Class 11 Commerce Economics - Measures Of Central Tendency

Ina retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
mangoes
50-52
53-55
56-58
59-61
62-64
No. of boxes
15
110
135
115
25
Mean number of mangoes kept in a packing box is 57.19
If true then enter $1$ and if false then enter $0$

Number of mangoes	Number of boxes $f_i$	Class mark $x_i$	$u_i=\frac {x_i-57}{3}$	$f_i\times u_i$
50-52	15	51	-2	-30
53-55	110	54	-1	-110
56-58	135	57 = a	0	0
59-61	115	60	1	115
62-64	25	63	2	50
Total	$n=400$			25

$a=57, h=2, n=400$ and

$\sum f_iu_i=25$ .
By step deviation method,

$mean=a+h\times \frac {1}{n}\times \sum f_iu_i=57+2\times \frac {1}{400}\times 25=57.19$

Find the mean of following distribution with step-deviation method to the nearest integer
Class 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40
Frequency 5 6 8 12 6 3

Class	$\displaystyle x_{i}$	$\displaystyle f_{i}$	Let A = $\displaystyle U_{i}=\frac{x_{i}-27.5}{5}$	$\displaystyle f_{i}u_{i}$
10 - 15	12.5	5	-3	-15
15 - 20	17.5	6	-2	-12
20 - 25	22.5	8	-1	-8
25 - 30	27.5	12	0	0
30 - 35	32.5	6	1	6
35 - 40	37.5	3	2	6
		$\displaystyle \sum f_{i}=40$		$\displaystyle \sum f_{i}u_{i}=-23$

$\displaystyle \Rightarrow \overline{x}=A+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right )h\: \: \Rightarrow \: \overline{x}=27.5+5\times \left ( \frac{-23}{40} \right )=24.625$

Find the mode of the given set of numbers-
$1, 1, 2, 4, 3, 2, 1, 2, 2, 4$

Arranging the numbers with the same values together, we get

$1, 1, 1, 2, 2, 2, 2, 3, 4, 4$
Mode of this data is

$2$ because it occurs more frequently than other observations.

Following are the margins of victory in the football matches of a league.
$1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 1, 1, 2, 3, 2, 6, 4, 2, 1, 1, 4, 2, 1, 5, 3, 3, 2, 3, 2, 4, 2, 1, 2, 3$ , find the mode of this data.

Let us put the data in a tabular form-
Looking at the table, we can quickly say that

$2$ is the mode since

$2$ has occured the highest number of times. thus, most of the matches have been won with a victory margin of

$2$ goals.

As the most no. of matches have been won with a victory margin of

$2$ goals, thus the mode of the given data is

$2$ .

Hence the correct answer is

$2$ .

1068366_403754_ans_8490786eeb384a95ab2d12a372a365ef.JPG

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.

To find the class mark

$\left( { x }_{ i } \right)$ for each interval, using the following relation:

$\left( { x }_{ i } \right) =\displaystyle\frac { Upper class limit + Lower class limit }{ 2 }$
Taking 17 as assumed mean

$\left( a \right)$ , calculating

${ d }_{ i }$ and

${ f }_{ i }{ d }_{ i }$ as:
From the table:

$\sum { { f }_{ i } } =40$

$\sum { { f }_{ i }{ d }_{ i } } =-181$
Mean,

$\bar { x } =a+\displaystyle\frac { \sum { { f }_{ i }{ d }_{ i } } }{ \sum { { f }_{ i } } }$

$=17+\left( \displaystyle\frac { -181 }{ 40 } \right)$

$=17-4.525$

$=12.475$

$\approx 12.48$
Therefore, the mean number of days

$= 12.48$ days for which a student was absent.
1027796_426993_ans_7eb716dfc8c640da8aade97115ca1241.png

Find the median of the data: $15, 22, 9, 20, 6, 18, 11, 25, 14$

Data in descending order is

$6,9,11,14,15,18,20,22,25$

Thus Median

$=15$

Find the mode for the following data:
$22, 36, 18, 22, 20, 34, 22, 42, 46,42$

Since mode has highest frequency

so, Mode

$=22$

Find the median of the data: $32, 30, 28, 31, 22, 26, 27, 21$

Arranging in descending order, we obtain

$32, 31, 30, 28, 27, 26, 22, 21$ .
There are

$8$ terms. Therefore, median is the average of the two middle terms, middle terms, which are

$27$ and

$28$ . Thus the median is

$\dfrac {(27 + 28)}{2} = 27.5$ .

