that median is

$46$ . Clearly

$46$ lies in the class

$40-50$ . So,

$40-50$ is the median class

$\displaystyle \therefore L =40,h=10,f=65$ and

$C=42+f_{1},N=229$
Median

$=$

$\displaystyle L +\frac{\frac{N}{2}-C}{f}\times h$

$\displaystyle \Rightarrow 46=40+\frac{\frac{229}{2}-\left ( 42+f_{1} \right )}{65}\times 10$

$\displaystyle \Rightarrow 46=40+\cfrac{145-2f_{1}}{13}$

$\displaystyle \Rightarrow 6=\cfrac{145-2f_{1}}{13}\Rightarrow 2f_{1}=67$

$\displaystyle \Rightarrow f_{1}=33.5$ or

$34$
Since

$\displaystyle f_{1}+f_{2}=79$

$\displaystyle \Rightarrow f_{2}=45$
Hence

$\displaystyle f_{1}=34$ and

$\displaystyle f_{2}=45$

Hence, the difference between frequenncies is

$11$ .

Calculate the median from the following distribution
Class 5-10 10-15 15-20 20-25 25-30
Frequency 4 7 10 9 4

Class	Frequency	Cumulative Frequency
5-10	4
10-15	7	11
15-20	10	21 $\mbox{(Median Class)}$
20-25	9	30
25-30	4	34
	34

$N = 34, \space \displaystyle\frac{N}{2} = 17$

The cumulative frequency greater than

$17$ is

$21$ and corresponding class is

$15-20$
Thus,

$15-20$ is the median class, such that

$\quad l = 15, \displaystyle\frac{N}{2} = 17, c\bar{f} = 11, f = 10, h = 5$

Using,

$\quad Median = l +\left(\displaystyle\frac{\displaystyle\frac{N}{2} - c\bar{f}}{f}\right)\times h = 15 + \left(\displaystyle\frac{17-11}{10}\right) \times 5 = 15 + \displaystyle\frac{6\times5}{10}$

$\quad Median = 18$

Calculate the mean deviation about median age for the age distribution of $100$ persons given below :
Age Number
16-20 5
21-25 6
26-30 12
31-35 14
36-40 26
41-45 12
46-50 16
51-55 9

Clearly, we can see that given data is not continuous. So, to convert into continuous frequency distribution, we need to subtract

$0.5$ from the lower limit and add

$0.5$ to the upper limit of each class interval

Class	$f_i$	Cumulative frequency	$x_i$	$\|x_i-M\|$	$f_i\|x_i-M\|$
15.5-20.5	5	5	18	20	100
20.5-25.5	6	11	23	15	90
25.5-30.5	12	23	28	10	120
30.5-35.5	14	37	33	5	70
35.5-40.5	26	63	38	0	0
40.5-45.5	12	75	43	5	60
45.5-50.5	16	91	48	10	160
50.5-55.5	9	100	53	15	135
Total	100				735

Here,

$N=100$

$\cfrac{N}{2}=\cfrac{100}{2}=50$

which occurs in the cumulative frequency

$63$ .

Hence, the median class is

$35.5-40.5$ .

Median

$M=l+\cfrac{\cfrac{N}{2}-C_{f_{-1}}}{f_m} \times h$

$=35.5+\cfrac{50-37}{26} \times 5$

$=35.5+\cfrac{13}{26}\times 5$

$\Rightarrow M= 35.5+2.5=38$

Mean deviation about the median is

$\displaystyle M.D.(M)=\frac{1}{N}\sum_{i=1}^{8}f_{i}\mid x_{i}-M \mid$

$=\displaystyle\frac{1}{100}\times 735=7.35$

Find the mean deviation about the median for the data
$\displaystyle X_{i}$ 15 21 27 30 35
$\displaystyle f_{i}$ 3 5 6 7 8

$x_i$	$f_i$	Cumulative frequency	$\|x_i-M\|$	$f_i\|x_i-M\|$
15	3	3	15	45
21	5	8	9	45
27	6	14	3	18
30	7	21	0	0
35	8	29	5	40
Total	29			148

Here

$N=29$ (odd)

Median

$=\left(\cfrac{N+1}{2}\right)^{th}\text{observation}$

i.e.

$M=\left(\cfrac{29+1}{2}\right)^{th} \text{observation}$

$\Rightarrow M=15^{th}\text{observation}$

Since,

$15$ occurs in the cumulative frequency

$21$ . The corresponding observation is

$30$ .

Hence,

$M=30$

Mean deviation about median

$=\cfrac{\sum{f_i|x_i-M|}}{\sum f_i}$

$=\cfrac{148}{29}$

$=5.1$

Find the mean deviation about the median for the data
$\displaystyle x_{i}$ 5 7 9 10 12 15
$\displaystyle f_{i}$ 8 6 2 2 2 6

$x_i$	$f_i$	Cumulative frequency	$\|x_i-M\|$ $\implies \|x_i-7\|$	$f_i\|x_i-M\|$
5	8	8	2	16
7	6	14	0	0
9	2	16	2	4
10	2	18	3	6
12	2	20	5	10
15	6	26	8	48
Total	26			84

Here,

$N=26$ (even)

So, Median

$M=\cfrac{(N/2) ^{th}\text{observation}+(N/2+1)^{th}\text{observation}}{2}$

$M=\cfrac{13^{th}\text{observation}+14^{th}observation}{2}$

$\Rightarrow M=\cfrac{7+7}{2}=7$ (Here,

$13$ and

$14$ both lies in the cumulative frequency

$14$ , so corresponding observation for both is

$7$ )

Mean deviation about median

$=\cfrac{\sum{f_i|x_i-M|}}{\sum f_i}$

$=\cfrac{84}{26}$

$=3.23$

Find the mean deviation about median for the following data:
Marks Number of girls
0-10 6
10-20 8
20-30 14
30-40 16
40-50 4
50-60 2

Marks	$f_i$	Cumulative frequency	$x_i$	$\|x_i-M\|$	$f_i\|x_i-M\|$
0-10	6	6	5	22.86	137.16
10-20	8	14	15	12.86	102.88
20-30	14	28	25	2.86	40.04
30-40	16	44	35	7.14	114.24
40-50	4	48	45	17.14	68.56
50-60	2	50	55	27.14	54.28
Total	50				517.32

Here,

$N=50$

$\cfrac{N}{2}=\cfrac{50}{2}=25$

Since,

$25$ occurs in the cumulative frequency

$25$ .

So, median class is

$20-30$ .

Median

$M=l+\cfrac{\cfrac{N}{2}-C_{f-1}}{f_m}\times h$

$=20+\cfrac{25-14}{14}\times 10$

$M=20+\cfrac{110}{14}=27.86$

Mean deviation about median

$=\cfrac{\sum{f_i|x_i-M|}}{\sum f_i}$

$=\cfrac{517.32}{50}$

$=10.35$

Find the median for the following frequency distribution table:
Class-interval $0-5$ $5-10$ $10-15$ $15-20$ $20-25$ $25-30$
Frequency $5$ $3$ $9$ $10$ $8$ $5$

$CI$	0-5	5-10	10-15	15-20	20-25	25-30
$F$	5	3	9	10	8	5
$CF$	5	8	17	27	35	40

$\cfrac{N}{2}=\cfrac{40}{2}=20$ lies in CF $27$ so, $15-20$ is Median class

So, Median $=L+\left(\cfrac{(n/2)-Cf}{F}\right)\times H$

$=15+\left(\cfrac{20-17}{10}\right)\times 5$

Median $16.5$

Marks scored by Hari in $5$ test (out of $25$ marks) are given:
$24, 22, 23, 23, 25$ . Find his average score

We observe that there are

$5$ scores, so that

$\sum x = 24 + 22 + 23 + 23 + 25 = 117$ . Hence the mean is given by

$\overline {X} = \dfrac {\sum x}{N} = \dfrac {117}{5} = 23.4$ .

Find the median for the following frequency distribution table:
Class Intervals $110-119$ $120-129$ $130-139$ $140-149$ $150-159$ $160-169$
Frequency $6$ $8$ $15$ $10$ $6$ $5$

$Ci$	110-119	120-129	130-139	140-149	150-159	160-169
$f$	6	8	15	10	6	5
$cf$	6	14(m)	29	39	45	50

$\cfrac{N}{2}=\cfrac{50}{2}=25$ lies between $cf=14$ and $cf = 29$

so, $130-139$ is Median class and $m=14$

Median $=l+\left(\cfrac{(N/2)-m}{f}\right)\times h$

$=130+\left(\cfrac{25-14}{15}\right)\times 9$

Median $=136.59$

If the mean of the following frequency distribution is $7.2$ find value of $'K'$ .
$x$ $2$ $4$ $6$ $8$ $10$ $12$
$f$ $4$ $7$ $10$ $16$ $K$ $3$

$\bar x=\cfrac { \sum { Fx } }{ \sum { F } }=\cfrac{260+10K}{40+K}=7.2$

$260+10K=288+7.2K$

$2.8K=28$

$K=10.$

1019754_570587_ans_0e305c4ad902453f98748ed7731555ba.png

The heights of boys and girls of IX class of a school are given below.
Height (cm) $135$ $140$ $147$ $152$ $155$ $160$
Boys $2$ $5$ $12$ $10$ $7$ $1$
Girls $1$ $2$ $10$ $5$ $6$ $5$
Compare the average heights of the boys and girls.

Height (cm)	135	140	147	152	155	160
Boys	2	5	12	10	7	1
Girls	1	2	10	5	6	5

Total number of Boys

$=2+5+12+10+7+1=37$

Sum of heights of all boys

$=135\times2+140\times5+147\times12+152\times10+155\times7+160\times1=5499$

Average height of boys

$=\dfrac{5499}{37}=148.62\ cm$

Total number of Girls

$=1+2+10+5+6+5=29$

Sum of heights of all girls

$=135\times1+140\times2+147\times10+152\times5+155\times6+160\times5=4375$

Average height of girls

$=\dfrac{4375}{29}=150.86\ cm$

Hence,

The average height of Girls is

$2\ cm$ greater than the average height of Boys.

Find the mean for the given frequency distribution table.
Class - Interval Frequency
$0 - 4$ $3$
$5 - 9$ $5$
$10 - 14$ $7$
$15 - 19$ $4$
$20 - 24$ $6$
$N = 24$

To find the mean, first we have to find the mid-point of each class interval. Mid-point of each class interval. Mid-point of

$0 - 4 = (0 + 4)/2 = 2$ ; mid-point of

$5 - 9 = (5 + 9)/2 = 7$ and the like. Denote the mid-point of the class interval by x. Write down the frequencies f corresponding to each class interval.

Class interval	Mid-point of CI	(x) Frequency (f)	$fx$
$0 - 4$	$2$	$3$	$6$
$5 - 9$	$7$	$5$	$35$
$10 - 14$	$12$	$7$	$84$
$15 - 19$	$17$	$4$	$68$
$20 - 24$	$22$	$6$	$132$
		$N = 25$	$\sum fx = 325$

Multiply f and x to get

$fx$ . Add all

$fx$ and find out

$\sum fx$ . Now the mean is calculated using the formula.

$\overline {X} = \dfrac {\text {sum of all scores}}{\text {total number of scores}} = \dfrac {\sum fx}{N}$
Thus we get

$\overline {X} = \dfrac {\sum fx}{N} = \dfrac {325}{25} = 13$ .
Therefore mean is

$13$ .

Calculate the median for an exclusive (continuous) distribution given below.
Class Interval Frequency
$10 - 20$ $11$
$20 - 30$ $13$
$30 - 40$ $13$
$40 - 50$ $9$
$50 - 60$ $4$

We first prepare the cumulative frequency table (see above).

Class Interval	Frequency (f)	Cumulative frequency $(f_{c})$
$10 - 20$	$11$	$11$
$20 - 30$	$13$	$24$
$30 - 40$	$13$	$37$
$40 - 50$	$9$	$46$
$50 - 60$	$4$	$50$

Here we see that the total number of observation is

$N = 50$ . Therefore

$(30 - 40)$ is the median class. We also observe that

$LRL = 30, f_{c} = 24, f_{m} = 13$ and

$i = 20 - 10 = 10$ . Now we are in a position to use the formula for median:

$median = LRL = \dfrac {(\dfrac{N}{2}) - fx}{fm}\times i = 30 + \dfrac {(25 - 24)}{13} \times 10 = 30 + \dfrac {10}{13} = 30.77$ approximately.

Find the median for the following grouped data.
Class - Interval Frequency
$1 - 5$ $4$
$6 - 10$ $3$
$11 - 15$ $6$
$16 - 20$ $5$
$21 - 25$ $2$
$N = 20$

Median is the middle score. Here actual scores are not given. So we have to use to different procedure to find the median. We have

$N = 20$ , an even number. Therefore, there are two middle scores: one corresponding to

$N/2 = 20/2 = 10th$ score and the other is

$11th$ score.
We have to locate the median between

$10th$ and

$11th$ scores. Since we are not given individual scores, we have to get some idea about

$10th$ and

$11th$ scores. For this it is required to find the cumulative frequency.
Observe the following table, to know how to find cumulative frequency.

Class Interval	Frequency (f)	Cumulative frequency $(f_{c})$
$1 - 5$	$4$	$4$	$4$
$6 - 10$	$3$	$7$	$4 + 3 = 7$ $
$11 - 15$	$6$	$13$	$7 + 6 = 13$
$16 - 20$	$5$	$18$	$13 + 5 = 18$
$21 - 25$	$2$	$20$	$18 + 2 = 20$
	$N = 20$

Weights of parcels in a transport office are given below.
Weight (kg) $50$ $65$ $75$ $90$ $110$ $120$
No of parcels $25$ $34$ $38$ $40$ $47$ $16$
Find the mean weight of the parcels.

Weight

$(x_i)$ Number of parcels

$(f_i)$

$f_ix_i$

-------------------------------------------------------------------------

50 25 1250

65 34 2210

75 38 2850

90 40 3600

110 47 5170

120 16 1920

---------------------------------------------------------------

Total :

$\sum f_i= 200$

$\sum f_ix_i = 17000$

$\Rightarrow Mean = \dfrac{\sum f_ix_i}{\sum f_i}$

$\Rightarrow Mean =\dfrac{17000}{200}$

$\Rightarrow Mean = 85$

The mean weight of parcel is

$85\ Kg$

The maximum bowling speed $(km/hr)$ of $33$ players at a cricket coaching centre is given below:
Bowling Speed $(km/hr)$ $85-100$ $100-115$ $115-130$ $130-145$
Number of players $9$ $11$ $8$ $5$
Find the modal bowling speed (in $km/hr$ ) of players.

Here, maximum frequency is

$11$
The corresponding class namely

$100-115$ is modal class.
We know that
Mode

$=L+\left[ \cfrac { { f }_{ m }-{ f }_{ 1 } }{ 2{ f }_{ m }-{ f }_{ 1 }{ -f }_{ 2 } } \right] h$
Here

$L=100,{f}_{m}=11,{f}_{1}=9,{f}_{2}=8,h=15$
Mode

$=100+\left[ \cfrac { 11-9 }{ 2\times 11-9-8 } \right] \times 15$

$=100+\left[ \cfrac { 2 }{ 22-17 } \right] \times 15$

$=100+\cfrac { 2 }{ 5 } \times 15$

$=100+6=106$

$\therefore$ The modal bowling speed of player is

$106\ km/hr$ .

Find the mode and median of the following frequency distribution:
$x$ $10$ $11$ $12$ $13$ $14$ $15$
$f$ $1$ $4$ $7$ $5$ $9$ $8$

$x$	$10$	$11$	$12$	$13$	$14$	$15$
$f$	$1$	$4$	$7$	$5$	$9$	$8$

for a frequency distribution,

So, total number of observation

$=\left( 1+4+7+5+9+8 \right) =34$

It is a even number.

So, median is the value of average of values at

${ \left( \dfrac { n }{ 2 } \right) }^{ th }$ and

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

$\therefore$ Here, median is the average of the values at

${ \left( \dfrac { 34 }{ 2 } \right) }^{ th }$ or

${ 17 }^{ th }$ position and

${ \left( \dfrac { 34+2 }{ 2 } \right) }^{ th }$ or

${ 18 }^{ th }$ position.

$x$	$10$	$11$	$12$	$13$	$14$	$15$
$f$	$1$	$4$	$7$	$5$	$9$	$8$
$position$	$1$	$1+4=5$	$5+7=12$	$12+5=17$	$17+9=26$	$26+8=34$

So, value of

$x$ at

${ 17 }^{ th }$ position is

$13$ and at

${ 18 }^{ th }$ position is

$14$ .

$\therefore \quad Median\quad =\quad \left( \dfrac { 13+14 }{ 2 } \right) =\dfrac { 27 }{ 2 } =13.5$ .

Mode for a frequency distribution is the value which is the most common frequent value.

Here

$14$ has frequency

$9$ which is highest. So, mode for the frequency distribution is

$14$ .

For a certain frequency distribution, the value of Mean is $101$ and Median is $100$ . Find the value of Mode.

By empirical formula
Mode

$=3\times$ Median

$-2\times$ Mean

$\Rightarrow$ Mode

$=3\times100-2\times101$

$\Rightarrow$ Mode

$=98$

The weight of coffee (in $gms$ ) in $70$ packets is given in the following table.
Weight (in $gms$ ) No. of Packets
$200-201$ $12$
$201-202$ $26$
$202-203$ $20$
$203-204$ $9$
$204-205$ $2$
$205-206$ $1$
Hence determine the modal weight of coffee packet.

Weight(in $gms$ )	No. of Packets
$200-201$	$12$
$201-202$	$26$
$202-203$	$20$
$203-204$	$9$
$204-205$	$2$
$205-206$	$1$

Here the maximum frequency

$f_m = 26$

The corresponding class

$201-202$ is the modal class.

$L = 201, f_m = 26, f_1 = 12, f_2 = 20, h=1$

Mode

$=L + \bigg(\cfrac{f_m - f_1}{2f_m - f_1-f_2}\bigg)h$

$= 201 + \bigg(\cfrac{26-12}{2\times26-12-20}\bigg)1$

$= 201 + \bigg(\cfrac{14}{20}\bigg)$

$= 201 + \cfrac7{10}$

$= 201.7$

Modal weight of the coffee is

$201.7\ gms.$

The marks obtained by $30$ students in a class assessment of $5$ marks is given below :
Marks 0 1 2 3 4 5
Number of Students 1 3 6 10 5 5
Calculate the mean, median and mode of the above distribution.

$x_f$		$fx$	$cf$
0	1	0	1
1	3	3	4
2	6	12	10
3	10	30	20
4	5	20	25
5	5	25	30
	$\sum { f } =30$	$\sum { fx } =90$

Mean

$=\dfrac { \sum { fx } }{ \sum { f } } =\dfrac { 90 }{ 30 }$

$=3$
Median

$=$ size of

$\left( \dfrac { N }{ 2 } \right)$ th

${ obs }^{ n }$

$=$ size of

$\left( \dfrac { 30 }{ 2 } \right)$ th

${ obs }^{ n }$

$=$ size of

$15$ th

${ obs }^{ n }$

$=3$
Mode

$= 3$ marks (as highest frequency is

$10$ .)

(Use a graph paper for this question.) The daily pocket expenses of $200$ students in a school are given below:
Pocket expenses (in Rs) $0 - 5$ $5 - 10$ $10 - 15$ $15 - 20$ $20 - 25$ $25 - 30$ $30 - 35$ $35 - 40$
Number of students (frequency) $10$ $14$ $28$ $42$ $50$ $30$ $14$ $12$
Draw a histogram representing the above distribution and estimate the mode from the graph.

The histogram for the given data is constructed. The class

$20-25$ has the highest frequency, hence it is the modal class.

Connect the endpoints from the modal class as shown in the histogram. The intersection points of these lines represents the mode, which comes out to be around

$21.5$ .

The table shows the distribution of the scores obtained by $160$ shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution.
Scores No. of shooters
$0-10$ $9$
$10-20$ $13$
$20-30$ $20$
$30-40$ $26$
$40-50$ $30$
$50-60$ $22$
$60-70$ $15$
$70-80$ $10$
$80-90$ $8$
$90-100$ $7$
Use your graph to estimate the median.

Scores	No. of shooters	c.f.
$0-10$	$9$	$9$
$10-20$	$13$	$22$
$20-30$	$20$	$42$
$30-40$	$26$	$68$
$40-50$	$30$	$98$
$50-60$	$22$	$120$
$60-70$	$15$	$135$
$70-80$	$10$	$145$
$80-90$	$8$	$153$
$90-100$	$7$	$160$

Since, no. of terms,

$n = 160$ which is even

$\therefore Median = \left (\dfrac {n}{2}\right )^{th} term$

$= \left (\dfrac {160}{2}\right )^{th} term = 80th\ term$
From the graph,

$m_{e} = 43.5$

Age (in years) $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
No. of patients $60$ $42$ $55$ $70$ $53$ $20$
The following table shows ages of $300$ patients getting medical treatment in a hospital on a particular day. Find the median.

Classes (Age in years)	Frequency $(f_i)$ (No. of patients)	Cumulative frequency less than type
10-20	60	60
20-30	42	102 $\rightarrow$ c.f
30-40	55 $\rightarrow$ f	157
40-50	70	227
50-60	53	280
60-70	20	300
Total	300 $\rightarrow$ N

Here total frequency,

$\sum f_i = N = 300$

$\therefore \dfrac N2 = \cfrac{300}2 = 150$

Cumulative frequency, which is just greater than

$150$ is

$157$ .

Therefore, corresponding class

$30-40$ is the median class.

$L = 30, N = 300, c.f. 102, f=55, h=10$

Median

$= L+\bigg(\cfrac N2-c.f.\bigg)\dfrac hf$

$= 30 + (150 - 102)\cfrac{10}{55}$

$= 30 + 48\times\cfrac{10}{55}$

$= 30 + \cfrac{96}{11} = 30 + 8.73 = 38.73$

Hence, median

$= 38.73$ years.

Find the mode of the following data.
1) $74, 81, 62, 58, 77, 74$ .
2) $43, 36, 27, 25, 36, 66, 20, 25$ .
3) $55, 51, 62, 71, 50, 32$ .
4) $24, 20, 27, 32, 20, 28, 20$ .

The mode of a data set is the number that occurs most. To easily find the mode, put the numbers in order from least to greatest and count how many times each number occurs. The number that occurs most is the mode.

$1)$

$74,81,62,58,77,74$

Arrange the numbers in ascending order

$58,62,74,74,77,81$

In these the number

$74$ occurs two times then

$74$ is mode.

$2)$

$43,36,27,25,36,66,20,25$

Arrange the numbers in ascending order

$20,25,25,27,36,36,43,66$

In these, the numbers

$25$ and

$36$ occurs two times then

$25,36$ are modes.

$3)$

$55,51,62,71,50,32$

Arrange the numbers in ascending order

$32,50,51,55,62,71$

In these, all the numbers occurs once then it has no mode.

$4)$

$24,20,27,32,20,28,20$

Arrange the numbers in ascending order

$20,20,20,24,27,28,32$

In these, the number

$20$ occurs three times then

$20$ is mode.

Calculate the arithmetic mean by assumed mean method for the data given below.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
No. of students $5$ $10$ $25$ $30$ $20$ $10$

Let the assumed mean be

$A = 35$

Marks	Mid-value $\left( x \right)$	No. of students $\left( f \right)$	$d=x-35$	$fd$
$0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$	$5$ $15$ $25$ $35$ $45$ $55$	$5$ $10$ $25$ $30$ $20$ $10$	$-30$ $-20$ $-10$ $0$ $10$ $20$	$-150$ $-200$ $-250$ $0$ $200$ $200$
		$\sum { f } =100$		$\sum { fd } =-200$

$\overline { x } =A+\dfrac { \sum { fd } }{ \sum { f } }$

$=35+\left( \dfrac { -200 }{ 100 } \right) =35-2=33$

Find the mode for the following frequency table.
x $5$ $10$ $15$ $20$ $25$ $30$
f $14$ $25$ $37$ $16$ $8$ $5$

The score whose frequency occurs more is the mode of data

Given table

$x$	5	10	15	20	25	30
$f$	14	25	37	16	8	5

In this table,

$37$ is the greatest frequency where

$x=15$ .

Hence, mode is

$15$ .

Obtain the mean of the following data.
$x$ $5$ $10$ $15$ $20$ $25$
$f$ $3$ $10$ $25$ $7$ $5$

$x$	$f$	$fx$
$5$ $10$ $15$ $20$ $25$	$3$ $10$ $25$ $7$ $5$	$15$ $100$ $375$ $140$ $125$
Total	$\sum { f } =50$	$\sum { f\left( x \right) } =755$

Mean

$=\dfrac { \sum { f\left( x \right) } }{ \sum { f } } =\dfrac { 755 }{ 50 } =15.1$
Mean

$= 15.1$

Find the median for marks of $50$ students.
Marks $20$ $27$ $34$ $43$ $58$ $65$ $89$
Number of students $2$ $4$ $6$ $11$ $12$ $8$ $7$

Marks $(x)$	Number of students $(f)$	Cumulative frequency
$20$	$2$	$2$
$27$	$4$	$(2+4=)6$
$34$	$6$	$(6+6=)12$
$43$	$11$	$(11+12=)23$
$58$	$12$	$(23+12=)35$
$65$	$8$	$(35+8=)43$
$89$	$7$	$(43+7=)50$

Here, the total frequency,

$N=\sum f =50$

$\therefore \displaystyle\frac{N}{2}=\frac{50}{2}=25$ .
The median is

$\left(\dfrac{N}{2}\right)^{th}$ value

$=25$ th value.
Now,

$25$ th value occurs in the cummulative frequency

$35$ , whose corresponding marks is

$58$ .
Hence, the median

$=58$ .

From the following table compute arithmetic mean by direct method.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
No. of students $5$ $10$ $25$ $30$ $20$ $10$

Marks	Midpoint $\left( x \right)$	No. of students $\left( f \right)$	$fx$
$0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$	$5$ $15$ $25$ $35$ $45$ $55$	$5$ $10$ $25$ $30$ $20$ $10$	$25$ $150$ $625$ $1050$ $900$ $550$
		$\sum { f } =100$	$\sum { f\left( x \right) } =3300$

$\overline { x } =\dfrac { \sum { f\left( x \right) } }{ \sum { f } } =\dfrac { 3300 }{ 100 } =33$

$\therefore$ Mean

$=33$

The mean mark of $100$ students was found to be $40$ . Later on, it was found that a score of $53$ was misread as $83$ . Find the correct mean corresponding to the correct score.

Given that the total number of students

$n = 100$ ,

$\overline { x } = 40$ . So,
Incorrect

$\sum { x } = \overline { x } \times n = 40 \times 100 = 4000$
Correct

$\sum { x } =$ Incorrect

$\sum { x } -$ wrong item

$+$ correct item.

$= 4000 - 83 + 53 = 3970$
Correct

$\overline { x } =\dfrac { correct\sum { x } }{ n }$

$=\dfrac { 3970 }{ 100 } =39.7$
Hence the correct mean is

$39.7$ .

The hearts of $60$ patients were examined through X-ray and the observations obtained are given. Find the median.
Diameter of heart(in mm) $130$ $131$ $132$ $133$ $134$ $135$
Number of patients $7$ $9$ $15$ $12$ $6$ $11$

Diameter of heart (in mm)	Number of patients	$cf$
$130$	$7$	$7$
$131$	$9$	$16$
$132$	$15$	$31$
$133$	$12$	$43$
$134$	$6$	$49$
$135$	$11$	$60$

Total	$60$

Here, $N=60$ (even)

So, Median $M=\cfrac{(N/2) ^{th}\text{observation}+(N/2+1)^{th}\text{observation}}{2}$

$M=\cfrac{30^{th}\text{observation}+31^{th}observation}{2}$

$\Rightarrow M=\cfrac{132+132}{2}=132$ (Here, $30$ and $31$ both lies in the cumulative frequency $31$ , so corresponding observation for both is $132$ )

Hence, the median is $132$

The height (in cm) of $50$ students in a particular class are given. Find the median.
Height (in cm) $156$ $155$ $154$ $153$ $152$ $151$ $150$
Number of students $8$ $4$ $6$ $10$ $12$ $3$ $7$

Height in cm	Number of students	$cf$
156	8	8
155	4	12
154	6	18
153	10	28
152	12	40
151	3	43
150	7	50
Total	50

Here N=50 Then $\dfrac{N}{2}=\dfrac{50}{2}=25$
$\text{N=50 is even, hence there will be two observations}\,\dfrac{n}{2}\,and \,\dfrac{n}{2}+1\,i.e,25th\,and \,26th$

Now, frequency greater than $25$ is $28$ and height for $28$ is $153$

Hence, median is $153$ .

The salary of $43$ employees are given in the following table. Find the median.
Salary (in Rs) $4000$ $5500$ $6000$ $8250$ $10,000$ $17,000$ $25,000$
Number of employees $7$ $5$ $4$ $3$ $13$ $8$ $3$

Salary (in Rs)	Number of Employees	$cf$
4000	7	7
5500	5	12
6000	4	16
8250	3	19
10000	13	32
17000	8	40
25000	3	43
Total	43

Here $N=43,$ then $\dfrac{N}{2}=\dfrac{43}{2}=21.5$

Now, frequency greater than $21.5$ is $32$ and salary is $10000$ for $cf=32$

Hence, the median is $10000$ .

Calculate the mode of the following data.
Size of item $10-15$ $15-20$ $20-25$ $25-30$ $30-35$ $35-40$ $40-45$ $45-50$
No. of items $4$ $8$ $18$ $30$ $20$ $10$ $5$ $2$

Size of the item	$f$
$10-15$	$4$
$15-20$	$8$
$20-25$	$18$
$25-30$	$30$
$30-35$	$20$
$35-40$	$10$
$40-45$	$5$
$45-50$	$2$

Modal class is

$25-30$ since it has the maximum frequency.
Lower limit of the modal class

$l=25$
Frequency of the modal class

$f=30$
Frequency of the preceding modal class

${f}_{1}=18$
Frequency of the class reducing the modal class

${f}_{2}=20$
Class width

$c=5$
Mode

$=l+\left(\dfrac{f-{f}_{1}}{2f-{f}_{1}-{f}_{2}}\right) \times c$

$=25+\left(\dfrac{30-18}{60-18-20}\right)\times 5=25+\dfrac{12\times 5}{22}$

$=25+\dfrac{60}{22}=25+2.73=27.73$
Mode

$= 27.73$

Find the median for the following distribution.
Value $1$ $2$ $3$ $4$ $5$ $6$
$f$ $1$ $3$ $2$ $4$ $8$ $2$

The given data is:

Value	$f$	$cf$
$1$	$1$	$1$
$2$	$3$	$4$
$3$	$2$	$6$
$4$	$4$	$10$
$5$	$8$	$18$
$6$	$2$	$20$
	$n=20$

Here,

$n$ is even.

We know that, for even number of observations, the median is the mean of

$\left(\dfrac n2 \right)^{th}$ and

$\left(\dfrac n2 +1\right)^{th}$ observation.

Hence, the median here will be the mean of

$\left(\dfrac {20}{2} \right)^{th}$ and

$\left(\dfrac {20}{2} +1\right)^{th}$ observation.

That is, the

$10^{th}$ and

$11^{th}$ observation.
From the table shown above, we can see that the tenth observation is

$4$ and the eleventh observation is

$5$ .
Hence median

$=\dfrac { 4+5 }{ 2 }$

$=\dfrac { 9 }{ 2 }$

$=4.5$

Hence, the median of the given data is

$4.5$ .

A shoe shop in Chennai sold hundred pairs of shoes of a particular brand in a certain day with the following distribution.
Size of shoe $4$ $5$ $6$ $7$ $8$ $9$ $10$
No. of pairs sold $2$ $5$ $3$ $23$ $39$ $27$ $1$
Find the mode of the following distribution.

In the given data, more number of size

$8$ pair are sold as it has the highest frequency

$=39$

Mode of the given data is shoe of size

$8$ .

Therefore, modal shoe size

$=8$

Calculate the median for the following data.
Marks $20$ $9$ $25$ $50$ $40$ $80$
No. of students $6$ $4$ $16$ $7$ $8$ $2$

Let us arrange marks in ascending order.

Marks	$f$	$cf$
$9$	$4$	$4$
$20$	$6$	$10$
$25$	$16$	$26$
$40$	$8$	$34$
$50$	$7$	$41$
$80$	$2$	$43$
	$n=43$

Here,

$n=43$ , which is odd
Position of median

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

$={ \left( \dfrac { 43+1 }{ 2 } \right) }^{ th }$ observation.

$={ 22 }^{ nd }$ observation.
The above table shows that all items from

$11^{th}$ to

$26^{th}$ have their value

$25$ . So, the value of

${ 22 }^{ nd }$ item is

$25$ .

$\therefore$ Median

$= 25$ .

Find the median for the following distribution.
Wages (Rupees in hundreds) $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
No. of workers $22$ $38$ $46$ $35$ $20$

Wages	$f$	$cf$
$0-10$	$22$	$22$
$10-20$	$38$	$60$
$20-30$	$46$	$106$
$30-40$	$35$	$141$
$40-50$	$20$	$161$
	$N=161$

Here,

$\dfrac { N }{ 2 } =\dfrac { 161 }{ 2 } =80.5$ . Median class is

$20-30$ .
Lower limit of the median class

$l=20$
Frequency of the median class

$f=46$
Cumulative frequency of the class preceding the median class

$m=60$
Width of the class

$c=10$
Median

$=l+\dfrac { \dfrac { N }{ 2 } -m }{ f } \times c$

$=20+\dfrac { 80.5-60 }{ 46 } \times 10=20+\dfrac { 10 }{ 46 } \times 20.5$

$=20+\dfrac { 205 }{ 46 } =20+4.46=24.46$

$\therefore$ Median

$=24.46$

Calculate the mode of the following data.
Size of shoe $4$ $5$ $6$ $7$ $8$ $9$ $10$
No. of pairs sold $15$ $17$ $13$ $21$ $18$ $16$ $11$

From the given data, frequency of size

$7$ shoe

$=21$

Number of pairs of

$7$ size shoe are sold more

Therefore, mode of the given data

$=7$ size shoe pair

modal size

$=7$

The following table gives the distribution of the average weekly wages of $800$ workers in a factory. Calculate the median for the data given below.
Wages
( in hundreds) $20-25$ $25-30$ $30-35$ $35-40$ $40-45$ $45-50$ $50-55$ $55-60$
No. of persons $50$ $70$ $100$ $180$ $150$ $120$ $70$ $60$

Wages	No. of persons( ${ f }_{ i }$ )	Cumulative Frequency (c.f.)
20-25	50	50
25-30	70	120
30-35	100	220(c)
35-40	180(f)	400
40-45	150	550
45-50	120	670
50-55	70	740
55-60	60	800=n

$\cfrac { n }{ 2 } =400$

median $=l+\cfrac { h }{ f } \times \left( \cfrac { n }{ 2 } -c \right)$

$l=35;h=5;f=180;c=220$

median $=35+\cfrac { 5 }{ 180 } \times \left( 400-220 \right)$

$=40$

Therefore, median $=40$

Calculate the median for the following data
Class interval $1 - 5$ $6 - 10$ $11 - 15$ $16 - 20$ $21 - 25$ $26 - 30$ $31 - 35$
Frequency $1$ $18$ $25$ $26$ $7$ $2$ $1$

Class	Frequency ( ${ f }_{ i }$ )	Cumulative Frequency
0.5-5.5	1	1
5.5-10.5	18	19
10.5-15.5	25	44
15.5-20.5	26	70
20.5-25.5	7	77
25.5-30.5	2	79
30.5-35.5	1	80=n

$\cfrac { n }{ 2 } =40$

$l=10.5;h=5;f=25;c=19$

median $=l+\cfrac { h }{ f } \times \left( \cfrac { n }{ 2 } -c \right)$

$=10.5+\cfrac { 5 }{ 25 } \times \left( 40-19 \right) \\ =10.5+\cfrac { 11 }{ 5 } \\ =12.7$

Find the median for the following data.
Age group $0-9$ $10-19$ $20-29$ $30-39$ $40-49$ $50-59$ $60-69$
No. of persons $4$ $6$ $10$ $11$ $12$ $6$ $1$

Age	No. of persons( ${ f }_{ i }$ )	Cumulative Frequency(c.f)
-0.5-9.5	4	4
9.5-19.5	6	10
19.5-29.5	10	20
29.5-39.5	11	31
39.5-49.5	12	43
49.5-59.5	6	49
59.5-69.5	1	50=n

Median is at $\cfrac { n }{ 2 } =25$ th value

$l=29.5;h=10;f=11;c=20$

median $=l+\cfrac { h }{ f } \times \left( \cfrac { n }{ 2 } -c \right)$

$=29.5+\cfrac { 10 }{ 11 } \times (5)\\$

$\approx 34$ years

Find the median for the following data.
Marks $11-15$ $16-20$ $21-25$ $26-30$ $31-35$ $36-40$
Frequency $7$ $10$ $13$ $26$ $9$ $5$

Since the table is given in terms of inclusive type we convert it into exclusive type.

Marks	$f$	$cf$
$10.5-15.5$	$7$	$7$
$15.5-20.5$	$10$	$17$
$20.5-25.5$	$13$	$30$
$25.5-30.5$	$26$	$56$
$30.5-35.5$	$9$	$65$
$35.5-40.5$	$5$	$70$
	$N=70$

$N=70$ ,

$\dfrac { N }{ 2 } =\dfrac { 70 }{ 2 } =35$
Median class is

$25.5-30.5$
Lower limit of the median class

$l=25.5$
Frequency of the median class

$f=26$
Cumulative frequency of the preceding median class

$m=30$
Width of the median class

$c=30.5-25.5=5$
Median

$=l+\dfrac { \dfrac { N }{ 2 } -m }{ f } \times c$

$=25.5+\dfrac { 35-30 }{ 26 } \times 5=25.5+\dfrac { 25 }{ 26 } =26.46$

The following is the distribution of the size of certain farms from a taluka (tehasil). Find median size of farms.
Size of Farms(in acres)
Number of Farms
$5-15$
$7$
$15-25$
$12$
$25-35$
$17$
$35-45$
$25$
$45-55$
$31$
$55-65$
$5$
$65-75$
$3$

Size of Farms(in acres)	Number of Farms	Cumulative Frequency
$5-15$	$7$	$7$
$15-25$	$12$	$19$
$25-35$	$17$	$36$
$35-45$	$25$	$61$
$45-55$	$31$	$92$
$55-65$	$5$	$97$
$65-75$	$3$	$100$

$n=\sum f = 100$

$\cfrac n2 = \cfrac{100}2 = 50$

Therefore, median class

$=35-45$ ,

$h=10$

Frequency preceding to the median class

$cf=36$

Frequency of the median class

$f=25$

Median

$=L + \cfrac hf\left(\cfrac n2 - cf\right)$

$= 35 + \cfrac{10}{25}(50 - 36)$

$= 35 + 5.6$

$= 40.6$

Find the median of the following frequency distribution
Class $4-8$ $8-12$ $12-16$ $16-20$ $20-24$ $24-28$
Frequency $9$ $16$ $12$ $7$ $15$ $1$

Class	Frequency	Cumulative frequency $(c.f.)$
$4-8$	$9$	$9$
$8-12$	$16$	$9+16=25$
$12-16$	$12$	$25+12=37$
$16-20$	$7$	$37+7=44$
$20-24$	$15$	$44+15=59$
$24-28$	$1$	$59+1=60$

Here

$n=60\Rightarrow$

$\cfrac{n}{2}=30$
Cumulative frequency greater than

$30$ is

$37$ and the class corresponding to it is

$12-16$

$\therefore$ Class

$12-16$ is the median class

$\therefore$ Lower limit

$(l)=12,\cfrac{n}{2}=30,c.f.=25,f=12$ and

$h=4$
Subtituting the values, we get

$M=l+\cfrac { \left( \cfrac { n }{ 2 } -c.f. \right) }{ f } \times h$

$M=12+\cfrac { 30-25 }{ 12 } \times 4\quad =12+\cfrac { 5 }{ 3 }$

$=12+1.67$

$=13.67$

Hence, Median is

$13.67$

In Hostel, one day reading hours of $20$ students was observed, whose result is mentioned in the table below. Form the table, find the Mode.
No. of reading hours 1-3 3-5 5-7 7-9 9-11
Student's strength in the hostel $7$ $2$ $8$ $2$ $1$

Here the largest frequency is

$8$ and the class corresponding to it is

$5-7$ .
So, the modal class is

$5-7$
Now, according to the table

$c=$ class length

$=2$

$l=$ lower limit

$=5$

${f}_{1}=$ freq. of the modal class

$=8$

${f}_{0}=$ freq. of the preceding class

$=2$

${f}_{2}=$ freq. of the succeeding class

$=2$
Using the formula to find mode, and substituting the values,

$Z=l+\left( \cfrac { { f }_{ 1 }-{ f }_{ 0 } }{ 2{ f }_{ 1 }-{ f }_{ 0 }-{ f }_{ 2 } } \right) \times c$

$=5+\left( \cfrac { 8-2 }{ 2(8)-2-2 } \right) \times 2=5+\cfrac { 6 }{ 12 } \times 2=5+1=6$

Hence, Mode

$=6$ .

Find the median of the following frequency distribution:
Class 0-100 100-200 200-300 300-400 400-500 500-600
Frequency $64$ $62$ $84$ $72$ $66$ $52$

Here

$n=400$ .

$\therefore$

$\cfrac{n}{2}=200$

Class	Frequency	Cumulative frequency( $c.f.$ )
0-100	64	64
100-200	62	64+62=126
200-300	84	126+84=210
300-400	72	210+72=282
400-500	66	282+66=348
500-600	52	348+52=400

Cumulative frequency greater than

$200$ is

$210$ , corresponds to the class

$200-300$ .
Class

$200-300$ is the median class

$l=200,\cfrac{n}{2}=200,cf=126,f=84$ and

$c=100$
Substituting the values in

$M=l+\cfrac { \left( \cfrac { n }{ 2 } -cf \right) }{ f } \times c$

$=200+\cfrac { 200-126 }{ 84 } \times 100$

$=200+\cfrac { 74\times 100 }{ 84 }$

$=200+88.10$

$=288.10$

Find the mean, mode and median of marks obtained by $20$ students in an examination. The marks are given below.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
No. of students $1$ $4$ $5$ $8$ $2$

$Marks$	$Mid-value$ $x_i$	$No.\,of\,frequency$ $f_i$	$c.f$	$f_ix_i$
$0-10$	$5$	$1$	$1$	$5$
$10-20$	$15$	$4$	$5$	$60$
$20-30$	$25$	$5$	$10$	$125$
$30-40$	$35$	$8$	$18$	$280$
$40-50$	$45$	$2$	$20$	$90$
		$\sum f_i=20$		$\sum f_ix_i=560$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{560}{20}=28$

$\Rightarrow$ Modal class

$=30-40$

So,

$l=30,\,f_0=5,\,f_1=8,\,f_2=2$ and

$h=10$

$\Rightarrow$

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

$=30+\dfrac{8-5}{2\times 8-5-2}\times 10$

$=30+3.33$

$\therefore$

$Mode=33.33$

$\Rightarrow$ Here,

$n=20$ , then

$\dfrac{n}{2}=\dfrac{20}{2}=10$

$\therefore$ Median class

$=20-30$

So,

$c.f=5,\,f=5,\,l=20$ and

$h=10$

$\Rightarrow$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

$=20+\dfrac{10-5}{5}\times 10$

$=20+10$

$\therefore$

$Median=30$

Find the mode for the following frequency distribution.
Class $4-8$ $8-12$ $12-16$ $16-20$ $20-24$ $24-28$
Frequency $9$ $6$ $12$ $7$ $15$ $3$

Here the maximum class frequency is

$15$ and the class corresponding to this frequency is

$20-24$ .
So, the modal class is

$20-24$ .
The lower limit of modal class

$20-24$ is

$l = 20$ .
Class size c = 4; and frequency of the modal class

$f_1 = 15.$
Frequency of the class preceding the modal class

$= f_0 = 7$ .
Frequency of the class succeeding the modal class

$f_2 = 3$ .

$z = l + \left( \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \right ) \times c$

$= 20 + \left( \dfrac{15 - 7}{2 \times 15 - 7 - 3} \right ) \times 4$

$= 20 + \dfrac{8}{20} \times 4$

$= 21.6$

$\therefore$ The mode for the given data is

$21.6$

The age (in years) of $150$ patients getting medical treatment in a hospital in a month are given below. Obtain its mode.
Age (yrs) $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
No. of patients $12$ $14$ $36$ $50$ $20$ $18$

Highest frequency=highest number of patients

$=50$

Therefore, modal class

$=40-50$

$l=40;h=10;{ f }_{ m }=50;{ f }_{ m-1 }=36;{ f }_{ m+1 }=20$

mode

$=L+\cfrac { { f }_{ m }-{ f }_{ m-1 } }{ 2{ f }_{ m }-{ f }_{ m-1 }-{ f }_{ m+1 } } \times n$

$=40+\cfrac { 14 }{ 14+30 } \times 10$

$=43.18$

$\approx 43.2$ years

For the following data obtain the mode.
Weight (in kg) $21-25$ $26-30$ $31-35$ $36-40$ $41-45$ $46-50$ $51-55$ $56-60$
No. of students $5$ $4$ $3$ $18$ $20$ $14$ $8$ $3$

Highest frequency=highest number of students in a class

$=20$

Given intervals are closed intervals. To make them open intervals, subtract

$0.5$ from lower boundary and add the same in the upper boundary.

Now, modal class

$=40.5-45.5$

$l=40.5;h=5;{ f }_{ m }=20;{ f }_{ m-1 }=18;{ f }_{ m+1 }=14$

mode

$=L+\cfrac { { f }_{ m }-{ f }_{ m-1 } }{ 2{ f }_{ m }-{ f }_{ m-1 }-{ f }_{ m+1 } } \times n$

$=40.5+\cfrac { 2 }{ 2+6 } \times 5$

$\\ =41.75$ kg

The cumulative frequency "more than type" is given in the following table.
Find the median of the following data.
Wages $\geq 0$ $\geq 50$ $\geq 100$ $\geq 150$ $\geq 200$ $\geq 250$
Cumulative Frequency 90 80 60 30 15 0

Wages	Frequency	Cumulative Frequency( $c.f$ )
$\geq 0$	$10$	$90$
$\geq 50$	$20$	$80$
$\geq 100$	$30$	$60$
$\geq 150$	$15$	$30$
$\geq 200$	$15$	$15$
$\geq 250$	$0$	$0$

$\dfrac{n}{2} = \dfrac{90}{2} = 45$ on the x-axis.
Therefore,

$\dfrac{100+150}{2}=125$ is the median.

Find the Median from the following table-
Class Interval $0-20$ $20-40$ $40-60$ $60-80$ $80-100$
Frequency $10$ $17$ $26$ $22$ $15$

Class Interval	Frequency $(f)$	Cumulative Frequency $(C.F.)$
$0-20$	$10$	$10$
$20-40$	$17$	$27$
$40-60$	$26$	$53$
$60-80$	$22$	$75$
$80-100$	$15$	$90$
	$\sum f = N = 90$

Here

$N = 90$

$Mid-point = \dfrac {N}{2} = \dfrac {90}{2} = 45$

$45th$ term is in the cumulative frequency

$53$ . Median class is

$40-60$ .
here

$l_{1} = 40$ ,

$l_{2} = 60$ ,

$f = 26$ ,

$C = 27$

$Md = l_{1} + \dfrac {\left (\dfrac {N}{2} - C\right )}{f}(l_{2} - l_{1})$

$= 40 + \left (\dfrac {\dfrac {90}{2} - 27}{26}\right ) (60 - 40)$

$= 40 + \dfrac {(45 - 27)}{26}\times 20$

$= 40 + \dfrac {18}{26} \times 20$

$= 40 + \dfrac {180}{26}$

$= 40 + 13.84$

$= 53.84$
Thus, the median

$= 53.84$ .

The table below shows the number of workers doing various jobs in a factory and their daily wages.
Daily wages (Rupees) Number of workers
225 4
250 7
270 9
300 5
350 3
400 2
What is the mean daily wage?

Daily wages (Rupees) $x$	Number of workers $f$	$fx$
225	4	900
250	7	1750
270	9	2430
300	5	1500
350	3	1050
400	2	800
Total	$\sum { f } =30$	$\sum { fx } =8430$

Mean daily wage

$=\cfrac { \sum { fx } }{ \sum { x } } =\cfrac { 8430 }{ 30 } =281$
Thus, mean daily wage of workers is Rs.

$281$ .

The table below shows the classification of people, participated in a medical camp, according to their weights.
Weight (in kilogram) Number of people
$20-30$ $16$
$30-40$ $21$
$40-50$ $28$
$50-60$ $24$
$60-70$ $11$
Calculate the mean weight.

Weight (in kilogram)	Number of people	Class mark	Total weight
$20 - 30$	$16$	$25$	$16\times 25 = 400$
$30 - 40$	$21$	$35$	$21\times 35 = 735$
$40 - 50$	$28$	$45$	$28\times 45 = 1260$
$50 - 60$	$24$	$55$	$24\times 55 = 1320$
$60 - 70$	$11$	$65$	$11\times 65 = 715$
Total	$100$		$4430$

$Mean\ weight = \dfrac {\text {Total weight}}{\text {Total number of people}} = \dfrac {4430}{100} = 44.3\ kg$ .

Find the mode of the given data:

Class 20-29 30-39 40-49 50-59 60-69
Frequency 15 20 50 30 10

Here the maximum class frequency is

$50$ and the class corresponding to this frequency is

$40-49$
So, the modal class is

$40-49$
The lower limit of modal class

$40-49$ is

$l = 40$
Class size

$c = 10$
Frequency of the modal class

$f_1 = 50$
Frequency of the class preceding the modal class

$f_0 = 20$
Frequency of the class succeeding the modal class

$f_2 = 30$

$\Rightarrow z = l + \left( \dfrac{f_1 - f_0}{2 f_1 - f_0 - f_2} \right ) \times c$

$= 40 + \left( \dfrac{50 - 20}{2 \times 50 - 20 - 30} \right ) \times 10$

$= 40 + \dfrac{30}{50} \times 10$

$= 40 + 6$

$= 46$

$\therefore$ The mode for the given data is 46

The table below shows the number of students in the Maths Club of a school, classified according to their heights.
Height (cm) Number of students
110-120 4
120-1330 24
130-140 20
140-150 32
150-160 20
Calculate the mean height.

Height (cm) C.I.	Number of students f	Mid value x	fx
110-120	4	115	460
120-130	24	125	3000
130-140	20	135	2700
140-150	32	145	4640
150-160	20	155	3100
Total	$\sum$ f = 100		$\sum$ fx = 13900

Mean height

$= \dfrac{\sum fx}{\sum x} = \dfrac{13900}{100} = 139$
Thus, the mean height of students in maths club is 139 cm.

The scores obtained by $50$ students in an examination is tabulated as shown below.
Score Number of students
below $10$ $3$
below $20$ $7$
below $30$ $13$
below $40$ $22$
below $50$ $32$
below $60$ $40$
below $70$ $46$
below $80$ $50$
Find the median score.

Score	Number of Students
Below $10$	$3$
Below $20$	$7$
Below $30$	$13$
Below $40$	$22$
Below $50$	$32$
Below $60$	$40$
Below $70$	$46$
Below $80$	$50$

Here median score is the score

$x$ for which the number of students is

$\dfrac{50}{2}=25$
The number of students

$y = 25$ occurs between the number of students

$y = 22$ having score

$x = 40$ and the number of students

$y = 32$ having score

$x = 50$ in the table.
So using proportionality assumption,

$\dfrac{x-40}{50-40}=\dfrac{25-22}{32-22}$

$x-40=3$

$x = 43$
This means that the median score is

$43$ .

Find the mode of the following frequency table:
Class Interval Frequency
$140-150$ $4$
$150-160$ $6$
$160-170$ $10$
$170-180$ $12$
$180-190$ $9$
$190-200$ $3$

Since the class,

$170-180$ has the maximum frequency.
Therefore class

$170-180$ is the modal class.
Here

${ l }_{ 1 }=170,h=180-170=10,{ f }_{ 1 }=12,{ f }_{ 0 }=10,{ f }_{ 2 }=9$

$\because$ Mode

$Z={ l }_{ 1 }+\left[ \cfrac { { f }_{ 1 }-{ f }_{ 0 } }{ 2{ f }_{ 1 }-{ f }_{ 0 }-{ f }_{ 2 } } \right] \times h$

$\Rightarrow Z=170+\left[ \cfrac { 12-10 }{ 2\times 12-10-9 } \right] \times 10$

$=170+\cfrac { 2\times 10 }{ 24-9 }$

$=170+4$

$=174$

Following are marks obtained in Mathematics by students of Class X.
Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60 60-70
No. of students $2$ $10$ $12$ $15$ $13$ $8$ $4$
Find the median of the marks obtained in Mathematics.

$C.I.$	$f$	$cf$
0 - 10	2	2
10 - 20	10	12
20 - 30	12	24
30 - 40	15	39
40- 50	13	52
50 - 60	8	60
60 - 70	4	64

Above frequency distribution table can be formed from the given data.

To find median, we need to find the median class first,

Total no of students

$(N)=64$

$\Rightarrow \dfrac{N}{2}=\dfrac{64}{2}=32$

From the

$cf$ column of the table, we can see that

$32$ is more than

$24$ and less than

$39$ .

So, median class will be

$30-40$

Now, we know that

$Median= l+\dfrac{\left(\dfrac{N}{2}-C\right)}{f}\times h$

where,

$l=$ lower limit of median class

$=30$ ,

$C=$ cumulative frequency of class preceding the median class

$=24$ ,

$f=$ frequency of the median class

$=15$ ,

$h=$ class size

$=10$

$\therefore \ Median=30+\dfrac{32-24}{15}\times 10$

$=30+\dfrac{8}{15}\times 10$

$=30+5.33=35.33$

Hence, the median marks obtained by the students are

$35.33$ .

If median of the following frequency distribution is 35 then find the value of x.
Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 2 3 x 6 5 3 2

Class interval	frequency	cumulitative frequency
$0-10$	$2$	$2$
$10-20$	$3$	$5$
$20-30$	$x$	$5+x$
$30-40$	$6$	$11+x$
$40-50$	$5$	$16+x$
$50-60$	$3$	$19+x$
$60-70$	$2$	$21+x$

Total frequency=

$21+x$

=> N=

$21+x$

=>

$\dfrac{N}{2}=\dfrac{21+x}{2}$

So; CF corresponding to this is

$11+x$

And class corresponds to this is

$30-40$

So,

$l=30$ ,

$F=5+x$ ,

$f=6$ and

$h=10$

$Median=l+\left( \dfrac{\dfrac{N}{2}-F}{f} \right)\times h$

=>

$35=30+\left( \dfrac{\dfrac{21+x}{2}-5-x}{6} \right)\times 10$

=>

$x=5$

Find mode for given data:
Class 20-29 30-39 40-49 50-59 60-69
Frequency 15 20 50 30 10

Class	20 - 29	30 - 39	40 - 49	50 - 59	60 - 69
Class Boundaries	19.5 - 29.5	29.5 - 39.5	39.5 - 49.5	49.5 - 59.5	59.5 - 69.5
Frequency	15	20	50	30	10

For the given data, Modal Class(Class with highest frequency)

$=39.5-49.5$

For the modal class

$=$ lower limit of modal class

$=39.5$

$=$ size of class interval

$=10$

$f_1=$ frequency of modal class

$=50$

$f_0=$ frequency of class preceding modal class

$=20$

$f_2=$ frequency of class succeeding modal class

$=30$

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

$Mode=39.5+\dfrac{50-20}{(2\times 50)-20-30}\times 10$

$Mode=39.5+\dfrac{30}{50}\times 10$

$Mode=45.5$

The table below shows the daily expenditure on food of $25$ households in a locality.
Daily expenditure (in Rs.) 100-150 150-200 200-250 250-300 300-350
Number of households: $4$ $5$ $12$ $2$ $2$
Find the mean daily expenditure on food by a suitable method.