Find the mode for the following data:
$4, 3, 1, 5, 3, 7, 9, 6$

Mode of any data is the values which has the most frequency,

Clearly in the given data

$3$ has occurred

$2$ times and none of the other number is repeated upto the frequency of

$3$

Hence the mode of above data is

$3$

Find the median of the data: $26, 31, 33, 37, 43, 8, 26, 33$

Arrange data is as follows:

$8,26,26,31,33,33,37,43$

so, median

$=\cfrac{31+33}{2}=32$

The following numbers are the sizes of shoes sold by a shop in a particular day. Find the mode.
$6, 7, 8, 9, 10, 6, 7, 10, 7, 6, 7, 9, 7, 6$

Given The following numbers are the sizes of shoes sold by a shop in a particular day.

$6,7,8,9,10,6,7,10,7,6,7,9,7,6$

Arrenge these number in ascending order

$6,6,6,6,7,7,7,7,7,8,9,9,10,10$

Then shoes number Frequency (f)

$6$

$4$

$7$

$5$

$8$

$1$

$9$

$2$

$10$

$2$

Frequency of shoes number

$7$ is more than any other shoes number, therefore mode is

$7$

Find the mode of the data: $5, 3, 3, 5, 7, 6, 3, 4, 3, 5, 8, 5$

Here

$3$ and

$5$ appear

$4$ times. Therefore modes are both

$3$ and

$5$ .

Find the mode for the data: $15, 20, 22, 25, 30, 20, 15, 20, 12, 20$

Here

$20$ appear maximum times (

$4$ times). Therefore, mode is

$20$ .

Calculate mean $(\overline{X})$ , when $\sum f_ix_i=100$ and $\sum f_i=20$ .

Mean

$(\overline{X})=\displaystyle\frac{\sum f_ix_i}{\sum f_i}$

$=\displaystyle\frac{100}{20}=5$ .

Find the mode of $5, 6, 9, 6, 12, 3, 6, 11, 6, 7$ .

Mode is the quantity which occurs mostly
Clearly,

$6$ occurs more times than any other items. Hence,

$6$ is mode of given data.

Find the mode of $22, 25, 21, 22, 29, 25, 34, 37, 30, 22, 29, 25$ .

Here

$22$ occurs

$3$ times and

$25$ also occurs

$3$ times.

$\therefore$ Both

$22$ and

$25$ are the modes for this data. We observe that there are two modes for the given data.

If $\sum f_{i}x_{i} = 75$ and $\sum f_{i} = 15$ , then find the mean $\overline {x}$

Given

$\sum f_{i}x_{i} = 75$ and

$\sum f_{i} = 15$
Mean

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

$= \dfrac {75}{15} = 5$

Calculate the arithmetic mean of given frequency distribution table.
Class-interval $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency $7$ $14$ $28$ $26$ $16$ $9$

class interval	frequency (f)	mean value (x)	$fx$
0 - 10	7	5	35
10 - 20	14	15	210
20 - 30	28	25	700
30 - 40	26	35	910
40 - 50	19	45	720
50 - 60	9	55	495
Sum	100		3070

$A.M. = \dfrac {\sum f_{i}x_{i}}{\sum f_{i}} = \dfrac {3070}{100} = 30.7$

Find the mode of following numbers $6, 3, 2, 6, 5, 6$ and $8$ .

The given numbers are $6, 3, 2, 6, 5, 6$ and $8$

Arrange the given numbers in increasing order : $2,3,5,6,6,6,8$

Mode is the number which occur most frequently in a set of numbers.

Since, $6$ is occurring $3$ times in the set

$\Rightarrow Mode = 6$

The following table gives the number of children of $150$ families in a village.
No. of children ( $x$ ) $0$ $1$ $2$ $3$ $4$ $5$
No. of families ( $f$ ) $10$ $21$ $55$ $42$ $15$ $7$
Find the average number of children per family.