Answer:-

Daily expenditure	No. of households ${f}_{i}$	${x}_{i}$	${f}_{i} {x}_{i}$
100-150	4	125	500
150-200	5	175	875
200-250	12	225	2700
250-300	2	275	550
300-350	2	325	650
	$\Sigma{{f}_{i}} = 25$		$\Sigma{{f}_{i} {x}_{i}} = 5275$

$\therefore \text{mean daily expenditure on food} = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{5275}{25} = 211$

The marks obtained by $50$ students of class $10$ out of $80$ marks are given in the following frequency distribution. Find the median.
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Frequency $2$ $5$ $8$ $16$ $9$ $5$ $3$ $2$

Class	$0 -10$	$10 - 20$	$20 -30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 -70$	$70 - 80$
Frequency	$2$	$5$	$8$	$16$	$9$	$5$	$3$	$2$
Cumulative Frequency	$2$	$7$	$15$	$31$	$40$	$45$	$48$	$50$

Total number of students

$=n=50$

$\dfrac{n}{2}=\dfrac{50}{2}=25$ ,

$25 < 31$ ,

$30-40$ is the median class.

$l$ is the lower limit of median class, so

$l=30$

$h$ is the size of class interval, so

$h = 40 - 30 = 10$

$f$ is the frequency of median class, so

$f = 16$

$F$ is the cumulative frequency of class preceding the median class, so

$F=15$

Median

$= l + \left (\dfrac{\dfrac{n}{2}-F}{f} \right )\times h$

Median

$= 30 + \left (\dfrac{\dfrac{50}{2}-15}{16} \right )\times 10$

Median

$= 30 + \left (\dfrac{10}{16} \right )\times 10$

Median

$=30 + \left (\dfrac{100}{16} \right )$

Median

$=36.25$

Hence, the median is

$=36.25$

Calculate the median of the following frequency distribution.
$class$ $f$
$5-10$ $49$
$10-15$ $133$
$15-20$ $63$
$20-25$ $15$
$25-30$ $6$
$30-35$ $7$
$35-40$ $4$
$40-45$ $2$
$45-50$ $1$

2150652_829900_ans_3e28fbd1296d42618ceca5af90fdf9f6.jpg

Compute the median for each of the following data:
(i)Marks No. of students (ii)Marks No. of students
Less than $10$ $0$ More than $150$ $0$
Less than $30$ $10$ More than $140$ $12$
Less than $50$ $25$ More than $130$ $27$
Less than $70$ $43$ More than $120$ $60$
Less than $90$ $65$ More than $110$ $105$
Less than $110$ $87$ More than $100$ $124$
Less than $130$ $96$ More than $90$ $141$
Less than $150$ $100$ More than $80$ $150$

$( i )$

Class interval	Frequency	Cumulative frequency
$0-10$	$0$	$0$
$10-30$	$10$	$10$
$30-50$	$15$	$25$
$50-70$	$18$	$43$
$70-90$	$22$	$65$
$90-110$	$22$	$87$
$110-130$	$9$	$96$
$130-150$	$4$	$100$
	$N=100$

We have,

$N=100,\,\dfrac{N}{2}=50$

The cumulative frequency just greater than

$\dfrac{N}{2}$ is

$65$ then median class is

$70-90$ such that,

$l=70,\,h=90-70=20,\,f=22,\,cf=43$

$Median=l+\dfrac{\dfrac{N}{2}-cf}{f}\times h$

$\Rightarrow$

$Median=70+\dfrac{50-43}{22}\times 20$

$\Rightarrow$

$Median=70+\dfrac{7\times 20}{22}$

$\Rightarrow$

$Median=70+6.36$

$\therefore$

$Median=76.36$

$(ii)$

Class interval	Frequency	Cumulative Frequency
$80-90$	$9$	$9$
$90-100$	$17$	$26$
$100-110$	$19$	$45$
$110-120$	$45$	$90$
$120-130$	$33$	$123$
$130-140$	$15$	$138$
$140-150$	$12$	$150$
$150-160$	$0$	$150$
	$N=150$

We have,

$N=150,$ then,

$\dfrac{N}{2}=\dfrac{150}{2}=75$

The cumulative frequency just more than

$\dfrac{N}{2}$ is

$90$ then the median class is

$110-120$ such that

$l=70,\,h=120-110=10,\,f=45,\,cf=45$

$Median=l+\dfrac{\dfrac{N}{2}-cf}{f}\times h$

$\Rightarrow$

$Median=110+\dfrac{75-45}{45}\times 10$

$\Rightarrow$

$Median=110+\dfrac{30\times 10}{45}$

$\Rightarrow$

$Median=110+6.67$

$\therefore$

$Median=116.67$

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is $24$ .
Age in years 0-10 10-20 20-30 30-40 40-50
No. of persons $5$ $25$ ? $18$ $7$

Let the missing frequency be

$x$ .

Age in years (x)	No. of persons (f)	C.F.
0-10	5	5
10-20	25	30
20-30	$x$	30 $+ x$
30-40	18	48 $+ x$
40-50	7	55 $+ x$
	$N = 55 + x$

As,

$N = 55 + x$

$\Rightarrow \cfrac{N}{2} = \cfrac{55 + x}{2}$

$Median = l + \cfrac{\cfrac{N}{2} + C}{f} \times h$

$median$ is given

$24$ , therefore median class is

$20-30.$

Therefore,

$l =$ lower limit of median class

$= 20$

$C = C.F.$ of the class preceding the median class

$= 30$

$h =$ higher limit - lower limit

$= 40 - 30 = 10$

$f =$ frequency of median class

$= x$

$\therefore \; median = 20 + \cfrac{\cfrac{55 + x}{2} - 30}{x} \times 10$

$24 = 20 + \cfrac{x - 5}{2x} \times 10$

$4 = \cfrac{x - 5}{x} \times 5$

$\Rightarrow 4x = 5x - 25$

$\Rightarrow x = 25$

Hence, the missing frequency is

$25.$

The following table gives the literacy rate (in percentage) of $35$ cities.
Class interval $45-55$ $55-65$ $65-75$ $75-85$ $85-95$
No. of cities $3$ $10$ $11$ $8$ $3$
Find the mean literacy rate.

Answer:-

Marks	No. of students ${f}_{i}$	${x}_{i}$	${f}_{i} {x}_{i}$
45-55	3	50	150
55-65	10	60	600
65-75	11	70	770
75-85	8	80	640
85-95	3	90	270
	$\Sigma{{f}_{i}} = 35$		$\Sigma{{f}_{i} {x}_{i}} = 2430$

We know that, arithmetic mean

$=\dfrac{\Sigma{f_i x_i}}{\Sigma{f_i}}$

$\therefore \text{Mean literacy rate} = \dfrac{2430}{35} = 69.43 \%$

The following is the distribution of height of students of a certain class in a certain city:
Height (in cms): 160-162 163-165 166-168 169-171 172-174
No. of students: $15$ $118$ $142$ $127$ $18$
Find the median height.

Class Interval (Exclusive)	Class Interval (Inclusive)	Class Interval Frequency	Cumulative Frequency
$160 - 162$	$159.5-162.5$	$15$	$15$
$163-165$	$162.5-165.5$	$118$	$133(F)$
$166-168$	$165.5-168.5$	$142(f)$	$275$
$169-171$	$168.5-171.5$	$127$	$402$
$172-174$	$171.5-174.5$	$18$	$420$
		$N = 420$

We have,

$N=420$ .

Then,

$\dfrac{N}{2}=\dfrac{420}{2}=210$

The cumulative frequency just greater than

$\dfrac{N}{2}$ is

$275$ then

$165.5-168.5$ is the median class such that,

$L=165.5,\,f=142,\,F=133$ and

$h=168.5-165.5=3$

$\Rightarrow$ Median

$= L + \left (\dfrac{\dfrac{N}{2}-F}{f} \right ) \times h$

Median

$= 165.5 + \left (\dfrac{210-133}{142} \right ) \times h$

Median

$= 165.5 + 1.63$

$\therefore$ Median

$= 167.13$

Following is the distribution of I.Q. of $100$ students. Find the median I.Q.
I.Q. $55-64$ $65-74$ $75-84$ $85-94$ $95-104$
$105-114$ $115-124$ $125-134$ $135-144$
No. of
students $1$ $2$ $9$ $22$ $33$ $22$ $8$ $2$ $1$

Class interval ( Exclusive )	Class interval ( inclusive )	Class interval frequency	Cumulative Frequency
$55-64$	$54.5-64.5$	$1$	$1$
$65-74$	$64.5-74.5$	$2$	$3$
$75-84$	$74.5-84.5$	$9$	$12$
$85-94$	$84.5-94.5$	$22$	$34$ $( F )$
$95-104$	$94.5-104.5$	$33$ $( f )$	$67$
$105-114$	$104.5-114.5$	$22$	$89$
$115-124$	$114.5-124.5$	$8$	$97$
$125-134$	$124.5-134.5$	$2$	$99$
$135-144$	$134.5-144.5$	$1$	$100$
		$N=100$

We have

$N=100$

$\dfrac{N}{2}=\dfrac{100}{2}=50$

The cumulative frequency just greater than

$\dfrac{N}{2}$ is

$67$ than the median class is

$94.5-104.5$ such that,

$L=94.5,\,F=33,\,h=104.5-94.5=10$

$\Rightarrow$

$Median=L+\dfrac{\dfrac{N}{2}-F}{f}\times h$

$\Rightarrow$

$Median=94.5+\dfrac{50-34}{33}\times 10=94.5+4.88=99.5$

$\therefore$

$Median=99.4$

To find out the concentration of ${SO}_{2}$ in the air (in parts per million, i.e., $ppm$ ), the data was collected for $30$ localities in certain city and is presented below:
Concentration of ${SO}_{2}$ (in ppm) Frequency
$0.00-0.04$ $4$
$0.04-0.08$ $9$
$0.08-0.12$ $9$
$0.12-0.16$ $2$
$0.16-0.20$ $4$
$0.20-0.24$ $2$
Find the mean concentration of ${SO}_{2}$ in the air.

Answer:-

Concentration of $S{O}_{2}$ (in ppm)	${x}_{i}$	Frequency $({f}_{i})$	${f}_{i} {x}_{i}$
0.00-0.04	0.02	4	0.08
0.04-0.08	0.06	9	0.54
0.08-0.12	0.10	9	0.90
0.12-0.16	0.14	2	0.28
0.16-0.20	0.18	4	0.72
0.20-0.24	0.22	2	0.44
		$\Sigma{f}_{i} = 30$	$\Sigma{{f}_{i} {x}_{i}} = 2.96$

$\text{mean concentration of } {S}{O}_{2} \text{ in the air} = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{2.96}{30} = 0.0986 \approx 0.099$

Calculate the median from the following data:
Rent In Rs. 15-25 25-35 35-45 45-55 55-65 65-75 75-85 85-95
No. of Houses $8$ $10$ $15$ $25$ $40$ $20$ $15$ $7$

Rent (in Rs.) (x)	No. of houses (f)	C.F.
15-25	8	8
25-35	10	18
35-45	15	33
45-55	25	58
55-65	40	98
65-75	20	118
75-85	15	133
85-95	7	140
	N = 140

As,

$N = 140 \Rightarrow \cfrac{N}{2} = \cfrac{140}{2} = 70$

$98$ is just greater than

$70$ , therefore median class is

$55 - 65$ .

$\text{Median} = l + \cfrac{\cfrac{N}{2} + C}{f} \times h$

Here,

$l =$ lower limit of median class

$= 55$

$C = C.F.$ of the class preceding the median class

$= 58$

$h =$ higher limit - lower limit

$= 65 - 55 = 10$

$f =$ frequency of median class

$= 40$

$\therefore \; \text{Median} = 55 + \cfrac{70 - 58}{40} \times 10 = 55 + 3 = 58$

A class teacher has the following absentee record of $40$ students of a class for the whole term. Find the mean number of days a student was absent.
Number of days: 0-6 6-10 10-14 14-20 20-28 28-38 38-40
Number of students: $11$ $10$ $7$ $4$ $4$ $3$ $1$

Answer:-

No. of days	No. of student ${f}_{i}$	${x}_{i}$	${f}_{i} {x}_{i}$
0-6	11	3	33
6 - 10	10	8	80
10 - 14	7	12	84
14 - 20	4	17	68
20 - 28	4	24	96
28 - 38	3	33	99
38 - 40	1	39	39
	$\Sigma {f}_{i} = 40$		$\Sigma {f}_{i} {x}_{i} = 499$

The mean number of days a student was absent

$= \cfrac{\Sigma{f}_{i} {x}_{i}}{\Sigma{f}_{i}} = \cfrac{499}{40} = 12.475$

Hence, the mean number of days a student was absent is

$12.475.$

The following table gives the frequency distribution of married women by age at marriage:
Age (in years) Frequency Age (in years) Frequency
$15-19$ $53$ $40-44$ $9$
$20-24$ $140$ $45-49$ $5$
$25-29$ $98$ $50-54$ $3$
$30-34$ $32$ $55-59$ $3$
$35-39$ $12$ $60$ and above $2$
Calculate the median and interpret the results.

$Age$	$Frequency$	$Cumulative\,frequency$
$14.5-19.5$	$53$	$53$
$19.5-24.5$	$140$	$193$
$24.5-29.5$	$98$	$291$
$29.5-34.5$	$32$	$323$
$34.5-39.5$	$12$	$335$
$39.5-44.5$	$9$	$344$
$44.5-49.5$	$5$	$349$
$49.5-54.5$	$3$	$352$
$54.5-59.5$	$3$	$355$
$59.5\,and\,above$	$2$	$357$
	$N=357$

$\Rightarrow$

$N=357$ then,

$\dfrac{N}{2}=\dfrac{357}{2}=178.5$

The cumulative frequency just greater than

$\dfrac{N}{2}$ is

$193$ , then the median class is

$20-24$ such that

$l=19.5,\,f=140,\,F=53,\,h=24.5-19.5=5$ .

$\Rightarrow$

$Median=l+\dfrac{\dfrac{N}{2}-F}{f}\times h$

$\Rightarrow$

$Median=19.5+\dfrac{178.5-53}{140}\times 5$

$\Rightarrow$

$Median=23.98$

Nearly half the women were married between the ages of

$15$ and

$25$ .

Calculate the median from the following data:
Marks below: 10 20 30 40 50 60 70 80
No. of students $15$ $35$ $60$ $84$ $96$ $127$ $198$ $250$

Marks (x)	No. of students (f)	C.F.
0-10	15	15
10-20	20	35
20-30	25	60
30-40	24	84
40-50	12	96
50-60	31	127
60-70	71	198
70-80	52	250
	N = 250

As,

$N = 250 \Rightarrow \cfrac{N}{2} = \cfrac{250}{2} = 125$

$127$ is just greater than

$125,$ therefore median class is

$50-60$ .

$Median = l + \cfrac{\cfrac{N}{2} + C}{f} \times h$

Here,

$l =$ lower limit of median class

$= 50$

$C = C.F.$ of the class preceding the median class

$= 96$

$h =$ higher limit - lower limit

$= 60 - 50 = 10$

$f =$ frequency of median class

$= 31$

$\therefore \; median = 50 + \cfrac{125 - 96}{31} \times 10 = 50 + 9.35 = 59.35$

For the following grouped frequency distribution find the mode:
Class: $3-6$ $6-9$ $9-12$ $12-15$ $15-18$ $18-21$ $21-24$
Frequency: $2$ $5$ $10$ $23$ $21$ $12$ $3$

We observe that the class

$12-15$ has a maximum frequency. Therefore, this is the modal class.

We have,

$l= 12, h= 3, f_1= 23, f_0= 10$ and

$f_2= 21$

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

$\Rightarrow \text{Mode}= 12 +\dfrac{23-10}{46-10-21}\times 3$

$\Rightarrow \text{Mode}= 12 +\dfrac{13}{15}\times 3$

$\Rightarrow \text{Mode}= 12+\dfrac{13}{5}$

$\Rightarrow \text{Mode}= 14.6$

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Runs scored Number of batsman Runs scored Number of batsman
$3000-4000$ $4$ $7000-8000$ $6$
$4000-5000$ $18$ $8000-9000$ $3$
$5000-6000$ $9$ $9000-10000$ $1$
$6000-7000$ $7$ $10000-11000$ $1$
find the mode of the data.

From the given data we may observe that maximum class frequency is

$18$ belonging to class interval

$4000 – 5000.$

So, modal class

$= 4000 – 5000$

Lower limit

$(l)$ of modal class

$= 4000$

Frequency

$(f_1)$ of modal class

$= 18$

Frequency

$(f_0)$ of class preceding modal class

$= 4$

Frequency

$(f_2)$ of class succeeding modal class

$= 9$

Class size

$(h) = 1000$

$\Rightarrow$

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

$\Rightarrow$

$Mode=4000+\dfrac{18-4}{2\times 18-4-9}\times 1000$

$\Rightarrow$

$Mode=4000+\dfrac{14000}{23}$

$\Rightarrow$

$Mode=4000+608.7=4608.7\,runs$

Compute the value of mode for the following frequency distribution.
Class: 100-110 110-120 120-130 130-140 140-150 150-160 160-170
Frequency: 4 6 20 32 33 8 2

2153039_1010137_ans_d3c5801a80dd43c18df47209483b65cb.jpg

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

$x_i$

$f_i$

$f_ix_i$

$2$

$3$

$6$

$4$

$2$

$8$

$6$

$3$

$18$

$10$

$1$

$10$


	$N= \sum f_i= 11$	$\sum f_ix_i= 2(p+ 5)$

We have,

$N = \sum f_i= 11, \sum f_ix_i= 2p+ 52$

$\therefore$ Mean =

$\dfrac{\sum f_ix_i}{N}= 6 \dfrac{2p+ 52}{11}\Rightarrow 66= 2p + 52 \Rightarrow 2p = 14 \Rightarrow p= 7$

A student noted the number of cars passing through a spot on a road for $100$ periods each of $3$ minutes and summarised it in the table given below. Find the mode of the data
Number of cars 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency $7$ $14$ $13$ $12$ $20$ $11$ $15$ $8$

From the given data, it can be observed that the maximum class frequency is

$20$ , belongings to

$40-50$ class intervals.

We know,

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

$l =$ lower limit of the modal class

$h =$ size of the class intervals

$f =$ frequency of the modal class

$f_1 =$ frequency of the class preceding the modal class

$f_2 =$ frequency of the class succeed in the modal class.

$\Rightarrow$ Modal class

$=40-50$ .

$\Rightarrow$

$l=40,\,f=20,\,f_1=12,\,f_2=11,\,h=10$

$\Rightarrow$

$Mode=40+\dfrac{20-12}{2\times 20-12-11}\times 10$

$\Rightarrow$

$Mode=40+\dfrac{80}{17}$

$\Rightarrow$

$Mode=40+4.71=44.7\,cars$

${1, 2, 3, 6, 8}$ is a set of five positive integers whose mean is 4 and median isWrite down two other sets of five positive integers, each having the same mean and median as this set.

We need 5 positive integers with same median, which means our 1st three numbers will be 1,2,3.
now mean should be 4
i.e

$\dfrac{(n_1+n_2+n_3+n_4+n_5)}{5} =4$

$=\dfrac{1+2+3+n_4+n_5}{5} = 4$ as

$6+n_4+n_5=20$

$n_4+n_5=14$
So, we will take one pair as 5,9 and other as 4and10

$\therefore$

$S={(1, 2, 3, 5, 9)}$

$={(1, 2, 3, 4, 10)}$

Compute the mode for the following frequency distribution:
Size of items: 0-4 4-8 8-12 12-16 16-20 20-24 24-28 28-32 32-36 36-40
Frequency: 5 7 9 17 12 10 6 3 1 0

Here, the maximum frequency is 17 and the corresponding class is

$12-16$ . So,

$12-16$ is the modal class.
We have,

$l= 12, h= 4, f= 17, f_1= 9$ and

$f_2= 12$

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

$\Rightarrow Mode = 12 + \dfrac{17-9}{34- 9 - 12}\times 4$

$\Rightarrow Mode = 12+ \dfrac{8}{13}\times 4=12 + \dfrac{32}{13}= 12 + 2.4= 14.4$

The table below
shows the daily expenditure on food of 30 households in a locality

Daily expenditure (in Rs.)

$100 -150$

$150 - 200$

$200 - 250$

$250 - 300$

$300 - 350$

No. of households

$6$

$7$

$12$

$3$

$2$

Find the mean and the median daily expenditure on food.

$C.I.$	$Mid-value$ $x_i$	$f_i$	$c.f.$	$x_if_i$
$100-150$	$125$	$6$	$6$	$750$
$150-200$	$175$	$7$	$13$	$1225$
$200-250$	$225$	$12$	$25$	$2700$
$250-300$	$275$	$3$	$28$	$825$
$300-350$	$325$	$2$	$30$	$650$
		$\sum f_i=30$		$\sum f_ix_i=6150$

We know that,

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}$

$\therefore \ \text{mean expenditure on food}=\dfrac{6150}{30}=205$

We also know that,

$Median=l+\dfrac{\dfrac{n}{2}-c.f}{f}\times h$

$n=\sum f_i=30$ , then

$\dfrac{n}{2}=\dfrac{30}{2}=15$

$\therefore$ Median class

$=200-250$

Then,

$l=200,\,c.f=13,\,f=12$ and

$h=50$

$\therefore \ \text{median expenditure on food}=200+\dfrac{15-13}{12}\times 50$

$=200+8.33$

$\therefore$

$Median=208.33$

Hence, the mean and median expenditure on food are

$205$ and

$208.33$ , respectively.

Calculate the median from the following data:
Marks: 0-10 10-30 30-60 60-80 80-90
No. of students: 5 15 30 8 2

Here, the class intervals are of unequal width. If the class intervals are of unequal width the frequencies need not be adjusted to make the class intervals equal.
Calculation of Median

Marks	No. of students (frequency)	Cumulative frequency
0-10	5	5
10-30	15	20
30-60	30	50
60-80	8	58
80-90	2	60
		$N=\sum f_i= 60$

Here,

$N= 60 \therefore N/2 = 30$
The cumulative frequency just greater than

$N/2 = 30$ is

$50$ and the corresponding class is

$30-60$ . Hence,

$30-60$ is the median class.

$\therefore l= 30 , f= 30, F= 20, h= 30$
Now, Median

$= l +\frac{\frac{N}{2}-F}{f}\times h$

$\Rightarrow Median =30+\dfrac{30-20}{30}\times 30 = 40$

Find mode for the following data.
Employees salary $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
No. of employees $10$ $12$ $18$ $8$ $9$ $5$ $6$ $4$

$Class$	$Frequency$
$0-10$	$10$
$10-20$	$12$
$20-30$	$18$
$30-40$	$8$
$40-50$	$9$
$50-60$	$5$
$60-70$	$6$
$70-80$	$4$

Here the maximum frequency is

$18$ then the corresponding class

$20-30$ is the modal class.

$l=20,\,h=30-20=10,\,f=18,\,f_1=12,\,f_2=8$

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

$=20+\dfrac{18-12}{2(18)-12-8}\times 10$

$=20+\dfrac{6}{16}\times 10$

$=20+\dfrac{15}{4}$

$=20+3.75$

$\therefore$

$Mode=23.75$

Weight of $40$ eggs were recorded as given below
Weight in grams $85-90$ $90-95$ $95-100$ $100-105$ $105-110$
Number of eggs $4$ $12$ $14$ $8$ $2$
Find the modal weight.

In the given frequency distribution,

$h= 5$

Class

$95-100$ has maximum frequency i.e,

$14$

Therefore,

$95-100$ is the model class in which

$l=95$ and

$f_{1} = 14$

Here,

$f_{0} = 12$ and

$f_{2} = 8$

Mode

$= l + \left(\dfrac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times h$

$= 95 + \left(\dfrac{14 - 12}{28 - 12 - 8} \right) \times 5$

$= 95 + \dfrac{10}{8}$

$= 95 + 1.25$

$= 96.25$

The model weight is equal to

$96.25$ grams.

The height (in cm) of the volly ball players from term $A$ and $B$ were recorded as
Team A: $180,178,176,181,190,175,187$
,Team B: $174,175,190,179,178,185,177.$
Which team had the greater average height?
Find the mediam of team $A$ and Team $B$

$Team\,A:$

Arranging team

$A$ members height in ascending order.

$\Rightarrow$

$175,176,178,180,181,187,190$

Number of players

$=7$

$\Rightarrow$ Average height of team

$A$

$=\dfrac{175+176+178+180+181+187+190}{7}=\dfrac{1267}{7}=181$

Number of players are in odd numbers.

$\therefore$

$Median=\left(\dfrac{n+1}{2}\right)^{th}player=4^{th}player$

$\therefore$

$Median=180$

$Team\,B:$

Arranging team

$A$ players height in ascending order.

$\Rightarrow$

$174,175,177,178,179,185,190$

Number of players

$=7$

$\Rightarrow$ Average height of team

$B=\dfrac{174+175+177+178+179+185+190}{7}=\dfrac{1258}{7}=179.71$

Number of players are in odd numbers.

$\therefore$

$Median=\left(\dfrac{n+1}{2}\right)^{th}player=4^{th}player$

$\therefore$

$Median=178$

$\therefore$ We can see team

$A$ players has greater average height.

Find the mean, median and mode of the given data
No. of Centuries 5 10 15 20 25
No. of Cricketers 56 23 39 13 8

mean is the average

$mean=\dfrac{5*56+10*23+15*39+20*13+25*8}{56+23+39+13+8}$

$=11.18$

median is the data in the middle when placed in ascending order

already data is in ascending order and the middle number would be seventieth number that is 10

$median =10$

mode is the most frequently occured number

there fore

$mode=5$

Find the mode for the following data: ( $4$ and $5$ )
Class $0-7$ $7-14$ $14-21$ $21-28$ $28-35$ $35-42$ $42-49$ $49-56$
Area $26$ $31$ $35$ $42$ $82$ $71$ $54$ $19$

$\Rightarrow$ Class interval

$28-35$ have higher frequency.

$\Rightarrow$ Modal class

$=28-35$

$\Rightarrow$

$l=28,\,f_0=42,\,f_1=82,\,f_2=71$ and

$h=7$

$\Rightarrow$

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

$=28+\dfrac{82-42}{2\times 82-42-72-}\times 7$

$=28+\dfrac{40}{50}\times 7$

$=28+5.6$

$=33.6$

Write the formula to find median from grouped frequency distribution

estimated median formula for grouped frequency distribution is

$L+(\dfrac { \dfrac n2-B }{ G } \times W)$

where

L is lower class boundary of the group containing the median

n is the total number of values

B is the cumulative frequency of the groups before the median group

G is the frequency of the median group

W is the group width

Weekly income of $600$ families is given below. Find the median.
Income in $Rs$ . $0-1000$ $1000-2000$ $2000-3000$ $3000-4000$ $4000-5000$ $5000-6000$
Frequency $250$ $190$ $100$ $40$ $15$ $5$

$CI$	0-1000	1000-2000	2000-3000	3000-4000	4000-5000	5000-6000
$f.i$	250	190	100	40	45	5
$c.fi$	250	440	540	580	595	600

$N=600$

Since,

$\frac{N}{2} = \frac{{600}}{2}$

$\,\,\,\,\,\, = 300$

We know that ,

$l=1000, i=1000, F=250, f=190$

$Median = l + \frac{{\frac{N}{2} - F}}{f} \times i$

$\,\,\,\,\,\,\, = 1000 + \frac{{\frac{{600}}{2} - 250}}{{190}} \times 1000$

$\begin{matrix} \, \, \, \, \, \, \, \, =1000+\frac { { 50 } }{ { 190 } } \times 1000 \\ \, \, \, \, \, \, \, \, =1000+\frac { { 5000 } }{ { 19 } } \\ \, \, \, \, \, \, \, \, =\frac { { 19000+5000 } }{ { 19 } } \\ \, \, \, \, \, \, \, =\frac { { 24000 } }{ { 19 } } \\ \end{matrix}$

Hence,

$Median = \frac{{24000}}{{19}}=1263.15$

Class Interval Frequency
$1 - 5$ $3$
$6 - 10$ $8$
$11 - 15$ $13$
$16 - 20$ $18$
$21 - 25$ $28$
$26 - 30$ $20$
$31 - 35$ $13$
$36 - 40$ $8$
$41 - 45$ $6$
$46 - 50$ $4$
Compute the mode of the given data.

$Class\ Interval$	$Frequency$
$0.5-5.5$	$3$
$5.5-10.5$	$8$
$10.5-15.5$	$13$
$15.5-20.5$	$18(f_0)$
$20.5-25.5$	$28(f_1)$
$25.5-30.5$	$20(f_2)$
$30.5-35.5$	$13$
$35.5-40.5$	$8$
$40.5-45.5$	$6$
$45.5-50.5$	$4$

Modal class

$=20.5-25.5$ with highest frequency.

Then we have,

$l=20.5,f_0=18,f_1=28,f_2=20,h=5$

$Mode=l+\dfrac{f_1-f_0}{2f_1-f_0-_f2}\times h$

$=20.5+\dfrac{28-18}{2\times 28-18-20}\times 5$

$=20.5+0.55$

$=21.05$

The following data gives the distribution of total monthly household expenditures of $200$ families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

$Expenditure$ (In Rs) Number of families
$1000-1500$ $24$
$1500-2000$ $40$
$2000-2500$ $33$
$2500-3000$ $28$
$3000-3500$ $30$
$3500-4000$ $22$
$4000-4500$ $16$
$4500-5000$ $7$

Here, the maximum class frequency is

$40$ and the class corresponding to it is

$1500-2000$ .

So, the modal class

$=1500-2000$ .

Thus, lower limit of the modal class

$l=1500$

Frequency of the modal class,

$\,f_1=40$

Frequency of the class preceding the modal class,

$\,f_o=24$

Frequency of the class succeeding the modal class,

$\,f_2=33$

and, class size

$h=2000-1500=500$

Now, substituting these values in the formula of mode, we get

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

$Mode=1500+\dfrac{40-24}{2\times 40-24-33} \times 500$

$Mode=1500+\dfrac{16}{80-57}\times 500$

$Mode=1500+\dfrac{8000}{23}=1500+347.83$

$Mode=1847.83$

Hence, the modal monthly expenditure of the families is

$Rs.1847.83$ .

Finding Mean:

$Expenditure$ $(in\,Rs)$	$Number\,of\,families$ $(f_i)$	$Class\,mark$ $(x_i)$	$d_i=x_i-3250$	$u_i=\dfrac{d_i}{500}$	$f_iu_i$
$1000-1500$	$24$	$1250$	$-2000$	$-4$	$-96$
$1500-2000$	$40$	$1750$	$-1500$	$-3$	$-120$
$2000-2500$	$33$	$2250$	$-1000$	$-2$	$-66$
$2500-3000$	$28$	$2750$	$-500$	$-1$	$-28$
$3000-3500$	$30$	$3250$	$0$	$0$	$0$
$3500-4000$	$22$	$3750$	$500$	$1$	$22$
$4000-4500$	$16$	$4250$	$1000$	$2$	$32$
$4500-5000$	$7$	$4750$	$1500$	$3$	$21$
$Total$	$\sum f_i=200$				$\sum f_iu_i=-235$

Here, we have

$\sum f_i=200,\,\sum f_iu_i=-235,\,h=500$ and Assumed mean

$A=3250$ .

Now,

$\overline{x}=A+\left(\dfrac{\sum f_iu_1}{\sum f_i}\right)\times h$

$\Rightarrow$

$\overline{x}=3250+\left(\dfrac{-235}{200}\right)\times 500$

$\Rightarrow$

$\overline{x}=3250-\dfrac{235\times 5}{2}=3250-\dfrac{1175}{2}$

$\Rightarrow$

$\overline{x}=3250-587.50=2662.50$

Hence, the mean monthly expenditure is

$Rs.2662.50$

The lengths of $40$ leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
Length (in mm)
Number of leaves
$118-126$
$127-135$
$136-144$
$145-153$
$154-162$
$163-171$
$172-180$
$3$
$5$
$9$
$12$
$5$
$4$
$2$
Find the median length of the leaves.

$CI$	$Frequency$	$c.f$
$117.5-126.5$	$3$	$3$
$126.5-135.5$	$5$	$8$
$135.5-144.5$	$9$	$17$
$144.5-153.5$	$12$	$29$
$153.5-162.5$	$5$	$34$
$162.5-171.5$	$4$	$38$
$171.5-180.5$	$2$	$40$
	$N=40$

$\dfrac{N}{2}=\dfrac{40}{2}=20$

$\therefore$ Median class

$=144.5-153.5$

$\Rightarrow$

$l=144.5,\,c.f=17,\,f=12$ and

$h=9$

$\Rightarrow$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

$=144.5+\dfrac{20-17}{12}\times 9$

$=144.5+\dfrac{3}{12}\times 9$

$=144.5+2.25$

$=146.75$

A student noted the number of cars passing through a spot on a road for $100$ periods each of $2$ minutes and complied data in the following table :
Number of cars $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Number of period $7$ $x$ $13$ $15$ $y$ $10$ $14$ $8$
Find the missing frequencies $x$ and $y$ , if the mode of the data is given to be $44$ .

Given:

Mode of the given data is

$44$

Number of periods is

$100$

$\implies 7+x+13+15+y+10+14+8=100$

$\implies x+y=33$

So,the modal class is

$40-50$

$L$ (the modal class lower limit)

$=40$

$f_1$ (frequency of the modal class)

$=y$

$f_0$ (frequency of the preceding modal class)

$=15$

$f_2$ (frequency of the succeeding modal class)

$=10$

h(class size)

$=10$

Mode

$=L + \Bigg(\cfrac{f_1-f_0}{2f_1-f_0-f_2}\Bigg)\times h$

$\implies 40 + \left(\cfrac{y-15}{2y-15-10}\right)\times 10 = 44$

$\implies \dfrac{10y-150}{2y-25}=4$

$\implies 10y-150=8y-100$

$\implies 2y=50$

$\implies y=25$

$\because x+y=33$

$\implies x+25=33$

$\implies x=8$

Hence

$x=8$ and

$y=25$

The age group and number of persons, who donated blood in a blood duration camp is given. Find the mode of ages of persons.
Age group(years) $20-25$ $25-30$ $30-35$ $35-40$
No. of persons $60$ $80$ $35$ $25$

The maximum class frequency =

$80$

The class corresponding to this frequency is

$25-30$

So,the modal class is

$25-30$

L(the modal class lower limit) =

$25$

$f_1$ (frequency of the modal class) =

$80$

$f_0$ (frequency of the preceding modal class) =

$60$

$f_2$ (frequency of the succeeding modal class) =

$35$

h(class size) =

$5$

Mode =

$L + \Bigg(\cfrac{f_1-f_0}{2f_1-f_0-f_2}\Bigg)\times h$

$\implies 25 + \cfrac{20}{13} = 25+1.5384 = 26.5834$

Find the mode of the following distribution table:
Class Interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80
Frequency 7 14 13 12 20 11 15 8

Formula for finding mode is given by,

Mode =

$\displaystyle L+\frac{f_{m}-f_{m-1}}{(f_{m}-f_{m-1})+(f_{m}-f_{m+1})}\times w$

Where, L= lower class boundary of modal class

(Model class = class with highest frequency)

$f_{m-1}=$ frequency of class before modal class

$f_{m}=$ frequency of modal class

$f_{m+1}=$ frequency of after modal class

w = group width

Here,

0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
7	14	13	12	20	11	15	8

Hence, Mode =

$\displaystyle 40+\frac{20-12}{(20-12)+(20-11)}\times 10$

$\displaystyle= 40+\frac{8}{8+9}\times 10$

$\displaystyle= 40+\frac{80}{17}= 40(1+\frac{2}{17})$

$\displaystyle= \frac{40\times 19}{17}= \frac{760}{17}$

Mode =

$\dfrac{760}{17}$

1055360_1174293_ans_cff31dad6e734dbca75c1e3bf5534d6b.png

Find the mode of the following frequency distribution of marks obtained by $50$ students.
Marks obtained $0 - 10$ $10 - 20$ $20 - 30$ $30 - 40$ $40 - 50$
No. of students $5$ $12$ $20$ $10$ $3$

$\Rightarrow$ Here, maximum frequency is

$20$ .

$\therefore$ Modal class

$=20-30$

So,

$l=20,\,f_0=12,\,f_1=20,\,f_2=10$ and

$h=10$

$\Rightarrow$

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

$=20+\dfrac{20-12}{2\times 20-12-10}\times 10$

$=20+\dfrac{8}{18}\times 10$

$=20+4.44$

$=24.44$

Calculate mean deviation about Median for the following data :
Class 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 6 7 15 16 4 2

$Class$	$Frequency$ $f_i$	$c.f$	$Class-mark$ $x_i$	$\|x_i-M\|$ $(x_i-28)$	$f_i\|x_i-M\|$
$0-10$	$6$	$6$	$5$	$23$	$138$
$10-20$	$7$	$13$	$15$	$13$	$91$
$20-30$	$15$	$28$	$25$	$3$	$45$
$30-40$	$16$	$44$	$35$	$7$	$112$
$40-50$	$4$	$48$	$45$	$17$	$68$
$50-60$	$2$	$50$	$55$	$27$	$54$
	$\sum f_i=50$				$\sum f_i\|x_i-M\|=508$

$n=50$ then,

$\dfrac{n}{2}=\dfrac{50}{2}=25$

$\therefore$ Median class

$=20-30$

We have,

$l=20,\,c.f=14,\,f=14$ and

$h=10$

$\Rightarrow$

$Median(M)=l+\dfrac{\dfrac{n}{2}-c.f}{f}\times h$

$=20+\dfrac{25-13}{15}\times 10$

$=20+8$

$=28$

$\Rightarrow$ Mean deviation about median

$=\dfrac{\sum f_i|x_i-M|}{\sum f_i}=\dfrac{508}{50}=10.16$

Mode of the following data.
Interval 0-20 20-40 40-60 60-80 80-100
Frequency 6 8 12 10 6

Here, maximum frequency is

$12$ .

$\therefore$ Modal class

$=40-60$

So,

$l=40,\,f_0=8,\,f_1=12,\,f_2=10$ and

$h=20$

$\Rightarrow$

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

$=40+\dfrac{12-8}{2\times 12-8-10}\times 20$

$=40+\dfrac{4}{6}\times 20$

$=40+13.33$

$=53..33$

The median of $230$ observations of the following frequency distribution is $46$ . Find $a$ and $b$ :
Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 12 30 a 65 b 25 18

We will add one more column of cumulative frequency as follows:

Class	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$12$	$30$	$a$	$65$	$b$	$25$	$18$
cumulative frequency	$12$	$42$	$42+a$	$107 +a$	$107 +a+b$	$132+a+b$	$150 +a +b$

Now the total number of observations will be equal to the cumulative frequency.

$\therefore$

$150+a+b =230$

Now we will use the median formula to form the second equation, as follows:

Median is

$l+h \left (\dfrac{n/2- cf}{f} \right )$

$h=10$

$l=40$

$cf=42+a$

$n/2= 230/2=115$

$f=65$

Now putting these values, we get:

Median

$= 46 =40+10 \left (\dfrac{115- (42+a)}{65} \right )$

$\Rightarrow a=34$

Putting this value of

$a$ in equation

$1$ , we get:

$b=80-a$

$b=46$

Thus the value of

$a$ is

$34$ and

$b$ is

$46$ .

Find median:
Wages per day $38$ $45$ $48$ $55$ $62$ $65$
Cumulative figure $14$ $22$ $29$ $39$ $45$ $47$

Here

$N =47$ which is odd.

$therefore = \dfrac {47}{2} = 23.5$

Here median is

$48$

1148200_1206693_ans_588af24b12a74f2e9cdd5885ca7b6873.jpg

Find the mode of given data.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
Frequency $20$ $24$ $40$ $36$ $20$

We use a formula to find mode of continuous data.

Mode

$=x_{1}+ \dfrac{f_{0}-f_{1}}{2f_{0}-f_{1}-f_{-1}} \times w$

$x_{i}=$ lower class of modal class

$f_{0}=$ frequency of modal class.

$f_{1}=$ frequency of class next to modal class

$f_{-1}=$ frequency of class previous to modal class

$w=$ width of modal class

Max freq

$=40 \therefore$ Modal class

$=20-30$

$x_{i}=20$

$f_{0}=40$

$f_{1}=24$

$f_{-1}=36$

$w=10$

$\therefore$ mode

$=20+ \dfrac{40-24}{2(40)-24-36} \times 10$

$=20+ \dfrac{16}{20} \times 10$

$=28$

$\therefore$ Mode

$=28$

Monthly expenditure of milk of $100$ families in a housing society are given in the following frequency distribution
Monthly expenditure(in) 0-175 175-350 350-525 525-700 700-875 875-1050 1050-1225
Number of families 10 14 15 21 28 7 5
Find the mode and the median for this distribution.

Solution -

M.Ex.(in)	0-175	175-350	350-525	525-700	700-875	875-1050	1050-1225
No.of Families	10	14	15	21	28	7	5

n = 100

Median

$= L+\frac{(n/2)-B}{G}\times W$

Median

$= 252+\frac{(50-39)}{21}\times 175$

Median

$= 616.667$

Where L = Lowers limit of median class

B = C.F of class before median class

G = Frequency of median class

W = class width

Mode

$L+ = \frac{fm-fm-1}{(fm-fm-1)+(fm-fm+1)}\times W$

Mode

$= 700+\frac{7}{21+7}\times 175$

Mode

$= 743.75$

Where L = lower limit of model class

fm-1 = Freq.of class before

fm+1 = Freq.of class after

fm = Freq.of class modal

W = class width

1100706_1188242_ans_a52a275426b34aef86d8e29be35be839.jpg

The following frequency distribution table shows the number of mango trees in a grove and their yield of mangoes, and also the cumulative frequencies. Find the median of the data.
Class
(No. of mangoes) Frequency
(No. of trees) Cumulative frequency
(less than)
$50-100$ $33$ $33$
$100-150$ $30$ $63$
$150-200$ $90$ $153$
$200-250$ $80$ $233$
$250-300$ $17$ $250$

Here,

$n=250$ then,

$\dfrac{n}{2}=\dfrac{250}{2}=125$

$\Rightarrow$ Median class

$=150-200$

$l=150,cf=63,f=90,h=50$

$\Rightarrow$

$Median=l+\dfrac{\dfrac{n}{2}-cf}{f}\times h$

$=150+\dfrac{125-63}{90}\times 50$

$=150+34.44$

$=184.44$

If the median for the following frequency distribution is 28.5, find the values of x and y.
Class Intervals frequency
0-10 5
10-20 x
20-30 20
30-40 15
40-50 y
50-60 5
Total 60

$Class-Interval$	$Frequency$ $f_i$	$c.f$
$0-10$	$5$	$5$
$10-20$	$x$	$5+x$
$20-30$	$20$	$25+x$
$30-40$	$15$	$40+x$
$40-50$	$y$	$40+x+y$
$50-60$	$5$	$45+x+y$
	$\sum f_i=45+x+y$

$N=60$ [ Given ]

Then,

$\dfrac{N}{2}=\dfrac{60}{2}=30$

$\therefore$ Median class

$=20-30$

Then,

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

$h=10$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

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

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

$17=25-x$

$\therefore$

$x=8$

$\sum f_i=60$

$45+x+y=60$

$45+8+y=60$

$\therefore$

$y=7$

$\therefore$

$x=8$ and

$y=7$

The following data gives the information of life of 300 electric bulbs (in Hours)

Life (in Hrs.)

0-20

20-40

40-60

60-80

80-100

100-120

No. of electric bulbs

31

36

38

42

82

71

Find the modal life of the electric bulbs.

$\begin{array}{l} Modal\, class\, is\, \, 80-100 \\ l=80,{ f_{ 1 } }=82,{ f_{ 0 } }=42,{ f_{ 2 } }=71,h=20 \\ modal\, life=l+\left( { \frac { { { f_{ 1 } }-{ f_{ 0 } } } }{ { 2{ f_{ 1 } }-{ f_{ 0 } }-{ f_{ 2 } } } } } \right) \times h \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =80+\left( { \frac { { 82-42 } }{ { 2\times 82-42-71 } } } \right) \times h \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =80+\frac { { 40\times 20 } }{ { 51 } } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =80+\frac { { 800 } }{ { 51 } } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =80+15.69 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =95.69 \end{array}$

The given distribution shows the number of wickets taken by bowlers in inter-school competitions:
Find the median.

No. of wickets	No. of bowlers $(f)$	$C.F$
$0-3$	$9$	$9$
$3-6$	$3$	$12$
$6-9$	$5$	$17$
$9-12$	$3$	$20$
$12-15$	$1$	$21$
	$N=21$

We have,

$\dfrac{N}{2}=\dfrac{21}{2}$

$=10.5$

So,

Median class

$=3-6$

So, we have

$l=3,\,c.f=9,\,f=3$ and

$h=3$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

$=3+\dfrac{10.5-9}{3}\times 3$

$=3+1.5$

$=4.5$

Monthly expenditures on milk in $100$ familes of a housing is given in the table.
Monthly expenditure (in Rs) $0-175$ $175-350$ $350-525$ $525-700$ $700-875$ $875-1050$ $1050-1225$
Number of familes $10$ $14$ $15$ $21$ $28$ $7$ $5$
Find the mean, mode and median for this distribution

$Monthly\,expenditure$	$Class-mark$ $x_i$	$No.\,of\,families$ $f_i$	$c.f$	$f_ix_i$
$0-175$	$87.5$	$10$	$10$	$875$
$175-350$	$262.5$	$14$	$24$	$3675$
$350-525$	$437.5$	$15$	$39$	$6562.5$
$525-700$	$612.5$	$21$	$60$	$12862.5$
$700-875$	$787.5$	$28$	$88$	$22050$
$875-1050$	$962.5$	$7$	$95$	$6737.5$
$1050-1225$	$1137.5$	$5$	$100$	$5687.5$
		$\sum f_i=100$		$=58450$

$\Rightarrow$

$Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{58450}{100}=584.5$

$N=100$ and

$\dfrac{N}{2}=\dfrac{100}{2}=50$

$\Rightarrow$ Median class

$=525-700$

$l=525,c.f.=39,f=21,h=175$

$Median=l+\dfrac{\dfrac{N}{2}-c.f.}{f}\times h$

$=525+\dfrac{50-39}{21}\times 175$

$=525+91.6$

$=616.6$

$\therefore$ Median

$=616.6$

$\Rightarrow$ Modal class

$=700-875$

$l=700,f_0=21,f_1=28,f_2=7,h=175$

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

$=700+\dfrac{28-21}{2\times 28-21-7}\times 175$

$=700+43.75$

$=743.75$

Determine the mean, mode and the median of the data:
$23,2,42,6,36,11,29,9,15$

Let us rewrite the given data in the ascending order,

$2,6,9,11,15,23,29,36,42$

$\begin{aligned}{}\text{Mean}& = \frac{{\left( {23 + 2 + 42 + 6 + 36 + 11 + 29 + 9 + 15} \right)}}{9}\\ &= \frac{{173}}{9}\\ &= 19.22\end{aligned}$

As there are

$9$ terms in total that is odd hence,

$\left(\dfrac{n+1}{2}\right)^{th}$ term will be the median of the data.

$\text{Median}=15$

By using the relation between mean, mode and median is,

$\begin{aligned}{}\text{Mode} + 2\text{Mean} &= 3\text{Median}\\\text{Mode}& = 3 \times 15 - 2 \times 19.22\\ &= 45 - 38.44\\ &= 6.56\end{aligned}$

For the following distribution:
Marks Below
$10$ Below
$20$ Below
$30$ Below
$40$ Below
$50$ Below
$60$
No. of students $3$ $12$ $27$ $57$ $75$ $80$
Find the median class.

$\therefore$ Median class is

$30.5-40.5$ with frequency

$30$

1380742_1243486_ans_e7b2b60ee1ba4477b26d71e02cc1b40b.png

During the medical checkup of 35 students of a class, their weights were recorded as followings:
Weight (in kg) Number of student
Less than 38
Less than 40
Less than 42
Less than 44
Less than 46
Less than 48
Less than 50
Less than 52
0
3
5
9
14
28
32
35
Draw a less than type 0 give for the given data.Hence obtain the weight from

1209880_1249178_ans_1197aa8f679b407bb3d8099d0b94f9bd.jpg

Find the mode of the data given below:

$Class$	$Frequency$
$19.5-29.5$	$15$
$29.5-39.5$	$20$
$39.5-49.5$	$50$
$49.5-59.5$	$30$
$59.5-69.5$	$10$

Here, maximum frequency is

$50$

$\therefore$ Modal class

$=39.5-49.5$

$l=39.5,\,f_0=20,\,f_1=50,\,f_2=30,h=10$

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

$=39.5+\dfrac{50-20}{(2\times 50)-20-30}\times 10$

$=39.5+6$

$=45.5$

The median of the data
Mid value 115 125 135 145 155 165 175 185 195
Frequency 6 25 48 72 116 60 38 22 3

$Mid-value$	$Class-Interval$	$Frequency$	$c.f$
$115$	$110-120$	$6$	$6$
$125$	$120-130$	$25$	$31$
$135$	$130-140$	$48$	$79$
$145$	$140-150$	$72$	$151$
$155$	$150-160$	$116$	$267$
$165$	$160-170$	$60$	$327$
$175$	$170-180$	$38$	$365$
$185$	$180-190$	$22$	$387$
$195$	$190-200$	$3$	$390$
		$N=390$

$N=390$ then,

$\dfrac{N}{2}=\dfrac{390}{2}=195$

$\therefore$ Median class

$=150-160$

$l=150,c.f=151,f=116,h=10$

$\Rightarrow$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

$=150+\dfrac{195-151}{116}\times 10$

$=150+3.79$

$=153.79$

Find the median if total students are $40$
Weight $45$ $46$ $47$ $48$ $49$ $50$ $51$ $52$ $53$
Median $6$ $2$ $3$ $4$ $7$ $4$ $7$ $4$ $3$

Median

$= \frac{1}{2}\left( {{{\frac{n}{2}}^{th}}obser + {{\left( {\frac{n}{2} + 1} \right)}^{th}}obser} \right)$

$\begin{array}{l} =\frac { 1 }{ 2 } \left( { { { 20 }^{ th } }+{ { 21 }^{ th } } } \right) \\ =\frac { 1 }{ 2 } \left( { 49+49 } \right) \\ =49 \end{array}$

1208117_1280094_ans_2414c5d82a2e401683452d9558e98729.JPG

Find the mean, median and mode of the following distribution : 8, 10, 7, 6 , 10, 11, 6, 13, 10

1714858_1412270_ans_bb4ff24cd5a94b229606c348e2ef0eda.jpg

Marks obtained by $50$ students in a unit test are given below:
Marks 0-20 20-40 40-60 60-80 80-100
Numbers of students 6 11 17 12 4
Calculate the median marks.

Marks obtained by

$50$ students in a unit test are given below:

Marks	0-20	20-40	40-60	60-80	80-100
Numbers of students	6	11	17	12	4

Cumulative frequency 6 17 34 46 50

Here,

$n=50$

Hence,

$\dfrac n2=\dfrac{50}{2}=25$

The median class lies in the cumulative frequency value

$34$ .

Hence, median class is

$40-60$ .

We know that,

Median,

$M = l+\left(\dfrac{\dfrac n2 - cf}{f}\right)\times h$

Where,

$l\rightarrow$ lower limit of the median class

$n\rightarrow$ total number of observations

$(\sum f)$

$cf\rightarrow$ cumulative frequency of the class preceding the median class

$f\rightarrow$ frequency of the median class

$h\rightarrow$ class width

Here,

$l=40, \ n=50,\ f=17,\ cf=17,\ h=10$

$\therefore \ M=40+\left[\dfrac{25-17}{17}\right]\times 10$

$\Rightarrow M=40+\dfrac{8\times 10}{17}$

$\rightarrow M=40+\dfrac{80}{17}$

$\Rightarrow M=40+4.71$

$\therefore \ M= 44.71$

Hence, the median marks for the given data is

$44.71$ .

Incomes of the families in a locality are given. Find the mode of the data.
Income (in Rs.) 1 - 200 201 - 400 401 - 600 600 - 800 801 - 1000
Number of families 7 10 16 12 3

Income $\left( \text{in Rs.} \right)$	Number of families
$1-200$	$7$
$201-400$	$10 = {f}_{0}$
$401-600$	$16 = {f}_{1}$
$601-800$	$12 = {f}_{2}$
$801-1000$	$3$
	$\sum{f} = 48$

As we know that,

Mode

$= l + \cfrac{{f}_{1} - {f}_{0}}{2 {f}_{1} - {f}_{0} - {f}_{2}} \times h$

Here,

$l =$ Lower limit of modal class

$= 401$

${f}_{1} =$ Frequency of class preceeding the modal class

$= 10$

${f}_{0} =$ Frequency of modal class

$= 16$

${f}_{2} =$ Frequency of class succeeding the modal class

$= 12$

$h =$ Class interval

$= 600 - 401 = 199$

Therefore,

Mode

$= 401 + \cfrac{16 - 10}{2 \times 16 - 10 - 12} \times 199 \\ = 401 + \cfrac{4}{10} \times 199 \\ = 401 + 79.6 = 480.6$

Hence the mode of the given data is

$480.6$ .

Size of the agriculture holding in a survey of 200 families is given in the following table:
Size of
Agriculture 0-5 5-10 10-15 15-20 20-25 25-30 30-35
Number of family 10 15 30 80 40 20 5
Complete median and mode size of the holdings.