Average (mean)

$=\dfrac {\sum x_{i}f_{i} } {\sum f_{i}}$

$\Rightarrow$ average fat number of children per family

$=\dfrac {(0×10)+(1×21)+(2×55)+(3×42)+(4×15)+(5×7)}{150}$

$\Rightarrow$ average fat number of children per family

$= \dfrac{0+21+110+126+60+35}{150}$

$=\dfrac {352}{150}=2.35$

If the mean of the following data is $15$ , find $p$ .
$x$ : $5$ $10$ $15$ $20$ $25$
$f$ : $6$ $p$ $6$ $10$ $5$

Mean =

$\dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ mean = $\dfrac {(5\times 6)+(10\times p)+(15 \times 6)+(20 \times 10)+(25 \times 5)}{6+p+6+10+5}$

$\Rightarrow$ mean =

$\dfrac {30+10p+90+200+125}{27+p}=15$

$\Rightarrow 405+15p=445+10p$

$\Rightarrow 5p=40$

$\Rightarrow p=8$

Find the mean of the following frequency distribution:

Class interval: 50-70 70-90 90-110 110-130 130-150 150-170
Frequency: $18$ $12$ $13$ $27$ $8$ $22$

Class interval	${x}_{i}$	Frequency $({f}_{i})$	${f}_{i} {x}_{i}$
50-70	60	18	1080
70-90	80	12	960
90-110	100	13	1300
110-130	120	27	3240
130-150	140	8	1120
150-170	160	22	3520
		$\Sigma{f}_{i} = 100$	$\Sigma{{f}_{i} {x}_{i}} = 11220$

$mean = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{11220}{100} = 112.20$

Find the missing value of $p$ for the following distribution whose mean is $12.58$ .
$x$ : $5$ $8$ $10$ $12$ $p$ $20$ $25$
$f$ : $2$ $5$ $8$ $22$ $7$ $4$ $2$

Mean =

$\dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ mean =

$\dfrac {(5\times 2)+(8\times 5)+(10 \times 8)+(12 \times 22)+(p \times 7)+(20 \times 4)+(25 \times 2)}{2+5+8+22+7+4+2}$

$\Rightarrow$ mean =

$\dfrac {10+40+80+264+7p+80+50}{2+5+8+22+7+4+2}=12.58$

$\Rightarrow \dfrac {629+7p}{50}=12.58$

$\Rightarrow 7p=105$

$\Rightarrow p=15$

Find the mean of the following data:
$x$ $19$ $21$ $23$ $25$ $27$ $29$ $31$
$f$ $13$ $15$ $16$ $18$ $16$ $15$ $13$

Mean =

$\dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ mean =

$\dfrac {(19×13)+(21×15)+(23×16)+(25×18)+(27×16)+(29×15)+(31×13)}{13+15+16+18+16+15+13}$

$\Rightarrow$ mean =

$\dfrac {247+1575+368+450+432+435+403}{13+15+16+18+16+15+13}$

$\Rightarrow mean =25$

If the mean of the following data is $20.6$ . Find the value of $p$ .
$x$ $10$ $15$ $p$ $25$ $35$
$f$ $3$ $10$ $25$ $7$ $5$

Mean =

$\dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ mean =

$\dfrac {(10\times 3)+(15\times 10)+(p \times 25)+(25 \times 7)+(35 \times 5)}{3+10+25+7+5}$

$\Rightarrow$ mean =

$\dfrac {30+150+25p+175+175}{3+10+25+7+5}=20.6$

$\Rightarrow 20.6=\dfrac {1030+25p}{50}$

$\Rightarrow 25p=500$

$\Rightarrow p=20$

Find the missing frequency ( $p$ ) for the following distribution whose mean is $7.68$
$x$ : $3$ $5$ $7$ $9$ $11$ $13$
$f$ : $6$ $8$ $15$ $p$ $8$ $4$

Mean =

$\dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ mean =

$\dfrac {(3\times 6)+(5\times 8)+(7 \times 15)+(9 \times p)+(11 \times 8)+(13 \times 4)}{8+6+15+p+8+4}$