$Size\,of\,Agriculture$	$No.\,of\,family$	$c.f$
$0-5$	$10$	$10$
$5-10$	$15$	$25$
$10-15$	$30$	$55$
$15-20$	$80$	$135$
$20-25$	$40$	$175$
$25-30$	$20$	$195$
$30-35$	$5$	$200$
	$N=200$

Here,

$N=200$ then,

$\dfrac{200}{2}=100$

$\therefore$ Median class

$=15-20$

We have,

$l=15,c.f=55,f=80,h=5$

$Median=l+\dfrac{\dfrac{N}{2}-c.f}{f}\times h$

$=15+\dfrac{100-55}{80}\times 5$

$=15+\dfrac{45}{80}\times 5$

$=15+2.81$

$=17.81$

Here, modal class

$=15-20$

We have,

$f_0=30,f_1=80,f_2=40,h=5$

$\Rightarrow$

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

$=15+\dfrac{80-30}{2\times 80-30-40}\times 5$

$=15+0.55$

$=15.55$

Find the median for the following frequency distribution table :
Class- interval 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30
frequency 5 3 9 10 8 5

Given distribution

Class-interval

$0-5$

$5-10$

$10-15$

$15-20$

$20-25$

$25-30$

Frequency

$5$

$3$

$9$

$10$

$8$

$5$

Now we can prepare a table for calculating median

Class-interval Frequency Cumulative frequency

$0-5$

$5$

$5-10$

$3$

$5+3=8$

$10-15$

$9$

$8+9=17$ (c.l)

$15-20$

$10$

$17+10=27$

$20-25$

$8$

$27+8=35$

$25-30$

$5$

$35+5=40$

$N=40$

$\Rightarrow N/2=40/2=20$

then the median class;

$15-20$

Lowest limit(l)

$=15$

Class width(h)

$=5$

Frequency(f)

$=10$

and c.f

$=$ preceding cumulative frequency of median class

$=$ median

$=l+\left(\dfrac{N/2-c.f}{f}\right)\times h$

$=15+\left(\dfrac{20-17}{10}\right)\times 15$

$=15+\dfrac{3\times 5}{10}=15+\dfrac{15}{10}$

Median

$=15+1.5=16.5$ .

The following are the marks of $9$ students in the class. Find the median
$34,32,48,38,24,30,27,21$ and $35$

First of all write the given data in ascending order

we get,

$21,24,27,30,32,34,35,38,48$

Number of observations

$=9$

$\text{Median}=\left(\dfrac{n+1}{2}\right)^\text{th}$ observation, when number of observations is odd

$[\because n=9]$

$\implies \text{Median}=\dfrac{10}{2}=5^{\text{th}}$ observation

$\implies \text{Median}=32$

The table below shows the members in "Stree-Sakti Kudambasree" Sorted according to their ages.
Age group Number of members
$20-30$ $4$
$30-40$ $8$
$40-50$ $10$
$50-60$ $7$
$60-70$ $4$
$70-80$ $2$
Total $35$
a. If the members are arranged in increasing order of gas, the member at what position is taken as media?
b. What is assumed to be age of the member at the 13th position?
c. Find the median of the ages.

Classes	Frequency (f)	Cumulative frequency
$20-30$	$4$	$4$
$30-40$	$8$	$12$
$40-50$	$10$	$22(N/2)$
$50-60$	$7$	$29$
$60-70$	$4$	$33$
$70-80$	$2$	$35$
Total	$35$

$\dfrac{35+1}{2}=18$

b.

$13^{th} position =40+\dfrac{1}{2}=40.5$

c. Median

$I_1+\dfrac{\dfrac{N}{2}-c}{f}\times h$

$I_1=$ lower limit of the median class.

$N$ = Total number of frequency,

$c$ = Above the cf of the median class

$f$ =frequency of the median class,

$h$ = class interval

$\dfrac{N}{2}=\dfrac{35}{2}=17.5$
Median class

$=40-50(\dfrac{N}{2}=17.5$ included class)

$I_1+40,c=12,f=10$
Hence median

$=I_1+\dfrac{\dfrac{N}{2}-c}{f}\times h=40+\dfrac{17.5-12}{10}\times 10$

$=40+5.5=45.5$

Heights of the pupils of a particular school are given. Draw greater than cumulative curve and find the median height from it.
Height (in cm) 90-100 100-110 110-120 120-130 130-140 140-150
Number of pupils 5 2 3 8 8 6

Height $\left( \text{in cm} \right)$	Number of pupils	Cumulative frequency
$90-100$	$5$	$5$
$100-110$	$2$	$7$
$110-120$	$3$	$10$
$120-130$	$8$	$18$
$130-140$	$8$	$26$
$140-150$	$6$	$32$
	$\sum{f} = 32$

$N = 32$

$\cfrac{N}{2}$

$= \cfrac{32}{2}$

$= 16$

Thus,

$120-130$ is the median class.

As we know that,

Median

$= l + \cfrac{\cfrac{N}{2} - C.f.}{f} \times h$

Here,

$l =$ Lower limit of median class

$= 120$

$C.f. =$ Cumulative frequency of class preceeding the median class

$= 10$

$f =$ frequency of median class

$= 8$

$h =$ Class interval

$= 130 - 120 = 10$

Therefore,

Median

$= 120 + \cfrac{16 - 10}{8} \times 10 \\ = 120 + \cfrac{6}{8} \times 10 \\ = 120 + 7.5 = 127.5$

Hence the median height is

$127.5$ .

Find the mean and median for the following data:
$i) 13 , 17 , 16 , 14 , 11 , 13 , 10 , 16 , 11 , 18 , 12 , 17$
$ii) 36 , 72 , 46 , 42 , 60 , 45 , 53 , 46 , 51 , 49$

10,11,11,12,13,13,14,16,16,17,17,18

median = (13 + 14)/2 = 13.5

mean =

$\dfrac{10+11+11+12+13+13+14+16+16+17+17+18}{12}$

mean = 14

mean deviation = 14-13.5 = 0.5

2) 36,42,45,46,46,49,51,53,60,72

median = (46+49)/2 = 47.5

$mean = \dfrac{36+42+45+46+46+49+51+53+60+72}{10}$

mean = 50

mean deviation = 50-47.5 = 2.5

Find the mode of the following data.

1484542_1567532_ans_af4320e2c3f84978ad7837d9c0efd4b8.jpg

Find the mean of the following distribution :
x : 10 12 20 25 35
f : 3 10 15 7 5

We can prepare a table following calculating mean by Direct method

$x$	$f$	$f_x$
10	3	30
12	10	120
20	15	300
25	7	175
35	5	175

$\sum f=40$

$\sum f_x=800$

W know

$mean(x)=\dfrac{\sum f_x}{\sum f}$

$=\dfrac{800}{40}$

$mean=20$

Find the mean of the following data:
$x_i$ $19$ $21$ $23$ $25$ $27$ $29$ $31$
$f_i$ $13$ $15$ $16$ $18$ $16$ $15$ $13$

Given data

$19$

$21$

$23$

$25$

$27$

$29$

$31$

$13$

$15$

$16$

$18$

$16$

$15$

$13$

We have to prepare table for calculating the mean

x f fx

$19$

$13$

$247$

$21$

$15$

$315$

$23$

$16$

$368$

$25$

$18$

$450$

$27$

$16$

$432$

$29$

$15$

$435$

$31$

$13$

$403$

$\sum fx=2650$

$\sum f=106$

we know that (direct method) Mean

$=\dfrac{\sum fx}{\sum f}$

$\therefore$ Mean

$=\dfrac{\sum fx}{\sum f}$

$=\dfrac{2650}{106}$

Mean

$=25$ .

The ages (in years) of 50 students of a class in a school are below:
Age (in years): 14 15 16 17 18
Numbers of students: 15 14 10 8 3
Find the mean age. (in years)

Given that

The ages(in years) of

$50$ students of a class in a school are below

Age(in years)

$14$

$15$

$16$

$17$

$18$

No. of students

$15$

$14$

$10$

$8$

$3$

Let's we have to prepare the table for calculating mean

Age(in years)(x) No. of students(f) fx

$14$

$15$

$210$

$15$

$14$

$210$

$16$

$10$

$160$

$17$

$8$

$136$

$18$

$3$

$54$

$N=\sum f=50$

$\sum fx=770$

We know

Mean

$=\dfrac{\sum fx}{\sum f}=\dfrac{\sum fx}{N}$

$=\dfrac{770}{50}$

Mean

$=15.4$

Hence, Mean age

$=15.4$ years

Find the mode of the following data:
Class: $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$ $120-140$
Frequency: $6$ $8$ $10$ $12$ $6$ $5$ $3$

$C.I$	$F_i$
$0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$ $120-140$	$6$ $8$ $10$ $12$ $6$ $5$ $3$

Highest frequency

$= 12$

Modal class

$= 60 - 80$

$l = 60, F_0 = 10, F_1 = 12, F_2 = 6, h = 20$

Mode

$= l + \left(\dfrac{F_1 - F_0}{2F_1 - F_0 - F_2} \right) \times h$

$= 60 + \dfrac{12 - 10}{2 \times 12 - 10 - 6} \times 20$

$= 60 + \dfrac{2}{8} \times 20 = 60 + 5 = 65$

Mode

$= 65$

Solve the following :
If $L = 10 , f_1 = 70 , f_0 = 58, f_2 = 42, h = 2$ , then find the mode by using formula.

Given

$L = 10 , f_1 = 70, f_0 = 58, f_2 = 42, h = 2$

Mode

$= L + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} . h$

$=10 + \dfrac{70 - 58}{2(70) - 58 - 42} \times 2$

$= 10 + \dfrac{12}{140 - 100} \times 2$

$= 10 + \dfrac{12}{40} \times 2$

$10 + \dfrac{12}{20}$

$= 10 + \dfrac{3}{5} = \dfrac{53}{5}$

Find the mode of the following frequency distribution:
Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency $12$ $35$ $45$ $25$ $13$

Here:

Maximum class frequency is

$45$ .

the class corresponding to Maximum class frequency is the modal class.

So, Modal class

$=30-40$

Lower limit

$=I=30$

Modal class size

$=h=10$

Frequency of class preceding the modal class

$=f_0=35$

Frequency of the modal class

$f_1=45$

Frequency of class succeeding the modal

$=f_2=25$

Mode formula is given by,

Mode

$=I+\left( \dfrac{f_1 -f_0}{2f_1-f_0-f_2}\right)\times h$

Mode

$=30+\left( \dfrac{45-35}{2(45)-35-25}\right)\times 10$

$=30+\dfrac{100}{30}$

$=33.33$

Find mode for given data:
Class $20-29$ $30-39$ $40-49$ $50-59$ $60-69$
Frequency $15$ $20$ $50$ $30$ $10$

class	continuous class	frequency
$20-29$	$19.5-29.5$	$15$
$30-39$	$29.5-39.5$	$(f_1)20$
$40-49$	$(l) 39.5-49.5$	$(f_0)50$
$50-59$	$49.5-59.5$	$(f_2)30$
$60-69$	$59.5-69.5$	$10$

Here observe that highest frequency i.e mode class

$39.5-49.5$

Lower limit

$=39.5$

class height

$h=10$

mode =

$l+\left(\dfrac{(f_0-f_1)}{2f_0-f_1-f_2}\times h\right)$

$39.5+ \left(\dfrac{50-20}{2\times 50-20-30}\right)10$

$39.5+\dfrac{30}{50}\times 10$

$39.5+\dfrac{30}{5}$

$39.5+6$

mode=

$45.5$

Calculate the mean for the following distribution :
x : 5 6 7 8 9
f : 4 8 14 11 3

To find a mean, we have to prepare a table

$x$	$f$	$f_x$
5	4	20
6	8	48
7	14	98
8	11	88
9	3	27

$\displaystyle \sum f=40$

$\displaystyle \sum f_x=281$

A the direct method of mean, we know mean

$=\dfrac{\sum f_x}{\sum f}$

$\therefore mean=\dfrac{281}{40}$

$=7.025$

The following tables shows the daily wages of works in a factory:
Daily wages (in $Rs$ ) $0-100$ $100-200$ $200-300$ $300-400$ $400-500$
Number of workers $40$ $32$ $48$ $22$ $8$

Find Median daily wages.

Daily Wages	Frequency $f_i$	$cf$
$0-100$	$40$	$40$
$100-200$	$32$	$72$
$200-300$	$48$	$120$
$300-400$	$22$	$142$
$400-500$	$8$	$150$
	$\displaystyle \sum f_i =150$

Where

$cf=$ cumulative frequency

Here

$N=150$

$\dfrac{N}{2}=75$

$cf$ just greater than $75$ to $120$

Median class $200-300$

So, $I=200,\ h=100,\ f=48$ and

$cf=cf$ of preceding class i.e $72$

Substitute all the value in the above formula, we get

Median $=200+\left\{100\times \dfrac{(75-72)}{48} \right\}$

$=200+6.25$

$=206.25$

Median of daily wages is $Rs.\ 206.25$

1558109_1714441_ans_ef937057c0b647b38bb807282fb11c51.JPG

Given below is the distribution of total household expenditure of $200$ manual workers in a city:
Expenditure (in Rs) $1000-\\ 1500$ $1500-\\ 2000$ $2000-\\ 2500$ $2500-\\ 3000$ $3000-\\ 3500$ $3500-\\ 4000$ $4000-\\ 4500$ $4500-\\ 5000$
No. of manual workers $24$ $40$ $31$ $28$ $32$ $23$ $17$ $5$
Find Mode.

Here:
Maximum class frequency

$=40$
The class corresponding to Maximum class frequency is the modal class.
So, Modal class

$=1500-2000$
Lower limit

$=L=1500$
Modal class size

$=h=500$
Frequency of class preceding the modal class

$=f_0=24$
Frequency of the modal class

$=f_1=40$
Frequency of class succeeding the modal

$=f_2=31$

Mode is given by.the following formula:

Mode

$=L+\left( \dfrac{f_1 -f_0}{2f_1 -f_0-f_2}\right) \times h$

Mode

$=1500+\left( \dfrac{40-24}{2(40)-24-31}\right)\times 500$

$=1500+\dfrac{16}{25}\times 500$

$=1820$

Compute the mode of the following data:
Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$
Frequency $25$ $16$ $28$ $20$ $5$

Here :

Maximum class frequency

$=28$

The class corresponding to Maximum class frequency is the modal class.

So, Modal class

$=40-60$

Lower limit

$=I=40$

Modal class size

$=h=20$

Frequency of class preceding the modal class

$=f_0=16$

Frequency of the modal class

$=f_1=28$

Frequency of class succeeding the modal

$=f_2=20$

Mode formula is given by,

Mode

$=I+\left( \dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

Mode

$=40+\left( \dfrac{28-16}{2(28)-16-20}\right)\times 20$

$=40+\dfrac{12}{20}\times 12$

$=47.2$

Find the mode of the following distribution:
Class interval $10-14$ $14-18$ $18-22$ $22-26$ $26-30$ $30-34$ $34-38$ $38-42$
Frequency $8$ $6$ $11$ $20$ $25$ $22$ $10$ $4$

Here:
Maximum class frequency

$=25$
The class corresponding to Maximum class frequency is the modal class.
So, Modal class

$=26-30$
Lower limit

$=I=26$
Modal class size

$=h=4$
Frequency of class preceding the modal class

$=f0=20$
Frequency of the modal class

$=f1=22$
Frequency of class succeeding the modal

$=f2=12$
Mode formula is given by,
Mode

$=I+\left( \dfrac{f_1- f_0}{2f_1-f_0-f_2}\right)\times h$
Mode

$=26+\left( \dfrac{25-20}{2(25)-20-22}\right)\times 4$

$=26+\dfrac{5}{8}\times 4$

$=28.5$

Calculate the median from the following frequency distribution:
Class $5-10$ $10-15$ $15-20$ $20-25$ $25-30$ $30-35$ $35-40$ $40-45$
Frequency $5$ $6$ $15$ $10$ $5$ $4$ $2$ $2$

Class	Frequency $(f_i)$	$c.f.$
$5-10$	$5$	$5$
$10-15$	$6$	$5+6=11$
$15-20$	$15$	$11+15=26$
$20-25$	$10$	$26+10=36$
$25-30$	$5$	$36+5=41$
$30-35$	$4$	$41+4=45$
$35-40$	$2$	$45+2=47$
$40-45$	$2$	$47+2=49$
	$\sum f_i=49$

Where

$c.f.=$ cumulative frequency
Median

$=L+\left\{h\times \dfrac {\left(\dfrac {N}{2}-c.f. \right)}{f}\right\}$
Here

$N=49$

$\dfrac{N}{2}=24.5$

$c.f.$ just greater than

$24.5$ is

$26$
Median class

$=15-20$
So,

$L=15,\ h=5,\ f=15$ and

$c.f.=c.f.$ of preceding class i.e

$11$
Substitute all the value in the above formula, we get
Median

$=15+\left(\dfrac{24.5-11}{15}\right)\times 5$
Median

$=15+4.5$

$=19.5$
Thus, median of frequency distribution is

$19.5$ .

Compute the median from the following data:
Marks $0-7$ $7-14$ $14-21$ $21-28$ $28-35$ $35-42$ $42-49$
Number of students $3$ $4$ $7$ $11$ $0$ $16$ $9$

Marks	No of students $fi$	$cf$
$0-7$	$3$	$3$
$7-14$	$4$	$7$
$14-21$	$7$	$14$
$21-28$	$11$	$25$
$28-35$	$0$	$25$
$35-42$	$16$	$41$
$42-49$	$9$	$50$
	Sum $=50$

Where

$cf=$ cumulative frequency
Median

$=L+\left\{h\times \dfrac {\left(\dfrac {N}{2}-cf \right)}{f}\right\}$
Here:

$N=50$

$N/2=25$
Here cumulative frequency is

$25$
Median class

$=21-28$
So,

$L=21,\ h=7,\ f=11$ and

$cf=cf$ of preceding class i.e

$14$
Substitute all the value in the above formula, we get
Median

$=21+\left\{7\times (25-14)/11 \right\}$

$=28$

In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age (in year) $0-15$ $15-30$ $30-45$ $45-60$ $60-75$
Number of patients $5$ $20$ $40$ $50$ $25$

Class-Interval Frequency

$Cf$

$0-15$

$5$

$15-30$

$20$

$25$

$30-45$

$40$

$65$

$45-60$

$50$

$115$

$60-75$

$25$

$140$

Where

$cf=$ cumulative frequency

Median

$=1+\left\{h\times \dfrac {\left(\dfrac {N}{2}-cf \right)}{f}\right\}$

Here:

$N=140$

$N/2=70$

$cf > 70$ is

$140$

Median class

$=45-60$

So,

$I=45,\ h=15,\ f=50$ and

$cf=cf$ of preceding class i.e.

$65$

Substitute all the value in the above formula, we get

Median

$=45+\left\{15\times (70-65)/50 \right\}$

$=45+1.5$

$=45.5$

Therefore , median age of diabetic patients is

$46.5$ years.

Find the mean, mode and median of the following data:
Marks obtained $25-35$ $35-45$ $45-55$ $55-65$ $65-75$ $75-85$
Number of students $7$ $31$ $33$ $17$ $11$ $1$

1669827_1714450_ans_a693d54771564243b107d990ef67bb3e.jpg

Given bellows is the number of units of electricity consumed in a week in a certain locality:
Consumption (in unit) $65-85$ $85-105$ $105-125$ $125-145$ $145-165$ $165-185$ $185-205$
Number of consumers $4$ $5$ $13$ $20$ $14$ $7$ $4$

Find Median value of Electricity Consumption.

Consumption (in unit)	Frequency $fi$	$cf$
$65-85$	$4$	$4$
$85-105$	$5$	$9$
$105-125$	$13$	$22$
$125-145$	$20$	$42$
$145-165$	$14$	$56$
$165-185$	$7$	$63$
$185-205$	$4$	$67$
	Sum $67$

Where

$cf=$ cumulative frequency
Median

$=1+\left\{h\times \dfrac {\left(\dfrac {N}{2}-cf \right)}{f}\right\}$
Here

$N=67$

$N/2=33.5$

$cf$ just greater than

$33.5$ is

$42$
Median class

$=125-145$
So,

$I=125,\ h=20,\ f=20$ and

$cf=cf$ of preceding class i.e

$22$
Substitute all the value in the above formula, we get
Median

$=125+\left\{20\times (33.5-22)/ 20\right\}$

$=125+11.5$

$=136.5$
Thus, median of electricity consumed is

$136.5$ .

Calculate the median from the following data:
Height (in $cm$ ) $135-140$ $140-145$ $145-150$ $150-155$ $155-160$ $160-165$ $165-170$ $170-175$
No of boys $6$ $10$ $18$ $22$ $20$ $15$ $6$ $3$

Where

$cf=$ cumulative frequency
Median

$=1+\left\{h\times \dfrac {\left(\dfrac {N}{2}-cf \right)}{f}\right\}$
Here

$N=100$

$N/2=50$

$cf$ just greater than

$50$ is

$56$
Median class

$=150-155$
So,

$I=150,\ h=5,\ f=22$ and

$cf=cf$ of preceding class i.e

$34$
Substitute all the value in the above formula, we get
Median

$=150+\left\{5\times (50-34)/ 22\right\}$

$=150+3.64$

$=153.64$

1558304_1714465_ans_cf51f30a144c40a3891a397eab97d30b.JPG

A survey reading the heights (in $cm$ ) of $50$ girls of a class was conducted and the following data was obtained:

Height (in $cm$ ) $120-130$ $130-140$ $140-150$ $150-160$ $160-170$ Total
Number of girls $2$ $8$ $12$ $20$ $8$ $50$
Find the mean, median and mode of the above data.

Assume mean

$A=145$
Class interval

$=h=20$

Class	Frequency $f_{i}$	Mid Point $x_{i}$	$u_{i}=\dfrac{x_{i}-A}{h}$	$f_{i}u_{i}$	$cf$
$120-130$	$2$	$125$	$-2$	$-4$	$2$
$130-140$	$8$	$135$	$-1$	$-8$	$10$
$140-150$	$12$	$145=A$	$0$	$0$	$22$
$150-160$	$20$	$155$	$1$	$20$	$42$
$160-170$	$8$	$165$	$2$	$16$	$50$
	$N=50$		$\displaystyle\sum f_{i}u_{i}=24$

Mean:

$\bar{x}=A+h\left(\dfrac{f_{i}u_{i}}{N}\right)$

$=145+10\times 24/50$

$=149.8$

Median

$N=50$

$N/2=25$

$cf$ just after

$25$ is

$42$
So,

$I=150, h=10, f=20$ and

$cf=cf$ of preceding class i.e.

$22$
Median

$=I+\left\{h\times \dfrac{\left(\dfrac{N}{2}-cf\right)}{f}\right\}$

$=150+\left\{10\times (25-22)/20\right\}$

$=151.5$

Mode:
We know, Mode

$=3$ Median

$-2$ Mean

$=3\times 151.5-2\times 149.5$

$=159.9$

Mean

$=149.8$ , Median

$=1515.5$ and Mode

$=154.9$

If the median of the following frequency distribution is $32.5$ , find the values of $f1$ and $f2$ .
Class interval $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ Total
Frequency $f_1$ $5$ $9$ $12$ $f_2$ $3$ $2$ $40$

Sum of all the frequencies

$=f1+5+9+12+f2+3+2=40$ (given)

$\Rightarrow f1+f2 =9....(1)$
Median is

$32.5$ lies in

$30-40$ , so the median class is

$30-40$

$N=40$

$\dfrac{N}{2}=20$

$I=30,\ h=10,\ f=12$

$cf=f_1+5+9=f_1+14$
We know,
Median

$=1+\left\{h\times \dfrac {\left(\dfrac {N}{2}-cf \right)}{f}\right\}$
Substitute all the values in the above formula, we get

$32.5=30+\left\{10\times \dfrac{(20-f_1-14)}{12}\right\}$

$\Rightarrow 32.5=30+\dfrac{(30-5f_1)}{6}$

$\Rightarrow 2.5=\dfrac{(30-5f_1)}{6}$
or

$f_1=3$
From

$(1)$

$\Rightarrow 3+f_2=9$

$\Rightarrow f_2=6$

Calculate the mean and the median of the numbers: $2, 1, 0, 3, 1, 2, 3, 4, 3, 5$ .

Arranging the numbers in ascending order :

$0, 1, 1, 2, 2, 3, 3, 3, 4, 5$

Here, $n = 10$ which is even

$Median = \dfrac{1}{2} ( n/2^{th} term + ((n/2)+1)^{th} term)$

$= \dfrac{1}{2}(10/2^{th} term + ((10/2)+1)^{th} term)$

$=\dfrac{1}{2}(5^{th} term + (5+1)^{th} term)$

$=\dfrac{1}{2}(5^{th} term + 6^{th} term)$

$=\dfrac{1}{2}(2+3)$

$= \dfrac{1}{2}\times 5$

$= 2.5$

Hence the median is $2.5$ .

$Mean =\dfrac {sum\,of \,all\,the\, observations}{number \,of \,observations}$

$= \dfrac{\Sigma x_i}{n}$

$= (0+1+1+2+2+3+3+3+4+5)/10$

$= \dfrac{24}{10}$

$= 2.4$

Hence the mean is $2.4$ .

The mean of the numbers 1, 7, 5, 3, 4, and 4 is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m $-1$ and median q. Find:
(i) p
(ii) q
(iii) the mean of p and q.

$(i)$ Given that, mean of $1, 7, 5, 3, 4, 4$ is $m$ .

We know that, $Mean=\dfrac{\displaystyle \sum x}{n}$
Here $\displaystyle \sum x =1+7+5+3+4+4=24$ and $n = 6$

$\therefore \ Mean(m) = \dfrac{24}{6}=4$

$\therefore \ m = 4$

Also given that, the numbers $3, 2, 4, 2, 3, 3, p$ have mean $m-1$ .

$\therefore \ m-1 = \dfrac{3+2+4+2+3+3+p}{7}$

$\Rightarrow 4-1 = \dfrac{(17+p)}{7}$ $[m=4]$

$\Rightarrow 3\times 7 = 17+p$

$\Rightarrow 21 = 17+p$

$\Rightarrow p = 21-17$

$\Rightarrow p = 4$

Hence, the value of $p$ is $4$ .

$(ii)$ Given that the numbers have median $q$ .

Arranging them in ascending order:

$2, 2, 3, 3, 3, 4, 4$

Here $n = 7$ , which is odd

$\therefore \ median = \left (\dfrac{n+1}{2}\right)^{th}\, term$

$\Rightarrow \left (\dfrac{7+1}{2}\right)^{th}\, term=4^{th} \ term$

$4^{th} \ term\rightarrow 3$

$\therefore \ q = 3$

So, value of $q$ is $3$ .

$(iii)$ We know that, mean of $p$ and $q$ is $\dfrac{p+q}{2}$

$p=4\ and \ q=3$

$\therefore\ mean=\dfrac{4+3}{2}=\dfrac{7}{2}$

$\Rightarrow 3.5$

Hence, the mean of $p$ and $q$ is $3.5$ .

At a shooting competition, the scores of a competitor were as given below :
Score
0
1
2
3
4
5
No. of shots
0
3
6
4
7
5
What was his total score?

We write the marks in cumulative frequency table.

Score x	No. of shots (f)	fx	Cumulative frequency
0	0	0	0
1	3	3	3
2	6	12	9
3	4	12	13
4	7	28	20
5	5	25	25
Total	$\Sigma f = 25$	$\Sigma fx = 80$

Total score $= \Sigma fx = 80$

Hence his total score is $80$ .

State the modal class.
Class interval
50-55
55-60
60-65
65-70
70-75
75-80
80-85
85-90
Frequency
5
20
10
10
9
6
12
8

Class mark ( $x_i$ ) = (upper limit + lower limit)/2

Let assumed mean (A) $= 67.5$

Class size (h) $= 5$

Class Interval	Frequency (fi)	Class mark ( $x_i$ )	$d_i = x_i – A$	$u_i = d_i/h$	$f_iu_i$
50-55	5	52.5	-15	-3	-15
55-60	20	57.5	-10	-2	-40
60-65	10	62.5	5	-1	-10
65-70	10	67.5	10	0	0
70-75	9	72.5	15	1	9
75-80	6	77.5	20	2	12
80-85	12	82.5	25	3	36
85-90	8	87.5	30	4	32
Total	$\Sigma f_i = 80$				$\Sigma f_iu_i = 24$

Modal class is the class with highest frequency.

Here the modal class is $55-60$ .

At a shooting competition, the scores of a competitor were as given below :
Score
0
1
2
3
4
5
No. of shots
0
3
6
4
7
5
What was his modal score?

We write the marks in cumulative frequency table.

Score x	No. of shots (f)	fx	Cumulative frequency
0	0	0	0
1	3	3	3
2	6	12	9
3	4	12	13
4	7	28	20
5	5	25	25
Total	$\Sigma f = 25$	$\Sigma fx = 80$

Here the frequency corresponding to 4 is maximum. $4$ occurs $7$ times.

Hence his modal score is $4$ .

For the given data below, calculate the mean, mode, and median.
$6,2,5,4,3,4,4,2,3$

Rearranging the data into ascending order, we get

$2,2,3,3,4,4,4,5,6$

The total numbers of observations are equal to

$9$ , which is odd. Hence,

$\left(\dfrac{9+1}{2}\right)^{th}=5^{th}$ term will be the median of the data.

Hence, the median is equal to

$4$ .

The data whose frequency is maximum is known as the mode of the data so,

$4$ occurs

$3$ times which is a maximum occurrence than any other data hence, the mode of the data is

$4$ .

Now,by using the relation between mean mode and median of a data,

$\text{Mean}=\dfrac{3\cdot \text{Median}- \text{Mode}}{2}$

$= \dfrac{3\cdot4-4}{2}$

$=4$

Hence, the mean, mode, and median of the data is

$4$ .

Find the median of the given data if the mean is 4.5
$5,7,7,8,x,5,4,3,1,2$

The given data is

$5,7,7,8,x, 5,4,3,1,2$

Now,

$\text{Mean}=\dfrac{\text{sum of observations}}{\text{number of observations}}\\$

$\therefore 4.5= \dfrac{5+7+7+8+x+5+4+3+1+2}{10}$

$\Rightarrow 45= x+42$

$\Rightarrow x= 45-42=3$

Now, rearranging the observations, we get

$1,2,3,3,4,5,5,7,7,8$

total number of observations

$=10$ , which is even

Therefore, median=

$\dfrac{4+5}{2}=\dfrac{9}{2}= 4.5$

A die was thrown 15 times and the outcomes recorded were
$5,3,4,1,2,6,4,2,2,3,1,5,6,1,2$
Find the mean, median and mode of the data.

Rearranging the data into ascending order, we get

$1,1,1,2,2,2,2,3,3,4,4,5,5,6,6$

We know that,

$\text{Mean}=\dfrac{\text{sum of observations}}{\text{number of observations}}\\$

$\Rightarrow\text{ Mean}=\dfrac{1+1+1+2+2+2+3+3+4+4+5+5+6+6}{15}$

$=\dfrac{47}{15}= 3.13$

Now, total number of observations

$=15$ , which is odd.

$\therefore$ Median

$=8^{th}$ term

$=3$

Now, the observation

$\ 2$ occurs the most i.e.,

$4$ times

$\therefore$ Mode

$=2$

Hence, Mean

$=3.13$ , Median

$=3$ and

$Mode$ =2$$.

Calculate the Mean, Median and Mode of the following data:
$5,10,10,12,13$
Are these three equal?

We know that,

$Mean=\dfrac{sum\ of \ obsrvations}{number\ of\ observations}$

The mode is the number that appears most frequently in a data set.

The median is the middle number in a sorted data.

Here,

Mean

$=\dfrac{5+10+10+12+13}{5}=\dfrac{50}{5}=10$

Mode

$= 10$ , as it occurs the most number of times.

Median

$= 10$ , as it is the middle number.

Hence, Mean

$=$ Median

$=$ Mode

$=10$

In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula
$\overline{x}=a+ \dfrac{f_i d_i}{f_i}$
where $a$ is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

No,
The statement is not correct. It is not necessary that assumed mean should be the mid-point of the class interval.

$a$ can be considered as any value which is easy to simplify it.

The monthly income of 100 families are given as below:
Income (in Rs) Number of families
0-5000 8
5000-10000 26
10000-15000 41
15000-20000 16
20000-25000 3
25000-30000 3
30000-35000 2
35000-40000 1
Calculate the modal income.

According to data given,

The highest frequency

$=41$

$41$ lies in the interval

$10000-15000$

Here,

$I=10000,f_m=41,f_1=26,f_2=16$ and

$h=5000$

$\therefore Mode=I+\left(\dfrac{f_m-f_1}{2f_m-f_1-f_2}\right)\times h$

$=10000+\left(\dfrac{41-26}{2\times 41-26-16}\right)\times h$

$=10000+\left(\dfrac{15}{82-42}\right)\times h$

$=10000+\left(\dfrac{15}{40}\right)\times h$

$=10000+15\times 125$

$=10000+1875$

$=11875$

The daily income of a sample of 50 employees are tabulated as follows:
Income (in Rs) 1-200 201-400 401-600 601-800
Number of employees 14 15 14 7
Find the mean daily income of employees.

Assumed mean

$a=300.5$ and

$d_i = (x_i-a)$

$\overline{x} = a+ \dfrac{\Sigma f_i d_i}{\Sigma f_i}$

$=300.5 + 2800/50$

$=356.5$

Hence, the average daily income of employees

$= Rs. 356.5$

Calculate the mean of the following data:
Class 4-7 8-11 12-15 16-19
Frequency 5 4 9 10

The given data is not continuous.
So, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.

Class	Class Marks ( $x_i$ )	Frequency ( $f_i$ )	$f_i x_i$
3.5-7.5	5.5	5	27.5
7.5-11.5	9.5	4	38
11.5-15.5	13.5	9	121.5
15.5-19.5	17.5	10	175
		$\Sigma f_i = 28$	$\Sigma f_i x_i = 362$

Mean,

$\overline{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}= \dfrac{362}{28} = 12.93$
Therefore, mean of the given data = 12.93.

In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula
$\overline{x}=a+ \dfrac{f_i d_i}{f_i}$
where $a$ is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

No, the statement is not correct. It is not necessary that assumed mean should be the mid-point of the class interval. a can be considered as any value which is easy to simplify it.

The weights (in kg) of 50 wrestlers are recorded in the following table:
Weight (in kg) 100-110 110-120 120-130 130-140 140-150
Number of wrestlers 4 14 21 8 3
Find the mean weight of the wrestlers.

Assumed mean, (a) = 125

Class width, (h) = 10

and total observations, (N) = 50

By step deviation method,

Mean

$\overline{x} = a+ \dfrac{\Sigma f_i d_i}{\Sigma f_i}$

Mean

$\overline{x}= 125 + \dfrac{(-80)}{50}$

$=125-15$

$=123.4 \ kg$

Hence, mean weight of wrestlers

$=123.4 \ kg$

Calculate the mean, the median and the mode of the following numbers:
$3, 7, 2, 5, 3, 4, 1, 5, 3, 6$

(a)

$Mean = \dfrac{3 + 7 + 2 + 5 + 3 + 4 + 1 + 5 + 3+ 6}{10} = \dfrac{39}{10} = 3.9$
Hence the mean

$= 3.9$
(b) Arranging the given data in ascending order, we get

$1, 2, 3, 3, 3, 4, 5, 5, 6, 7$
Total numbers of observations (items) = 10 (even)
There are two middle items - 3 and 4
Their average

$= \dfrac{3 + 4}{2} = \dfrac{7}{2} = 3.5$
(c) In the given numbers 3 is repeated more than any other number

$\therefore Mode = 3$

Calculate the mean, the median and the mode of the numbers:
$1, 3, 2, 6, 2, 3, 1, 3$

(a) Mean

$= \dfrac{1 + 3 + 2 + 6 + 2 + 3 + 1 + 3}{8}$

$= \dfrac{21}{8} = 2.625$
Hence mean

$= 2.625$
(b) Arranging the given data in ascending order, we get

$1, 1, 2, 2, 3,3, 3, 6$
Total number of observations (items) = 8(even)
There are two middle items:- 2 and 3
Their average =

$\dfrac{2 + 3}{2} = \dfrac{5}{2} = 2.5$
Hence the median of the given numbers

$= 2.5.$
(c) In the given numbers,

$3$ is repeated more number of time than any other number.

$\therefore Mode = 3.$

Will the median class and modal class of grouped data always be different? Justify your answer.

The median class and modal class of grouped data is not always different, it depends on the data given.

Find the mean and the median of the following data:
$5, 3, 12, 0, 7, 11, 4, 3, 9$

Mean

$= \dfrac{5 + 3 + 12 + 0 + 7 + 11 + 4 + 3 + 9}{9} = \dfrac{54}{9} = 6$
Arranging the given data in ascending order:

$0, 3, 3, 4, 5, 7, 9, 11, 12$
Middle term is

$5$
Hence, median of a given data is

$5.$

Find the mean and the median of :
$10,12,12,15,15,17,18,18,18$ and $19$

Given
Numbers are

$10,12,12,15,15,17,18,18,18$ and

$19$
the mean can be calculated as below
Mean

$=$ (Sum of numbers)

$/$ (Number of numbers)

$=(10+12+12+15+15+17+18+18+18+19)/10$

$=154/10$
We get,

$=15.4$
Hence, the mean is

$15.4$
Total numbers are

$10$
Since, total number of numbers is even
Therefore, the median will be the half of the sum of middle numbers
The median can be calculated as below
Median

$=$ (Sum of two middle numbers)

$/2$

$(15+17)/2$

$=32/2$
We get,

$=16$
Therefore, the median is

$16$

Find the mean and the median of :
$1,3,4,5,9,9,$ and $11$

Given
Numbers are

$1,3,4,5,9,9,$ and

$11$
Hence, the mean can be calculated as below
Mean

$=$ (Sum of numbers)

$/$ (Number of numbers)

$=(1+3+4+5+9+9+11)/7$

$=42/7$
We get,

$=6$
Hence, the mean is

$6$
Total numbers are

$7$
Since, total number of numbers is odd
Therefore, the median will be the middle number
Here, middle number is

$5$
Therefore, the median is

$5$

Find the mean and the median of :
$1.2,1.9,2.2,2.6$ and $2.9$

Given
Numbers are

$1.2,1.9,2.2,2.6$ and

$2.9$
The mean can be calculated as below
Mean

$=$ (Sum of numbers)

$/$ (Number of numbers)

$=(1.2+1.9+2.2+2.6+2.9)/5$

$=10.8/5$
We get,

$=2.16$
Hence, the mean is

$2.16$
Total numbers are

$5$
Since, total number of numbers is odd
Therefore, the median will be the middle number
Here, middle number is

$2.2$
Therefore, the median is

$2.2$

Find the mean and the median of :
$5,8,10,11,13,16,19$ and $20$

Given
Numbers are

$5,8,10,11,13,16,19$ and

$20$
the mean can be calculated as below
Mean

$=$ (Sum of numbers)

$/$ (Number of numbers)

$=(5+8+10+11+13+16+19+20)/8$

$=102/8$
We get,

$=12.75$
Hence, the mean is

$12.75$
Total numbers are

$8$
Since, total number of numbers is even
Therefore, the median will be the half of the sum of middle numbers
The median can be calculated as below
Median

$=$ (Sum of two middle numbers)

$/2$

$(11+13)/2$

$=24/2$
We get,

$=12$
Therefore, the median is

$12$

The mean of $1, 7, 5, 3, 4$ and $4$ is $m.$ The numbers $3, 2, 4, 2, 3, 3$ and $p$ have mean $m 1$ and median $q.$ Find $p$ and $q$

Mean of

$1, 7, 5, 3, 4$ and

$4 = (1 + 7 + 5 + 3 + 4 + 4)/ 6 = 24/6 = 4$
So,

$m = 4$
Now, given that
The mean of

$3, 2, 4, 2, 3, 3$ and

$p = m -1 = 4 – 1 = 3$
Thus,

$17 + p = 3 \times n$ …. ,where

$n = 7$

$17 + p = 21$

$p = 4$
Arranging the terms in ascending order, we have:

$2, 2, 3, 3, 3, 3, 4, 4$
Mean

$= 4^{th}$ term

$= 3$
Hence,

$q = 3$

Find the median of the given data.

Let

$\bar{x}$ be the mean of n number of observations

$x_1, x_2, x_3,...,x_n$
Mean

$= \frac{x_1 + x_2 + x_3 +... +x_n}{n}$
Therefore,
Mean of given data

$= \frac{30 + 32 + 24 + 34 + 26 + 28 + 30 + 35 + 33 + 25}{10}$

$= \frac{297}{10}$

$= 29.7$
Consider the given data:
30, 32, 24, 34, 26, 28, 30, 35, 33, 25
Let us rewrite the above data in ascending order.
24, 25, 26, 28, 30, 30, 32, 33, 34, 35
There are 10 observations, which is even.
Therefore, median

$= \frac{1}{2} \left [ \left ( \frac{n}{2} \right )^{th} term + \left ( \frac{n}{2} + 1 \right )^{th} term \right ]$

$= \frac{1}{2} \left [ \left ( \frac{10}{2} \right )^{th} term + \left ( \frac{10}{2} + 1 \right )^{th} term \right ]$

$= \frac{1}{2} \left [ (5)^{th} term + ( 5 + 1)^{th} term \right ]$

$= \frac{1}{2}[5^{th} term + 6^{th} term]$

$= \frac{1}{2} [30 + 30]$

$= \frac{1}{3} [60]$

$= 30$

The marks of $20$ students in a test were as follows:
$2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19$ and $20.$
Calculate:
(i) the mean (ii) the median (iii) the mode

Arranging the terms in ascending order:

$2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20$
Number of terms

$= 20$

$∑ x = 2 + 6 + 8 + 9 + 11 + 11 + 12 + 13 + 13 + 14 + 14 + 15 + 15 + 15 + 15 + 16 + 16 + 18 + 19 + 20 = 257$
(i) Mean

$= \dfrac{∑x}{ ∑n} = \dfrac{257}{ 20} = 12.85$
(ii) Median

$=\dfrac{ (10^{th}\ term + 11^{th}\ term)}{ 2}$

$= \dfrac{(13 + 14)}{2}$

$= \dfrac{27}{ 2} = 13.5$
(iii) Mode

$= 15$ since it has maximum frequencies i.e.

$3$

A frequency distribution table for the production of oranges of some farm owners in given below. Find the mean production of oranges by 'assured mean' method.
Production (Thousand rupees) 25 - 30 30 - 35 35 - 40 40 - 45 45 - 50
No of farm owners 20 25 15 10 10

Class (Production in thousand rupees)	Class Mark $x_{i}$	$d_{i} = x_{i} - A$	Frequency (Number of farm owners) $f_{i}$	Frequency * deviation $f_{i} * d_{i}$
25 - 30	27.5	-10	20	-200
30 - 35	32.5	-5	25	-125
35 - 40	37.5 = A	0	15	0
40 - 45	42.5	5	10	50
45 - 50	47.5	10	10	100
			$\sum f_{i} = 80$	$\sum f_{i} d_{i} = -175$

Required Mean

$= A + \dfrac{\sum f_{i} d_{i}}{\sum f_{i}}$

$=37.5 - \dfrac{175}{80}$

$= 37.5 - 2.19$

$= 35.31$ thousand rupees

$= Rs \;35310$
Hence, the mean production of oranges is Rs 35310.

Find the mean and the median of all the positive factors of 72.

Let us find the factors of 72:
72 = 1 x 72
= 2 x 36
= 3 x 24
= 4 x 18
= 6 x 12
= 8 x 9
= 9 x 8
= 12 x 6
= 18 x 4
= 24 x 3
= 36 x 2
= 72 x 1
Hence, the data becomes,
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Mean of the above data set

$= \frac{1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 36 + 72}{12}$
$$ = \frac{195}
{12}$$

$=16.25$
Since the number of observations is 12, which is even, median is given by
Median

$= \frac{1}{2} \left [ \left ( \frac{n}{2} \right )^{th} term + \left ( \frac{n}{2} + 1 \right )^{th} term \right ]$

$= \frac{1}{2} \left [ \left ( \frac{12}{2} \right )^{th} term + \left ( \frac{12}{2} + 1 \right )^{th} term \right ]$

$= \frac{1}{2} [6^{th} term + 7^{th} term]$

$= \frac{1}{2} [8 + 9]$

$= \frac{1}{2} \times 17$

$= 8.5$

Find the mean and median of the data:
35, 48, 92, 76, 64, 52, 51, 63 and 71.
If 51 is replaced by 66, what will be the new median?

Let

$\bar{x}$ be the mean of n number of observations

$x_1, x_2, x_3,...,x_n$
Mean

$= \frac{x_1 + x_2 + x_3 +... +x_n}{n}$
Therefore,
Mean of given data

$= \frac{35 + 48 + 92 + 76 + 64 + 52 + 51 + 63 + 71}{9}$

$= \frac{552}{9}$

$= 61.33$
Rewriting the given data in ascending order:
35, 48, 51, 52, 63, 64, 71, 76, 92
Therefore, median

$= \left ( \frac{n + 1}{2} \right )^{th}$ observation

$\Rightarrow$ Median

$= \left ( \frac{9 + 1}{2} \right )^{th}$ observation

$\Rightarrow$ median

$= \left ( \frac{10}{2} \right )^{th}$ observation

$\Rightarrow$ Median

$= 5^{th}$ observation

$\Rightarrow$ Median

$= 63$
If 51 is replaced by 66, the new set of data in ascending order is
35. 48, 52, 63, 64, 66, 71, 76, 92
Since median

$= 5^{th}$ observation,
We have, new median = 64

Find the mean and median of the data:
35, 48, 92, 76, 64, 52, 51, 63 and 71.

Let

$\bar{x}$ be the mean of n number of observations

$x_1, x_2, x_3,...,x_n$
Mean

$= \frac{x_1 + x_2 + x_3 +... +x_n}{n}$
Therefore,
Mean of given data

$= \frac{35 + 48 + 92 + 76 + 64 + 52 + 51 + 63 + 71}{9}$

$= \frac{552}{9}$

$= 61.33$
Rewriting the given data in ascending order:
35, 48, 51, 52, 63, 64, 71, 76, 92
Therefore, median

$= \left ( \frac{n + 1}{2} \right )^{th}$ observation

$\Rightarrow$ Median

$= \left ( \frac{9 + 1}{2} \right )^{th}$ observation

$\Rightarrow$ median

$= \left ( \frac{10}{2} \right )^{th}$ observation

$\Rightarrow$ Median

$= 5^{th}$ observation

$\Rightarrow$ Median

$= 63$

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetimes
(in hours) 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120
Frequency 10 35 52 61 38 29
Determine the modal lifetimes of the components:

Class interval which has mode = 60 - 80
Here,

$l = 60, f_{1} = 61, f_{0} = 52, f_{2} = 38, h = 20$

$\therefore \textbf{Mode} = l + \left [ \dfrac{f_{1} - f_{0}}{2 \times f_{1} - f_{0} - f_{2}} \right ] \times h$

$= 60 + \left \{ \dfrac{61 + 52}{122 - 52 - 38} \right \} \times 20$

$= 60 + \dfrac{9 \times 20}{32}$

$= 60 + \dfrac{45}{8}$

$= 60 + 5.63$

$\therefore \textbf{Mode} = 65.63$ Hours

The following table shows the ages of the patients admitted in a hospital during a year :
Ages
(in years) 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65
No. of patients 6 11 21 23 14 5
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency?

(i) Here the maximum class interval is 35 - 45. This class interval has mode

$l = 35, f_{1} = 23, f_{0} = 21`, f_{2} = 14, h = 0$

$\therefore \textbf{Mode} = l + \left [ \dfrac{f_{1} - f_{0}}{2 \times f_{1} - f_{0} - f_{2}} \right ] \times h$

$= 35 + \left \{ \dfrac{23 + 21}{46 - 21 - 14} \right \} \times 10$

$= 35 + \dfrac{20}{11}$

$\therefore$ Mode = 36.8 years

(ii) Mean

$(\bar {X})$ : Step Deviation Method:

Age (in years)	N0. of patients. $f_{1}$	Mid- point. $x_{1}$	$u_{1} =$ $\dfrac{a_{1} - 30}{10}$	$f_{1} \times u_{1}$
5 - 15	6	10	-2	-12
15 - 25	11	20	-1	-11
25 - 35	21	30 = 1	0	0
35 - 45	23	40	1	23
45 - 55	14	50	2	28
55 - 65	5	60	3	15
	$n = 80$			43

Here,

$a = 30$

$h = 10, n = 80. \sum _{1}u_{1} = 43$
By step deviation method.

$\bar {X} = a + h \times \dfrac{1}{n} \times \sum f_{1}u_{1}$

$= 30 + 10 \times \dfrac{1}{80} \times (43)$

$= 30 + 5.37$

$= 35.37$

$\therefore \textbf{Mode} = 36.8$

$\therefore \textbf{Mode} = 35.7$
Hence Age group 36 .8 patients are admitted more in number.
Mean age group patients are 35.57

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. interpret the two measures.
No. of Students
per Teacher No. of
States/U.T
15 - 20 3
20 - 25 8
25 - 30 9
30 - 35 10
35 - 40 3
40 - 45 0
45 - 50 0
50 - 55 2

(i) Class interval which has mode 30 - 35
Here,

$l = 30, f_{1} = 10, f_{0} = 9, f_{2} = 3, h = 5$

$\therefore \textbf{Mode} = l + \left [ \dfrac{f_{1} - f_{0}}{2 \times f_{1} - f_{0} - f_{2}} \right ] \times h$

$= 30 + \left \{ \dfrac{10 - 9}{2 \times 10 - 9 - 3} \right \} \times 5$

$= 30 + \left \{ \dfrac{10 - 9}{2 - 9 - 3} \right \} \times 5$

$= 30 + \dfrac{5}{8}$

$\therefore \textbf{Mode} = 30.6$

(ii) Mean

$(\bar{X})$ : Step Deviation Method:

No. of Students	No. of States $f_{1}$	Mid- point, $x_{1}$	$u_{1} =$ $\dfrac{x_{1} - 32.5}{5}$	$f_{1} \times u_{1}$
15 - 20	3	17.5	-3	-9
20 - 25	8	22.5	-2	-16
25 - 30	9	27.5	-1	-9
30 - 35	10	32.5 = 1	0	0
35 - 40	3	37.5	1	3
40 - 45	0	42.5	2	0
45 - 50	0	47.5	3	0
50 - 55	2	52.5	4	8
	$n = 35$			-23

Here,

$a = 32.5, h = 5, \sum f_{1} u_{1} = -23$

By Step Deviation Method:

$\bar{X} = a +h \times \dfrac{1}{n} \times \sum f_{1} u_{1}$

$= 32.5 + 5 \times \dfrac{1}{35} \times (-23)$

$= 32.5 - \dfrac{23}{7}$

$= 32.5 - 3.3$

$= 29.2$

$\therefore$ Mode = 30.6

$\therefore$ Mean = 29.2

We conclude that more number of states has a teacher-student ratio about 30.6
Mean of Age ratio is 29.2.

Write the modal class for the given frequency distribution
Class
interval 1 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30
Frequency 5 12 09 15 14 3

Modal class is 15 - 20. It has highest frequency.

Write the median class for the given frequency distribution.
Class mark 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35
Cumulative
frequency 5 12 18 25 40

Cumulative frequency

$> \dfrac{N}{2} = \dfrac{40}{2} = 20$ is

$25$ .
Median class is

$25 - 30$

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, and the mean monthly expenditure:
Expenditure (?) No. of families
1000 - 1500 24
1500 - 2000 40
2000 - 2500 33
2500 - 3000 28
3000 - 3500 30
3500 - 4000 22
4000 - 4500 16
4500 - 5000 7

1979546_1868941_ans_9fdf6d13646b40e780310d81b88d591b.jpeg

find the mean from mode for data
$x_{i}$ 10 20 30 40 50 60 70 80
$f_{i}$ 2 8 16 26 20 16 7 5

term $(x_{i})$	Frequency $(f)_{i}$	$\|x_{i} - z\|$	$f_{i} \|x_{i} - z \|$
10	2	30	60
20	8	20	160
30	16	10	160
40	26	0	0
50	20	10	200
60	16	20	320
70	7	30	210
80	5	40	200
	$\sum f_{i} = 100$		$\sum f_{i} \| x_{i} - z \| = 130$

Mode

$(z) = 40$
Mean deviation =

$\dfrac{\sum f_{i} | x_{i} - z|}{N}$

$=\dfrac{\sum f_{i} | x_{i} - z |}{\sum f_{i} }$

$\dfrac{1310}{100}$

$=13.1$

find the mean from mode for data
$x{i}$ 3 4 5 6 7 8
$fv$ 2 4 6 3 2 1

Term $(x{i})$	Frequency $(f{i})$	$\|x{i} - z\|$	$f{i} \| x{i} - z \|$
3	2	2	4
4	4	1	4
5	6	0	0
6	3	1	3
7	2	2	4
8	1	3	3
	$\sum f{i} = 18$		$\sum f{i} \| x{i} - z \| = 18$

Mode

$(z) = 5$
and

$N = \sum f{i} = 18$
Mean deviation =

$\dfrac{\sum f{i} | x{i} - z |}{\sum f{i}}$

$\dfrac{18}{18} = 1$

find the mean from mode for data
Income (daily) Number
0-10 4
10-20 8
20-30 9
30-40 10
40-50 7
50-60 5
60-70 4
70-80 3

Class Interval	Frequency $f$	Mid Point $x_{i}$	Frequency Mid Point $f_{i} x_{i}$	$\|x_{i} - \bar{x} \|$	$f_{i} \| x_{i} - \bar {x} \|$
0-10	4	5	20	30.8	123.2
12-20	8	15	120	20.8	166.4
20-30	9	25	225	10.8	97.2
30-40	10	35	350	0.8	8.0
40-50	7	35	351	9.2	64.4
50-60	5	55	275	19.2	96.0
60-70	4	65	260	29.2	116.8
70-80	3	75	22.5	39.2	117.6
	N = 50		$\sum f_{i} x_{i}= 1790$		$\sum f{i} \| x{i} - \bar{x}\| = 789.6$

Mean

$(\bar {x}) = \dfrac{\sum f{i}x{i}}{N} = \dfrac{1790}{50} = 35.8$
Mean deviation =

$\dfrac{\sum f{i} |x{i} - \bar{x}}{N} = \dfrac{789.6}{20} = 15.7912$

find the mean from mode for data
Height (cm) 95-10 105-115 115-125 125-135 135-145 145-155
Number 9 13 26 30 12 10

Class	Frequency $(f_{i})$	$f_{i} x_{i}$	$\|x_{i} - \bar{x}\|$	$\|x_{i} - \bar{x}\|$	$f_{i} \| xv - \bar {x} \|$
95-105	9	100	900	25.3	227.7
105-125	26	120	3120	5.3	137.85
125-135	30	130	3900	4.7	141.0
135-145	12	140	1680	14.7	176.4
145-155	10	150	1500	24.7	247.0

Mean =

$\dfrac{\sum f{i}x{i}}{N} = \dfrac{12530}{100} = 125.3$
Mean deviation =

$\dfrac{\sum f{i} |x{i} - \bar{x}}{N} = \dfrac{1128.8}{100} = 11.28$

find the mean from mode for data
Age Number
16-20 5
21-35 6
26-30 12
31-35 14
36-40 26
41-45 12
46-50 16
51-55 9

Age group

Frequency

(fi)(fi) th term
Median class = class included 50.5 th term

find the mean from mode for data
Marks Number
10-20 3
20-30 4
30-40 7
40-50 8
50-60 2
60-70 1

Class	Frequency $(f_{i})$	Cumulative frequency	mid point $x_{i}$	$$\|x_{i} - M x_{i} - 35.5 $$	$f_{i} \| x_{i} - M \|$
10-20	3	3	15	22.86	68.58
20-30	4	7	25	12.86	51.04
30-40	7	14	35	2.86	20.02
40-50	8	22	45	7.14	57.42
50-60	2	24	55	17.54	34.22
60-70	1	25	65	27.54	27.14
	N = 25				$\|sum f_{i} \|x_{i} - \bar{x} \| = 258.58$

Calculate of Median
=

$\dfrac{N + 1}{2}$ th term

$= \dfrac{25 + 1}{2}$ th term = 13th term

Runs scored by players of a cricket team are as follows:
57, 17, 26, 91, 115, 26, 83, 41, 57, 0, 26
Find their arithmetic mean, median and mode.

(i)For arithmetic mean:

$\because$ Arithmetic mean

$= \frac{Sum\ of\ observations}{No.\ of\ observations}$

$= \frac{Sum\ of\ runs}{No.\ of\ players}$

$= \frac{57 + 17+26+91+115+26+83+41+57+0+26}{11}$

$= \frac{539}{11} = 49$
Thus, arithmetic mean = 49 runs
(ii)For median: Arranging the data ascending order 0, 17, 26,26,26,41,57,57,83,91,115

$\Rightarrow$ Median

$= (\frac{n+1}{2})^{th}$ term

$= (\frac{11+1}{2})^{th}$ term = 6 th term = 41

$\therefore$ Required median = 41

(iii) For mode: Since 26 has maximum frequency so mode = 26
Thus arithmetic mean = 49, Median = 41 and mode = 26

Following frequency distribution represents electricity cost 0f 68 users in a colony. Find mean, median and mode from these data.
Monthly consumption 65-85 85-105 105-125 125-145 145-165 165-185 185-205
Number of users 4 5 13 20 14 8 4

1977672_1879300_ans_e43feece4f8f42f093bed7c1ed39813f.jpg

The following number of goals were scored by a football team in a series of $10$ matches : $2, 3, 4, 5,1, 3, 3, 4, 3$
Find the mean, median and mode of these scores.

The number of goals scored by a football team in a series of 10 matches are as under

$2, 3, 4, 5, 0. 1, 3, 3, 4, 3$
(i) Mean

$\bar{(x)} = \frac{\sum x}{n} = \frac{2 + 3 + 4 + 5 + 0 + 1 3 + 3 + 4 + 3}{10} = \frac{28}{10} = 2.8$
(ii) Median : Arranging the given data in ascending order, we have

$0, 1, 2, 3, 3, 3, 3, 4, 4, 5$
Here

$n = 10$ which is even, so there will be two median class
i. e.

$\left ( \frac{n}{2} \right )$ th,

$\left ( \frac{n}{2} + 1 \right )$ th terms
Here,

$\frac{n}{2} = \frac{10}{2} = 5$

$\therefore$ Median

$= \frac{[5th + 6th]}{2} = \frac{3 + 3}{2} = 3$
(iii) Mode : The number

$3$ occurs most frequently in the data i.e. 4 times. Hence, Mode

$= 3$

Find the mode from the following data:
Variable 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 7 12 18 25 16 14 8

Solution is given by:
2157595_2031923_ans_8320d3dd8abe4e9f8fc5879df2962668.PNG

Find the mean for the following frequency distribution:

Class

Frequency (f)

0-10

5

10-20

8

20-30

15

30-40

16

40-50

6

Solution is given by:
2157603_2031931_ans_23f64ce480d84b28b74a3640edb9784c.PNG

Find out missing frequency Iif the arithmetic mean of the following series is8.
X 20 30 40 50 60 70
F 5 ? 15 10 8 5

X	F	F*X
20	5	100
30	f	30f
40	15	600
50	10	500
60	8	480
70	5	350
	43+f	2030+30f

Mean = Total of F*X / Total of F
44.8 = 2030 + 30f / 43 + f
44.8 (43 + f ) = 2030 + 30f
1926.4 + 44.8f = 2030 + 30f
44.8 f - 30f = 2030 - 1926.4
14.8 f = 103.6
f = 103.6 / 14.8 =7

Calculate the arithmetic mean of the marks secured by students of a class in their weekly test.
Marks 0-5 5-10 10-15 15-20 20-25
No. of students 2 8 6 10 4

2145750_2024952_ans_054fec2e476243fc84aade67af081540.png

The following frequency distribution gives the monthly consumption of electricity of $70$ customers of a certain locality:
Monthly consumptions of Electricity
(in units) Frequency
No. of consumers
$65 - 85$ $4$
$85 - 105$ $5$
$105 - 125$ $13$
$125 - 145$ $22$
$145 - 165$ $14$
$165 - 185$ $8$
$185 - 205$ $4$
Hence find mean and mode

Monthly consumption (units) Class interval	Class mark $(x_{i})$	No. of consumers Frequency $(f_{i})$	$(f_{i}x_{i})$
$65 - 85$	$75$	$04$	$300$
$85 - 105$	$95$	$05$	$475$
$105 - 125$	$115$	$13$	$1495$
$125 - 145$	$135$	$22$	$2970$
$145 - 165$	$155$	$14$	$2170$
$165 - 185$	$175$	$08$	$1400$
$185 - 205$	$195$	$04$	$780$
		$\sum f_{i} = 70$	$\sum f_{i}x_{i} = 9590$

$\therefore$ Mean

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

$= \dfrac {9590}{70}$

$= 137$

$\therefore$ Mean of monthly consumption of electricity is

$137$ units.
Mode:

Original Class	True Class Boundaries	Frequency
$65 - 85$	$64.5 - 84.5$	$4$
$85 - 105$	$84.5 - 104.5$	$5$
$105 - 125$	$104.5 - 124.5$	$13\rightarrow f_{1}$
$125 - 145$	$124.5 - 144.5$	$22\rightarrow f_{m}$
$145 - 165$	$144.5 - 164.5$	$14\rightarrow f_{2}$
$165 - 185$	$164.5 - 184.5$	$8$
$185 - 205$	$184.5 - 204.5$	$4$

Here, maximum frequency

$f_{m} = 22$ .
The corresponding class namely,

$124.5 - 144.5$ is 'modal' class.
Mode

$= L + \left [\dfrac {f_{m} - f_{1}}{2f_{m} - f_{1} - f_{2}}\right ]h$
Where

$L:$ lower limit of modal class

$= 124.5$

$f_{m} :$ maximum frequency

$= 22$

$f_{1}:$ frequency of premodal class

$= 13$

$f_{2} :$ frequency of postmodal clas

$= 14$

$h:$ width of modal class

$= 20$

$\therefore$ Mode

$= L + \left [\dfrac {f_{m} - f_{1}}{2f_{m} - f_{1} - f_{2}}\right ] h$

$= 124.5 + \left [\dfrac {22 - 13}{2(22) - 13 - 14}\right ] 20$

$= 124.5 + \left [\dfrac {9}{17}\right ] 20$

$=135.09$

The distribution below gives the weights of $30$ students of a class. Find the median weight of the students.
Weight (in $kg$ )
$40-45$
$45-50$
$50-55$
$55-60$
$60-65$
$65-70$
$70-75$
Number of students
$2$
$3$
$8$
$6$
$6$
$3$
$2$

Weight (in kg)	Number of students (Frequency)	Cumulative Frequency
40 - 45	2	2
45 - 50	3	5
50 - 55	8	13
55 - 60	6	19
60 - 65	6	25
65 - 70	3	28
70 - 75	2	30 $= n$

We have,

$n = 30$

$\dfrac n2 = 15$

The cumulative frequency just greater than

$\dfrac n2$ is

$19$ and the corresponding class is

$55 – 60$ .

Thus,

$55 – 60$ is the median class such that

$\dfrac{n}{2} = 15, l = 55, f = 6, cf = 13$ , and

$h = 5$

Substituting these values in the formula

Median,

$M = l+\left(\dfrac{\dfrac n2 - cf}{f}\right)\times h$

$M = 55+\left(\dfrac{15-13}{6}\right)\times 5$

$M = 55+\dfrac26\times5 = 55+1.67 = 56.67$

Hence, the median weight

$= 56.67$ kg .

The following frequency distribution gives the monthly consumption of electricity of $70$ consumers of a locality.
Monthly Consumption(Units) No. of Consumers
$65-85$ $04$
$85-105$ $05$
$105-125$ $13$
$125-145$ $22$
$145-165$ $14$
$265-185$ $08$
$185-205$ $04$
Find its mean and median.

Mean:

Monthly consumption (units)\Class interval	Class mark $({x}_{i})$	No. of consumers Frequency ( ${f}_{i}$ )	${f}_{i}{x}_{i}$
$65-85$	$75$	$04$	$300$
$85-105$	$95$	$05$	$475$
$105-125$	$115$	$13$	$1495$
$125-145$	$135$	$22$	$2970$
$145-165$	$155$	$14$	$2170$
$165-185$	$175$	$08$	$1400$
$185-205$	$195$	$04$	$780$
Total		$\sum { { f }_{ i } } =70$	$\sum { { f }_{ i }{ x }_{ i } } =9590$

$\therefore$

$Mean=\overline { x } =\cfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\cfrac { 9590 }{ 70 }$

$=137$

$\therefore$ Mean of monthly consumption of electricity is

$137$ units.

Monthly consumption (Units)	Frequency $({f}_{i})$	Cum. Frequency (less than type)
$65-85$	$04$	$04$
$85-105$	$05$	$09$
$105-125$	$13$	$22\rightarrow c.f$
$125-145$ Median class	$22\rightarrow f$	$44$
$145-165$	$14$	$58$
$165-185$	$08$	$66$
$185-205$	$04$	$70$

Here, total frequency

$=\sum { { f }_{ i } } =N=70$

$\therefore$

$\cfrac{N}{2}=\cfrac{70}{2}$

$=35$
Cumulative frequency is just greater than or equal to

$35$ is

$44$ .

$\therefore$ Corresponding class, namely,

$125-145$ is median class.

$\therefore$ Median

$=L+(\cfrac{N}-c.f)\cfrac{h}{f}$
Here,

$L=125,c.f=22,h=20,f=22$

$\therefore$ Median

$=125+(35-22)\cfrac{20}{22}$

$=125+(13)\cfrac{10}{11}$

$=125+\cfrac{130}{11}$

$=125+11.8$

$=136.8$

$\therefore$ Median monthly consumption of electricity

$=136.8$ units.

Obtain the mean number of bags sold by a shopkeeper on $6$ consecutive days from the following table.
Days Monday Tuesday Wednesday Thursday Friday Saturday
No. of bags sold $55$ $32$ $30$ $25$ $10$ $20$

Mean

$=\cfrac { 55+32+30+25+10+20 }{ 6 } \\ =\cfrac { 172 }{ 6 } \\ \approx 29$

The following table shows the age distribution of cases of a certain disease reported during a year in a particular city.
Age in years $5-14$ $15-24$ $25-34$ $35-44$ $45-54$ $55-64$
No. of cases $6$ $11$ $12$ $10$ $7$ $4$
Obtain the mean, median and mode of the above data.

Convert closed intervals to open intervals.

Age	Mid Point ( ${ x }_{ i }$ )	Cases ( ${ f }_{ i }$ )	Cumulative Frequency (c.f.)
4.5-14.5	9.5	6	6
14.5-24.5	19.5	11	17
24.5-34.5	29.5	12	29
34.5-44.5	39.5	10	39
44.5-54.5	49.5	7	46
54.5-64.5	59.5	4	50=n

$N=50;h=10$

mean

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

$=\cfrac { 9.5(6)+19.5(11)+29.5(12)+39.5(10)+49.5(7)+59.5(4) }{ 50 } =32.1$

$\cfrac { N }{ 2 } =25,l=24.5,c=17,f=12$

median

$=l+\cfrac { h }{ f } \times \left( \cfrac { n }{ 2 } -c \right)$

$=24.5+\cfrac { 10 }{ 12 } \times 8$

$=31.167$

$\approx 31.2$ years

mode

$=L+\cfrac { { f }_{ m }-{ f }_{ m-1 } }{ 2{ f }_{ m }-{ f }_{ m-1 }-{ f }_{ m+1 } } \times n$

$l=24.5;h=10;{ f }_{ m }=12;{ f }_{ m-1 }=11;{ f }_{ m+1 }=10$

mode

$=24.5+\cfrac { 1 }{ 3 } \times 10$

$=27.83$ years

Find the median for the following frequency table.
Value $12$ $13$ $15$ $19$ $22$ $23$
Frequency $4$ $2$ $4$ $4$ $1$ $5$

Values are already in ascending order

Total frequency=

$20$

There are two mid-values

${ 10 }^{ th }$ &

${ \quad 11 }^{ th }$ i.e,

$15$ and

$19$

Median=

$\cfrac { 15+19 }{ 2 } =17$

The median of 230 observations isFind $a$ and $b$ .
Class 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 12 30 $a$ 65 $b$ 25

Class	Frequency $(f)$	Cumulative Frequency $(c.f)$
$0-10$	$12$	$12$
$20-30$	$30$	$42$
$30-40$	$a$	$42+a$
$40-50$	$65$	$107+a$
$50-60$	$b$	$107+a+b$
$60-70$	$25$	$132+a+b$
Total $(N)$	$230$

Given: Median

$=46$ and

$\cfrac N2 = \cfrac {230}{2} = 115$

Hence, median class

$\Rightarrow 40-50$

So,

$L=40, h=10$ ,

$f=65$ and

$c.f = 42+a$

Median

$=L+ \cfrac hf \left(\cfrac N2 - c.f\right)$

$46 = 40 + \cfrac {10}{65}\left(115 - 42-a\right)$

$\cfrac {6 \times 65}{10} = 73 - a$

$\therefore a = 73-39 = 34$

And,

$132+34+b = 230$

$b=64$

Find the mode for the following distribution:
Class-interval Frequency
0-10 5
10-20 8
20-30 7
30-40 12
40-50 28
50-60 20
60-70 10
70-80 10

1203844_1249087_ans_6ef0c5e0450348ec9246516d60fd44e6.jpg

1)Numbers $50,42,35,2 x + 10,2 x - 8,12,11,8$ are written in descending order and their median is $25$ find $x$ .
2)Find the median of data: $15,6,16,8,22,21,9,18,25$ .
3)Find the mean,median and mode of the data: $10,15,18,10,10,20,10,20,15,21,15,25$ .

Numbers

$50, 42, 35, 2x+10, 2x-8, 12, 11, 8$

Medium is means middle.

So median is

$25$ (given)

$\Rightarrow \dfrac{2x+102x-8}{2}=25$

$=\dfrac{4x+2}{2}=25$

$2x+1=25$

$x=12$

Median of

$15, 6, 16, 8, 22, 21, 9, 18, 25 ?$

for median data should be arrenged in any order

So,

$6, 8, 9, 15, 16, 18, 21, 22, 25$

median is

$16$

Find mean, mode and median

$, -10, 15, 18, 10, 10, 20, 10, 20$

mode is

$-10$ (most frequintly used)

mean

$\dfrac{10+15+18+10+10+20+10+20+15+21+15+25}{12}$

$=\dfrac{189}{12}=15.75$

meadian

$= 15$

The following table gives the numbers of branches and number plants in a garden of a school.
No. of branches $2$ $3$ $4$ $5$ $6$
No. of plants $14$ $21$ $28$ $20$ $17$
Calculate the mean, median and mode of the above data.

Mean

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

$=\cfrac { 14(2)+21(3)+28(4)+20(5)+17(6) }{ 100 }$

$=\cfrac{28+63+112+100+102}{100}$

$=\cfrac { 405 }{ 100 } \approx 4$

Therefore, mean of given data is approximately

$4$ branches.

Since plants with

$4$ branches are highest in number which is highest frequency of all the items, the modal number of branches will be

$4$ .
Mode

$=4$

For a given data,

Relation between mean, mode and median is given by :

$\text{mode} = 3 \times \text{median} - 2 \times \text {mean}$

$4= 3\times \text{median}-2 \times 4$

$\text{median} = 4$ .

Therefore, for the given data, mean

$=$ median

$=$

$mode$

$=4$ $.

The table below classifies $60$ students in a class according to their heights.
Height (cm) Number of students
140-145 5
145-150 8
150-155 12
155-160 16
160-165 11
165-170 5
170-175 3
Find the median of the amount paid.

Height (cm

Number of students

Below 145

Below 150

5+8=13

Below 155

13+12=25

Below 160

25+16=41

Below 165

41+11=52

Below 170

52+5=57

Below 175

57+3=60

Height (cm)	Number of students
Below 145	5
Below 150	5+8=13
Below 155	13+12=25
Below 160	25+16=41
Below 165	41+11=52
Below 170	52+5=57
Below 175	57+3=60

Here

$N=\cfrac { 60 }{ 2 } =30$
Median is the value of

$x$ such that

$=30$
From the table, position

$y=30$ is in between

$y=25$ and

$y=41$

$y=25$ corresponds to

$x=155$ and

$y=41$ corresponds to

$x=160$ .
Using proportionately assumption, we have

$\cfrac { x-55 }{ 160-155 } =\cfrac { 30-25 }{ 41-25 }$

$\Rightarrow \cfrac { x-155 }{ 160-155 } =\cfrac { 5 }{ 16 }$

$\Rightarrow x-155=\cfrac { 25 }{ 16 } =1.56$

$\Rightarrow x=155+1.56=156.56$
Thus, median of amount paid is

$156.56$

Find the median for the following data.
Height (ft) $5-10$ $10-15$ $15-20$ $20-25$ $25-30$
No. of trees $4$ $3$ $10$ $8$ $5$

Height	Frequency $({ f }_{ i })$	Cumulative Frequency
5-10	4	4
10-15	3	7
15-20	10	17
20-25	8	25
25-30	5	30
	$\sum { { f }_{ i } } =30$

Median is at $\cfrac { n }{ 2 } =15$ th value

l $=15$ ; h $=5$ ; f $=10$ ; c $=7$

Median $=l+\cfrac { h }{ f } \times (\cfrac { n }{ 2 } -c)$

$=15+\cfrac { 1 }{ 2 } \times (15-7)\\ =19$

The heights of the student of the college are given below:

Height of
students

(in cm)

No. of students

160-163

22

164-167

80

168-171

98

172-175

145

176-179

104

180-183

43

154-187

5

Calculate the height of students such that 75% students are below it.

2041636_1063878_ans_0ffea1fffaa341ef8d2bfd9d4a7dfab3.jpg

The following table shows the distribution of weights of $100$ candidates appearing for a competition. Determine the model weight.
Weight $(in\ kg)$ $50-55$ $55-60$ $60-65$ $65-70$ $70-75$ $75-80$
Number of candidates $13$ $18$ $45$ $16$ $06$ $02$

Weight	No. of candidates
$50-55$	$13$
$55-60$	$10$
$60-65$	$45$
$65-70$	$16$
$70-75$	$06$
$75-80$	$02$

The model weight is

$60-65$

We know,

$mode=l+\frac{f-f_1}{2f-f_1-f_2}\times h$

$f=45$

$f_2=16$

$f_1=10$

$h=5$

$l=60$

$=60+\dfrac{45-10}{2\times 45-16-10}\times 5$

$=60+\dfrac{27}{56}\times 5$

$=60+\dfrac{135}{56}$

$=60+2.4$

$=62.4$

Find the median class of the following distribution.
Weight (in $kg$ ) Number of students
$45-47$ $7$
$47-49$ $5$
$49-51$ $8$
$51-53$ $12$
$53-55$ $10$
$55-57$ $2$
$57-59$ $1$

$\begin{array}{l} N=45 \\ \Rightarrow \frac { N }{ 2 } =\frac { { 45 } }{ 2 } =22.5 \\ \Rightarrow \, \, commulative\, \, frequency\, \, just\, \, more\, \, than\, \, 22.5\, \, is\, \, \left( { 32 } \right) \\ \Rightarrow and\, \, its\, \, { { int } }erval\, \, \left( { { 5^{ 1 } }-{ 5^{ 3 } } } \right) \\ \Rightarrow \therefore l=51,\, \, \, f=12,\, \, \, F=20,\, \, \, i=2 \\ \Rightarrow Median\, \, \left( M \right) =l+\frac { { \left( { \frac { N }{ 2 } -F } \right) } }{ f } \times i \\ =51+\frac { { \left( { 22.5-20 } \right) } }{ { 12 } } \times 2 \\ =51+\frac { { \left( { 2.5 } \right) } }{ 6 } \\ =51+0.42 \\ Median\left( M \right) =51.42\, \, \, Ans. \end{array}$

1205034_1295873_ans_ef0f2dae3a40454a872cbfbc28512a9a.PNG

1205034_1295873_ans_ef0f2dae3a40454a872cbfbc28512a9a.PNG

Calculate the median for the following data frequency
C.I 0-20 20-40 40-60 60-80 80-100 100-120 120-140
Freq. 6 8 10 12 6 5 3

1484548_1540474_ans_539dde3d21d746ba969e65aadf7df3b8.jpg

The runs scored in a cricket match by 11 players are as follows:
$6,15,120,50,100,80,10,15,8,10,10$
Find the mean, mode and median of this data.

2039858_1677442_ans_e4df39cb4ff9426fb4607521dbd8068d.JPG

Calculate the mode for the following frequency distribution table:

1484544_1568059_ans_62fe6ca856764ca39233119b6ab6bc98.jpg

Compute the mode for the following frequency distribution.

Class

Frequency

0-3

3

3-6

4

6-9

5

9-12

2

12-15

1

15-18

6

18-21

7

21-24

3

24-27

5

27-30

2

1484622_1544541_ans_5ff6ddc1cd28458c82203f4a53d997c6.jpg

Calculate the mean and median for the following data:
Marks 10 11 12 13 14 16 19 20
Number of students: 3 5 4 5 2 3 2 1
Using empirical formula, find its mode.

2039860_1677452_ans_7c0b545d6aff478f926bb729e0438209.JPG

The marks of $15$ students in an examination are:
25, 19, 17, 24, 23, 29, 31, 40, 19, 20, 22, 26, 17, 35,Find the median score.

Given that the marks of 15 students in an examination are

$25, 19, 17, 24, 23, 29, 31, 40, 19, 20 ,22, 26 , 17, 35, 21$

first of all arrange the marks in ascending order

$17, 17, 19, 19, 20 , 21, 22, 23, 24, 25, 26, 29, 31, 35, 40$

Here observe that no of student off i.e.

$n = 15$

So median is

$\dfrac{n + 1}{2} th$ term i.e. middle term

median

$= \dfrac{n + 1}{2} th$ term

$= \dfrac{15 + 1}{2} th$ term

$= \dfrac{16}{2} th$ term

$= 8^{th}$ term

Median

$= 23$

$\therefore$ median scole student marks

$=23$

Find the mode and median of the data: $13,16,12,14,19,12,14,13,14$
By using the empirical relation also find the mean.

2039857_1677427_ans_57962ebe34664396aeb65f8e0910f91e.JPG

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :
Determine the median number of letters in the surnames. Find the mean number of letters in surnames. Also, find the modal size of the surnames.

(i)

$n=100, \therefore \dfrac{n}{2}=50$
Class interval having median is

$(7-10)$

$\text{l=7, n=100, f=40, cf=36, h=3}$

$\begin{aligned} \therefore \quad \text { Median } &=l+\left[\dfrac{\dfrac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right] \times \mathrm{h} \\ &=7+\left(\dfrac{50-36}{40}\right) \times 3 \\ &=7+\dfrac{21}{20} \end{aligned}$ :
Median

$=8.05$ Letters
(ii) Class interval which has mode is

$(7-10)$
Maximum frequency,

$l=7, f<\operatorname{sub}>1</ \operatorname{sub}>=40, f<\operatorname{sub}>0</ \operatorname{sub}>=30, f<\operatorname{sub}>2</ \operatorname{sub}>=16, h=3$

$\therefore \operatorname{Mode}=l+\left[\dfrac{f_{1}-f_{0}}{2 \times f_{1}-f_{0}-f_{2}}\right] \times h$

$=7+\left\{\dfrac{40-30}{80-30-16}\right\} \times 3$

$=7+\dfrac{30}{34}$ : Mode

$=7.88$
(iii) Mean

$\bar{X}$ ) : Step Deviation Method:

$\begin{array}{|c|c|c|c|c|}\hline No. of & No. of & Midpoint & \mathrm{u}_{1}= \dfrac{\mathrm{x}_{1}-8.5}{3} & \mathrm{f}_{1} \times \mathrm{u}_{1} \\letters & surnames, \mathrm{f}_{1} & X_1 & \\\hline 1-4 & 6 & 2.5 & -2 & -12 \\ 4-7 & 30 & 5.5 & -1 & -3.0 \\ 7-10 & 40 & 8.5=\mathrm{a} & 0 & 00 \\ 10-13 & 16 & 11.5 & 1 & 16 \\ 13-16 & 4 & 14.5 & 2 & 8 \\ 16-19 & 4 & 17.5 & 3 & 12 \\\hline & \mathrm{n}=100 & & & \sum f_1u_1=-6 \\\hline\end{array}$
Here,

$a=8.5 . \quad h=3, \quad n=100, \quad \Sigma f_1u_{1}=-6$
from Step Deviation Method,

$\begin{aligned} \bar{X} &=a+h \times \dfrac{1}{n} \times \sum f_{1} u_{1} \\ &=8.5+3 \times \dfrac{1}{100} \times(-6) \\ &=8.5-\dfrac{18}{100} \\=& 8.5-0.18 \\=& 8.32 \\ \therefore M e a n &=8.32 \end{aligned}$

$\therefore$ (i) Median

$=8.05$ letters
(ii) Mode

$=7.88$
(iii) Mean

$=8.32 .$

1841201_1869697_ans_2ad2e67e888c48b3b1869cb9cfaed9ef.PNG

Which is the median class and why?

Median class is the class which contains

${ \left( \frac { N }{ 2 } \right) }^{ th }$ term, where

$N$ is total frequency of the data.

In this problem,

$N=200$

$\therefore { \frac { N }{ 2 } =\frac { 200 }{ 2 } }$

$\therefore { \frac { N }{ 2 } =100 }$

In the column of cumulative frequency,

$100$ appears in the class

$70-74$

Thus, the median class is

$70-74$ .

Which is the model class? Why?

Model class is the class which contains highest frequency.

In given problem, highest frequency is

$82$ which belongs to the class

$75-79$

Thus, model class is

$75-79$

Size	$38$	$39$	$40$	$41$	$42$	$43$	$44$
Number of persons(wearing it)	$27$	$40$	$51$	$16$	$14$	$8$	$6$

Temperature in $^oC$	$29$	$32.4$	$34.6$	$36.9$	$38.7$	$40$
Number of days	$7$	$2$	$6$	$4$	$8$	$3$

Marks	Below 10	Below 20	Below 30	Below 40	Below 50
No. of students	8	20	45	58	70

Shirt size(cm)	38	36	40	42	44	37	34	41
Numbers of students	7	12	18	6	2	14	3	8

Size (in cms)	80-85	85-90	90-95	95-100	100-105	105-110	110-115
No. of shirts sold	33	27	85	155	110	45	15

No. of bulbs produced (Thousands)	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
No. of factories	$12$	$35$	$20$	$15$	$8$	$7$	$8$

Class interval	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$	$100 - 120$	$120 - 140$
Frequency	$6$	$8$	$10$	$12$	$6$	$5$	$3$

Weekly wages (in Rs.)	$50-55$	$55-60$	$60-65$	$65-70$	$70-75$	$75-80$	$80-85$	$85-90$
No. of workers	$5$	$20$	$10$	$10$	$9$	$6$	$12$	$8$

Marks in Mathematics	$45$	$60$	$80$	$90$	$95$	$100$
Number of students	$5$	$15$	$6$	$4$	$6$	$4$

Size of shoes	$6$	$7$	$8$	$9$	$10$
Number of student	$10$	$15$	$9$	$4$	$2$

Difference in ages (in years)	0-2	2-4	4-6	6-8	8-10	10-12
No. of couples	1	2	8	5	3	1

Age(in years )	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

Weight	No of students
$40-45$	$2$
$45-50$	$3$
$50-55$	$8$
$55-60$	$6$
$60-65$	$6$
$65-70$	$3$
$70-75$	$2$

Daily wages(Rupees)	$100-110$	$110-120$	$120-130$	$130-140$	$140-150$	$150-160$
No. of workers	$37$	$38$	$40$	$33$	$28$	$24$

Height(in cm)	30-40	40-50	50-60	60-70	70-80	80-90
No. of plants	4	3	8	11	6	2

Height (in inches)	$60$	$61$	$62$	$63$	$64$	$65$	$66$
Number of children	$0$	$10$	$6$	$4$	$18$	$10$	$2$

monthly consumption	no. of consumers
Below 85	4
Below 105	9
Below 125	22
Below 145	42
Below 165	56
Below 185	64

$\text {Family} \,\text {size}$	$1-3$	$3-5$	$5-7$	$7-9$	$9-11$
$\text {Number}\, \,of\, \text {families}$	$7$	$8$	$2$	$2$	$1$

Lifetimes (in hours)	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$
Frequency	$10$	$35$	$52$	$61$	$38$	$29$

Class	25-29	30-34	35-39	40-44	45-49	550-54	555-59
Frequency	14	22	16	6	5	3	4

Statistics - Class 10 Maths - Extra Questions

For a certain frequency distribution, if median =156= 156 and mode =180= 180, then obtain the value of mean

Find the mean of the data whose median is 6464 and mode is 4545.

The demand of different shirt sizes, as obtained by a survey, is given. Calculate the Mode.Size3838393940404141424243434444Number of persons(wearing it)272740405151161614148866

Find the mode for the following table.Temperature in oC^oC292932.432.434.634.636.936.938.738.74040Number of days772266448833

Find the mode for the following frequency table.Wages(Rs.)250250300300350350400400450450500500Number of workers101015151616121211111313

The Mean and Median of a Uni-modal grouped data are 36.536.5 and 38.538.5 respectively. Find the Mode of the data.

Four integers are added to a group of integers 3,4,5,53, 4, 5, 5 and 88 and the mean, median, and mode of the data increases by 11 each. What is the greatest integer in the new group of integers?

Observations of a data are 16,72,0,55,65,55,10,16, 72, 0, 55, 65, 55, 10, and 4141. Chaitanya calculated the mode and median without taking the zero into consideration. Did Chaitanya do the right thing?

If the mean of the following distribution be 5050, then find the values of f1{ f }_{ 1 } and f2{ f }_{ 2 }. Class 0-20 20-40 40-60 60-80 80-100 Total Frequency 17 28 f1{ f }_{ 1 } 24 f2{ f }_{ 2 } 120

Find the mode of the following data:MarksBelow 10Below 20Below 30Below 40Below 50No. of students820455870

Answer the following :The median of 230 observation isFind a and b.Class10-2020-3030-4040-5050-6060-7080-90Frequency1230a65b2518

Find the mode for the following frequency distribution.Class0−1000-100100−200100-200200−300200-300300−400300-400400−500400-500500−600500-600Frequency646462627777626266665454

Find the mode of the following frequency distribution.Age (years)1413121510Numbers of students48729Shirt size(cm)3836404244373441Numbers of students71218621438

Find the median of the following data.xx10102020303040405050ff2233223311

The distribution of sale of shirts sold in a month in a departmental store is as under. Calculate the model class of shirts sold.Size (in cms)80-8585-9090-9595-100100-105105-110110-115No. of shirts sold3327851551104515

Calculate mode of the following data.Marks 0-200-20 20-4020-4040-6040-60 60-8060-80 80-100 80-100Number of students88101012126633

The mode of the following data isFind the missing frequency \chi\chi in it.CLASS0-1010-2020-3030-4040-5050-6060-70FREQUENCY810\chi\chi161267

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 2424.Class0-100-1010-2010-2020-3020-3030-4030-4040-5040-50Frequency552525??181877

The sum of squares of deviations for 1010 observations taken from mean 5050 is 250250. The co-efficient of variations is

The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.No. of bulbs produced (Thousands)30-4030-4040-5040-5050-6050-6060-7060-7070-8070-8080-9080-9090-10090-100No. of factories1212353520201515887788

Find the mean of the following data 30, 32, 24, 34,28, 30, 35, 33, 2530, 32, 24, 34,28, 30, 35, 33, 25(i) Show that the sum of the deviations of all the given observations from the mean is zero.(ii) Find the median of the given data.

Find the median of the following data:Class interval0 - 200 - 2020 - 4020 - 4040 - 6040 - 6060 - 8060 - 8080 - 10080 - 100100 - 120100 - 120120 - 140120 - 140Frequency668810101212665533

Finding median graphically for the following grouped data ?Class Interval Frequency 0-5035-100610-151115-200820-2506

The following table given the weekly wages (in Rs.) of workers in a factory:Weekly wages (in Rs.)50-5550-5555-6055-6060-6560-6565-7065-7070-7570-7575-8075-8080-8580-8585-9085-90No. of workers552020101010109966121288Calculate:the modal class.

If mean = 60mean = 60 and median = 50median = 50, then find mode using empirical relationship.

Find the mode when median is 12 and mean is 16 of a data.

Find the modeMarks1-56-1011-1616-2021-25No. of Students71063224

Find the mode of the following distributionMarks in Mathematics 45456060808090909595100100Number of students55151566446644

Find the mode of the following distributionSize of shoes 667788991010Number of student10101515994422

The following table shows the difference of the ages 2020 couples. Find the mean by direct method.Difference in ages (in years)0-22-44-66-88-1010-12No. of couples128531

The following table shows the ages of the patients admitted in a hospital during a year:Age(in years )5-1515-2525-3535-4545-5555-65Number of patients6112123145Find the mode and the mean of the data given above. Compare and interpret the two measure of central tendency.

Find the median :WeightNo of students40-4540-452245-5045-503350-5550-558855-6055-606660-6560-656665-7065-703370-7570-7522

Find the mode of Class0-50-55-105-1010-1510-1515-2015-2020-2520-25Frequency223310101144

The following table shows the number of workers in a factory and their daily wages. Find the median of the daily wages.Daily wages(Rupees)100-110100-110110-120110-120120-130120-130130-140130-140140-150140-150150-160150-160No. of workers 373738384040333328282424

Find the mode of following data:CI5-1515-2525-3535-4545-5555-65f6112123145

Find the mode of the following distribution:Height(in cm)30-4040-5050-6060-7070-8080-90No. of plants4381162

Find the mode of the following distributionHeight (in inches)6060616162626363646465656666Number of children00101066441818101022

Find the median marks for the following.Marks25-3525-3535-4535-4545-5545-5555-6555-6565-7565-7575-8575-85No. of students77313130301818121222

Find the median of the following data:monthly consumption no. of consumersBelow 85 4Below 1059Below 12522Below 14542Below 16556Below 18564

Find the mode of the following frequency distribution.Class0-1010-2020-3030-4040-5050-6060-70Frequency81010161267

Write the formula to find the median of a grouped data and explain the alphabet in it.

Find the median of the following data3,1,5,6,3,4,53,1,5,6,3,4,5

If the mode of a distribution is 88 and mean is also 88, then find median.

If median = 137 units and mean = 137.05 units, then find the mode.

Find the mean of the following frequency distribution:Class0-2020-4040-6060-8080-100Frequency1518212917

The following data gives the information on the observed lifetimes (in hours) of 225225 electrical components:Lifetimes (in hours)0-200-2020-4020-4040-6040-6060-8060-8080-10080-100100-120100-120Frequency101035355252616138382929Determine the modal lifetimes of the components.

Write the modal class for the following frequency distribution:Class interval10-2010-2020-3020-3030-4030-4040-5040-5050-6050-6060-7060-70Frequency333338386565525219194848

A data constitutes of heights (in cm) of 5050 children. What do you understand by mode for the data?

Find the mean of the following frequency distribution:Class 25-2930-3435-3940-4445-49550-54555-59Frequency1422166534

Find the median class of the following distribution:Class0-100-1010-2010-2020-3020-3030-4030-4040-5040-5050-6050-6060-7060-70Frequency444488101012128844

Find the mean of the following frequency distribution:Class interval50-6060-7070-8080-9090-100Frequency86121113

Find the mean of the following frequency distribution:Class interval15-2525-3535-4545-5555-65Frequency6035221815

Find the mean of the following data:Classes0-100100-200200-300300-400400-500Frequency6915128

Find the mean of the following data:Classes0-1010-2020-3030-4040-50Frequency35953

A hockey player has scored the following number of goals in 99 matches.$$5, 4, 0, 2, 2, 4, 4, 3, 3$$.Find the mean, median, and mode of the data.

The arithmetic mean of the following frequency distribution is 25.Determine the value of p:Class0-1010-2020-3030-4040-50Frequency78p154

Class Interval0-66-1212-1818-2424-30Frequency681097If mean of above distribution is 15.45 15.45 , If true then enter 11 and if false then enter 00

Calculate the missing frequency ff from the following distribution, it is being given that the median of the distribution is 2424.Class0-1010-2020-3030-4040-50Frequency525f187

Find the median for the following frequency distribution:Class Interval0-1010-2020-3030-4040-5050-60Frequency691421910

Find the median for the following frequency distribution:Class Interval25-3535-4545-5555-6565-75Frequency2025574Give answer upto the nearest integer value.

The.following table gives the distribution of IQ of 100 students. Find the median IQ.IQFrequency75-84885-941195-10426105-11431115-12418125-1344135-1442

Find the median from the following data:Class groupsFrequency110-1206120-13025130-14048140-15072150-160116160-17060170-18038180-19022190-2003mean is 153.8If true then enter 11 and if false then enter 00

The following table gives the marks obtained by 50 students in a class test :MarksNo. of students11-15216-20321-25626-30731-351436-401241-45446-502Find the median

Find the value of, the missing frequency, If the mean of the following distribution is 67.Class IntervalFrequency25-351035-45645-55455-65f65-75475-851285-9526Mean is 23.71If true then enter 11 and if false then enter 00

Calculate the mode for the following frequency distribution.ClassFrequency0-10510-20820-30730-401240-502850-602060-701070-8010mode is 4.67If true then enter 11 and if false then enter 00

Compute the mode of the following data:Class IntervalFrequency1-536-10811-151316-201821-252826-302031-351336-40841-45646-504Mode 23.2223.22.If true then enter 11 and if false then enter 00

The following frequency distribution gives the monthly consumption of electricity of 6868 consumers of a locality. Find the median, mean and mode of the data and compare them.Monthly consumption (in units)Number of consumers65-85485-1055105-12513125-14520145-16514165-1858185-2054

The following number of goals were scored by a team in a series of 1010 matches:22,33,44,55,00,11,33,33,44,33Find the mean, median and mode of these scores.

The following table shows the ages of the patients admitted in a hospital during a year.Age (in years)No. of patients5-15615-251125-352135-452345-551455-655Find the mode and the mean of the data given above. Compare and interpret the measures of central tendency.

If the mode of the following distribution is given by MM, then find the value of a nearest integer.Daily Wages31 - 3637 - 4243 - 4849 - 5455 - 6061 - 66No.of workers612201594

The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospitalAge (in Years)5 - 1415 - 2425 - 3435 - 4445 - 5455 - 64No. of Cases6112123145Find the nearest integer to the modal age.

The daily wages (in rupees) of 100100 workers in a factory are given belowDaily wages (in Rs.)125125130130135135140140145145150150160160180180No. of workers662020242428281515442211Find the median wages of a worker (in the nearest integer value).

Calculate the median for the following distribution classClass0 - 1010 - 2020 - 3030 - 4040 - 5050 - 60Frequency51020785

If the median of the following frequency distribution is 4646. find the absolute difference of missing frequenciesVariable10-2010-2020-3020-3030-4030-4040-5040-5050-6050-6060-7060-7070-8070-80TotalFrequency12123030f_{1}f_{1}6565f_{2}f_{2}25251818229229

Calculate the median from the following distributionClass5-1010-1515-2020-2525-30Frequency471094

Calculate the mean deviation about median age for the age distribution of 100100 persons given below :AgeNumber16-20521-25626-301231-351436-402641-451246-501651-559

Find the mean deviation about the median for the data \displaystyle X_{i} \displaystyle X_{i} 1521273035 \displaystyle f_{i} \displaystyle f_{i} 35678

Find the mean deviation about the median for the data \displaystyle x_{i} \displaystyle x_{i} 579101215 \displaystyle f_{i} \displaystyle f_{i} 862226

Find the mean deviation about median for the following data:MarksNumber of girls0-10610-20820-301430-401640-50450-602

For a certain frequency distribution, if median $= 156$ and mode $= 180$ , then obtain the value of mean

Find the mean of the data whose median is $64$ and mode is $45$ .

The demand of different shirt sizes, as obtained by a survey, is given. Calculate the Mode.
Size $38$ $39$ $40$ $41$ $42$ $43$ $44$
Number of persons(wearing it) $27$ $40$ $51$ $16$ $14$ $8$ $6$

Find the mode for the following table.
Temperature in $^oC$ $29$ $32.4$ $34.6$ $36.9$ $38.7$ $40$
Number of days $7$ $2$ $6$ $4$ $8$ $3$

Find the mode for the following frequency table.
Wages(Rs.) $250$ $300$ $350$ $400$ $450$ $500$
Number of workers $10$ $15$ $16$ $12$ $11$ $13$

The Mean and Median of a Uni-modal grouped data are $36.5$ and $38.5$ respectively. Find the Mode of the data.

Four integers are added to a group of integers $3, 4, 5, 5$ and $8$ and the mean, median, and mode of the data increases by $1$ each. What is the greatest integer in the new group of integers?

Observations of a data are $16, 72, 0, 55, 65, 55, 10,$ and $41$ . Chaitanya calculated the mode and median without taking the zero into consideration. Did Chaitanya do the right thing?

If the mean of the following distribution be $50$ , then find the values of ${ f }_{ 1 }$ and ${ f }_{ 2 }$ .

Class 0-20 20-40 40-60 60-80 80-100 Total

Frequency 17 28 ${ f }_{ 1 }$ 24 ${ f }_{ 2 }$ 120

Find the mode of the following data:
Marks Below 10 Below 20 Below 30 Below 40 Below 50
No. of students 8 20 45 58 70

Answer the following :
The median of 230 observation isFind a and b.
Class 10-20 20-30 30-40 40-50 50-60 60-70 80-90
Frequency 12 30 a 65 b 25 18

Find the mode for the following frequency distribution.
Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$ $500-600$
Frequency $64$ $62$ $77$ $62$ $66$ $54$

Find the mode of the following frequency distribution.
Age (years) 14 13 12 15 10
Numbers of students 4 8 7 2 9
Shirt size(cm) 38 36 40 42 44 37 34 41
Numbers of students 7 12
18 6 2 14 3 8

Find the median of the following data.
$x$ $10$ $20$ $30$ $40$ $50$
$f$ $2$ $3$ $2$ $3$ $1$

The distribution of sale of shirts sold in a month in a departmental store is as under. Calculate the model class of shirts sold.

Size (in cms) 80-85 85-90 90-95 95-100 100-105 105-110 110-115
No. of shirts sold 33 27 85 155 110 45 15

Calculate mode of the following data.
Marks $0-20$ $20-40$
$40-60$ $60-80$
$80-100$
Number of students $8$ $10$ $12$ $6$ $3$

The mode of the following data isFind the missing frequency $\chi$ in it.
CLASS 0-10 10-20 20-30 30-40 40-50 50-60 60-70
FREQUENCY 8 10 $\chi$
16 12 6 7

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is $24$ .
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
Frequency $5$ $25$ $?$ $18$ $7$

The sum of squares of deviations for $10$ observations taken from mean $50$ is $250$ . The co-efficient of variations is

The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.
No. of bulbs produced (Thousands) $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$
No. of factories $12$ $35$ $20$ $15$ $8$ $7$ $8$

Find the mean of the following data $30, 32, 24, 34,28, 30, 35, 33, 25$
(i) Show that the sum of the deviations of all the given observations from the mean is zero.
(ii) Find the median of the given data.

Find the median of the following data:
Class interval $0 - 20$ $20 - 40$ $40 - 60$ $60 - 80$ $80 - 100$ $100 - 120$ $120 - 140$
Frequency $6$ $8$ $10$ $12$ $6$ $5$ $3$

Finding median graphically for the following grouped data ?
Class Interval Frequency
0-5 03
5-10 06
10-15 11
15-20 08
20-25 06

The following table given 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 modal class.

If $mean = 60$ and $median = 50$ , then find mode using empirical relationship.

Find the mode
Marks 1-5 6-10 11-16 16-20 21-25
No. of Students 7 10 6 32 24

Find the mode of the following distribution
Marks in Mathematics $45$ $60$ $80$ $90$ $95$ $100$
Number of students $5$ $15$ $6$ $4$ $6$ $4$

Find the mode of the following distribution
Size of shoes $6$ $7$ $8$ $9$ $10$
Number of student $10$ $15$ $9$ $4$ $2$

The following table shows the difference of the ages $20$ couples. Find the mean by direct method.
Difference in ages (in years) 0-2 2-4 4-6 6-8 8-10 10-12
No. of couples 1 2 8 5 3 1

The following table shows the ages of the patients admitted in a hospital during a year:
Age(in years ) 5-15 15-25 25-35 35-45 45-55 55-65
Number of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measure of central tendency.

Find the median :
Weight No of students
$40-45$ $2$
$45-50$ $3$
$50-55$ $8$
$55-60$ $6$
$60-65$ $6$
$65-70$ $3$
$70-75$ $2$

Find the mode of
Class $0-5$ $5-10$ $10-15$ $15-20$ $20-25$
Frequency $2$ $3$ $10$ $1$ $4$

The following table shows the number of workers in a factory and their daily wages. Find the median of the daily wages.
Daily wages(Rupees) $100-110$ $110-120$ $120-130$ $130-140$ $140-150$ $150-160$
No. of workers $37$ $38$ $40$ $33$ $28$ $24$

Find the mode of following data:

CI 5-15 15-25 25-35 35-45 45-55 55-65
f 6 11 21 23 14 5

Find the mode of the following distribution:
Height(in cm) 30-40 40-50 50-60 60-70 70-80 80-90
No. of plants 4 3 8 11 6 2

Find the mode of the following distribution

Height (in inches) $60$ $61$ $62$ $63$ $64$ $65$ $66$
Number of children $0$ $10$ $6$ $4$ $18$ $10$ $2$

Find the median marks for the following.
Marks $25-35$ $35-45$ $45-55$ $55-65$ $65-75$ $75-85$
No. of students $7$ $31$ $30$ $18$ $12$ $2$

Find the median of the following data:
monthly consumption no. of consumers
Below 85 4
Below 105 9
Below 125 22
Below 145 42
Below 165 56
Below 185 64

Find the mode of the following frequency distribution.
Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 8 10 10 16 12 6 7

Find the median of the following data
$3,1,5,6,3,4,5$

If the mode of a distribution is $8$ and mean is also $8$ , then find median.

Find the mean of the following frequency distribution:
Class 0-20 20-40 40-60 60-80 80-100
Frequency 15 18 21 29 17

The following data gives the information on the observed lifetimes (in hours) of $225$ electrical components:
Lifetimes (in hours) $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$
Frequency $10$ $35$ $52$ $61$ $38$ $29$
Determine the modal lifetimes of the components.

Write the modal class for the following frequency distribution:
Class interval $10-20$
$20-30$ $30-40$ $40-50$ $50-60$ $60-70$
Frequency $33$ $38$ $65$ $52$ $19$ $48$

A data constitutes of heights (in cm) of $50$ children. What do you understand by mode for the data?

Find the mean of the following frequency distribution:
Class 25-29 30-34 35-39 40-44 45-49 550-54 555-59
Frequency 14 22 16 6 5 3 4

Find the median class of the following distribution:
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
Frequency $4$ $4$ $8$ $10$ $12$ $8$ $4$

Find the mean of the following frequency distribution:
Class interval 50-60 60-70 70-80 80-90 90-100
Frequency 8 6 12 11 13

Find the mean of the following frequency distribution:
Class interval 15-25 25-35 35-45 45-55 55-65
Frequency 60 35 22 18 15

Find the mean of the following data:
Classes 0-100 100-200 200-300 300-400 400-500
Frequency 6 9 15 12 8

Find the mean of the following data:
Classes 0-10 10-20 20-30 30-40 40-50
Frequency 3 5 9 5 3

A hockey player has scored the following number of goals in $9$ matches.$$5, 4, 0, 2, 2, 4, 4, 3, 3$$.Find the mean, median, and mode of the data.

The arithmetic mean of the following frequency distribution is 25.Determine the value of p:
Class
0-10
10-20
20-30
30-40
40-50
Frequency
7
8
p
15
4

Class Interval
0-6
6-12
12-18
18-24
24-30
Frequency
6
8
10
9
7
If mean of above distribution is $15.45$ , If true then enter $1$ and if false then enter $0$

Calculate the missing frequency $f$ from the following distribution, it is being given that the median of the distribution is $24$ .
Class
0-10
10-20
20-30
30-40
40-50
Frequency
5
25
f
18
7

Find the median for the following frequency distribution:
Class Interval
0-10
10-20
20-30
30-40
40-50
50-60
Frequency
6
9
14
2
19
10

Find the median for the following frequency distribution:

Class Interval
25-35
35-45
45-55
55-65
65-75
Frequency
20
25
5
7
4
Give answer upto the nearest integer value.

The.following table gives the distribution of IQ of 100 students. Find the median IQ.
IQ
Frequency
75-84
8
85-94
11
95-104
26
105-114
31
115-124
18
125-134
4
135-144
2

Find the median from the following data:
Class groups
Frequency
110-120
6
120-130
25
130-140
48
140-150
72
150-160
116
160-170
60
170-180
38
180-190
22
190-200
3
mean is 153.8
If true then enter $1$ and if false then enter $0$

The following table gives the marks obtained by 50 students in a class test :
Marks
No. of students
11-15
2
16-20
3
21-25
6
26-30
7
31-35
14
36-40
12
41-45
4
46-50
2
Find the median

Find the value of, the missing frequency, If the mean of the following distribution is 67.
Class Interval
Frequency
25-35
10
35-45
6
45-55
4
55-65
f
65-75
4
75-85
12
85-95
26
Mean is 23.71
If true then enter $1$ and if false then enter $0$

Calculate the mode for the following frequency distribution.
Class
Frequency
0-10
5
10-20
8
20-30
7
30-40
12
40-50
28
50-60
20
60-70
10
70-80
10
mode is 4.67
If true then enter $1$ and if false then enter $0$

Compute the mode of the following data:
Class Interval
Frequency
1-5
3
6-10
8
11-15
13
16-20
18
21-25
28
26-30
20
31-35
13
36-40
8
41-45
6
46-50
4
Mode $23.22$ .
If true then enter $1$ and if false then enter $0$

The following frequency distribution gives the monthly consumption of electricity of $68$ consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption (in units)
Number of consumers
65-85
4
85-105
5
105-125
13
125-145
20
145-165
14
165-185
8
185-205
4

The following number of goals were scored by a team in a series of $10$ matches:
$2$ , $3$ , $4$ , $5$ , $0$ , $1$ , $3$ , $3$ , $4$ , $3$
Find the mean, median and mode of these scores.

The following table shows the ages of the patients admitted in a hospital during a year.
Age (in years)
No. of patients
5-15
6
15-25
11
25-35
21
35-45
23
45-55
14
55-65
5
Find the mode and the mean of the data given above. Compare and interpret the measures of central tendency.

If the mode of the following distribution is given by $M$ , then find the value of a nearest integer.
Daily Wages 31 - 36 37 - 42 43 - 48 49 - 54 55 - 60 61 - 66
No.of workers 6 12 20 15 9 4

The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital
Age (in Years) 5 - 14 15 - 24 25 - 34 35 - 44 45 - 54 55 - 64
No. of Cases 6 11 21 23 14 5
Find the nearest integer to the modal age.

The daily wages (in rupees) of $100$ workers in a factory are given below
Daily wages (in Rs.) $125$ $130$ $135$ $140$ $145$ $150$ $160$ $180$
No. of workers $6$ $20$ $24$ $28$ $15$ $4$ $2$ $1$
Find the median wages of a worker (in the nearest integer value).

Calculate the median for the following distribution class
Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60
Frequency 5 10 20 7 8 5

If the median of the following frequency distribution is $46$ . find the absolute difference of missing frequencies
Variable $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Total
Frequency $12$ $30$ $f_{1}$ $65$ $f_{2}$ $25$ $18$ $229$

Calculate the median from the following distribution
Class 5-10 10-15 15-20 20-25 25-30
Frequency 4 7 10 9 4

Calculate the mean deviation about median age for the age distribution of $100$ persons given below :
Age Number
16-20 5
21-25 6
26-30 12
31-35 14
36-40 26
41-45 12
46-50 16
51-55 9

Find the mean deviation about the median for the data
$\displaystyle X_{i}$ 15 21 27 30 35
$\displaystyle f_{i}$ 3 5 6 7 8

Find the mean deviation about the median for the data
$\displaystyle x_{i}$ 5 7 9 10 12 15
$\displaystyle f_{i}$ 8 6 2 2 2 6

Find the mean deviation about median for the following data:
Marks Number of girls
0-10 6
10-20 8
20-30 14
30-40 16
40-50 4
50-60 2

Find the median for the following frequency distribution table:
Class-interval $0-5$ $5-10$ $10-15$ $15-20$ $20-25$ $25-30$
Frequency $5$ $3$ $9$ $10$ $8$ $5$

Marks scored by Hari in $5$ test (out of $25$ marks) are given:
$24, 22, 23, 23, 25$ . Find his average score

Find the median for the following frequency distribution table:
Class Intervals $110-119$ $120-129$ $130-139$ $140-149$ $150-159$ $160-169$
Frequency $6$ $8$ $15$ $10$ $6$ $5$

If the mean of the following frequency distribution is $7.2$ find value of $'K'$ .
$x$ $2$ $4$ $6$ $8$ $10$ $12$
$f$ $4$ $7$ $10$ $16$ $K$ $3$

IQ	Frequency
75-84	8
85-94	11
95-104	26
105-114	31
115-124	18
125-134	4
135-144	2

Class groups	Frequency
110-120	6
120-130	25
130-140	48
140-150	72
150-160	116
160-170	60
170-180	38
180-190	22
190-200	3

Age (in years)	No. of patients
5-15	6
15-25	11
25-35	21
35-45	23
45-55	14
55-65	5

Age (in Years)	5 - 14	15 - 24	25 - 34	35 - 44	45 - 54	55 - 64
No. of Cases	6	11	21	23	14	5

Variable	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	Total
Frequency	$12$	$30$	$f_{1}$	$65$	$f_{2}$	$25$	$18$	$229$

Age	Number
16-20	5
21-25	6
26-30	12
31-35	14
36-40	26
41-45	12
46-50	16
51-55	9

Class Intervals	$110-119$	$120-129$	$130-139$	$140-149$	$150-159$	$160-169$
Frequency	$6$	$8$	$15$	$10$	$6$	$5$

Class - Interval	Frequency
$0 - 4$	$3$
$5 - 9$	$5$
$10 - 14$	$7$
$15 - 19$	$4$
$20 - 24$	$6$
	$N = 24$

Weight (kg)	$50$	$65$	$75$	$90$	$110$	$120$
No of parcels	$25$	$34$	$38$	$40$	$47$	$16$

Bowling Speed $(km/hr)$	$85-100$	$100-115$	$115-130$	$130-145$
Number of players	$9$	$11$	$8$	$5$