$\Rightarrow$ mean =

$\dfrac {18+40+105+9p+88+52}{8+6+15+p+8+4}=7.68$

$\Rightarrow 314.88+7.68p=303+9p$

$\Rightarrow 11.88=1.32p$

$\Rightarrow p=9$

The arithmetic mean of the following data is $14$ . Find the value of $k$ .
$x$ : $5$ $10$ $15$ $20$ $25$
$f$ : $7$ $k$ $8$ $4$ $5$

Arithmetic Mean =

$\dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ Arithmetic mean = $\dfrac {(5\times 7)+(10\times k)+(15 \times 8)+(20 \times 4)+(25 \times 5)}{7+k+8+4+5}$

$\Rightarrow$ arithmetic mean =

$\dfrac {35+10k+120+80+125}{7+k+8+4+5}=14$

$\Rightarrow 336+14k=360+10k$

$\Rightarrow 4k=24$

$\Rightarrow k=6$

Find the mean median and mode of $20,15,25,7,25,17,5$

$20,15,25,7,25,17,5$

Arranging in descending order

$25,25,20,17,15,7,5$

$Mean = \dfrac{{sum\,of\,observations}}{{number\,of\,observations}}$

$=\dfrac{{25 + 25 + 20 + 17 + 15 + 7 + 5}}{7}$

$= \dfrac{{114}}{7}$

$= 16.29$

Median

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

$= \left( {\dfrac{{7 + 1}}{2}} \right)th\,observation$

$= 4th\,observation$

$= 17$

$Mode = 25$

Find a if the mean mark is $1$ .
Mark 0 1 2 3 4
No. of students a 4 3 2 1

Mean

$= \dfrac{a\times0+4\times1+3\times2+2\times3+1\times4}{a+4+3+2+1}$

$\Rightarrow 1\times(a+10) = 4+6+6+4$

$\Rightarrow a+10=20$

$\Rightarrow a=10$

Calculate mean with step Deviation method from the following data.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
No. of student $4$ $6$ $10$ $20$ $10$

Class inreval	Mid values $(x_i)$	Frequency $(f_i)$	$d_i=x_i-A$	$u_i=\frac{d_i}{h}$	$F_iu_i$

0-10	5	4	-20	-2	-8
10-20	15	6	-10	-1	-6
20-30	25=A	10	0	0	0
30-40	35	20	10	1	20
40-50	45	10	20	2	20
		N=50			Total=26

$h=10,A=25,N=50, \sum f_iu_i=26$

Mean

$(\overline x)=A+h[\frac{1}{N} \sum f_iu_i]$

put the values

$Mean=30.2$

Find the mode of the following data:
$120, 110, 130, 110, 120, 140, 130, 120, 140, 120.$

Let us first form the frequency table for the given data as given below:

Value

$x_i$ : 110 120 130 140

Frequency

$(f_i)$ : 2 4 2 2

We observe that the value

$120$ has the maximum frequency.

Hence, the mode or modal value is

$120.$

Following are the lives in hours of $15$ pieces of the components of aircraft engine. Find the median.
$715,724,725,710,729,745,694,699,696,712,734,728,716,705,719$

Arrange the numbers in ascending order -

$694, 696, 699, 705, 710, 712, 715, 716, 79, 724, 725, 728, 729, 734,745$

Total number of pieces

$(n)= 15$ (odd)

As we know that,

median = middle no.

$\rightarrow$ (if

$n$ = odd)

$\therefore$ median

$= 716$

X C.F.
Less than 10 12
10-20 19
20-30 28
30-40 30
40-50 40
Mean= ?
Median = ?

$Mean=\dfrac{12+19+28+30+40}{5}=\dfrac{129}{5}=25.8$

$Median=28$ (middle number when data is arranged in increasing Order)

The arithmetic mean for the following distribution is $53$ . Find the value of $p$ .
$Class$ $0-20$ $20-40$ $40-60$ $60-80$ $80-100$
$Frequency$ $15$ $18$ $21$ $p$ $17$

Class	Mid values $(x_i)$	Frequency $(f_i)$	$f_ix_i$

0-20	10	17	170
20-40	30	P	30p
40-60	50	32	1600
60-80	70	24	1680
80-100	90	19	1710
		Total=92+p

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

$50=\dfrac{5160+30p}{92+p}$

Therfore,

$p=28$

If the median of the scores $\frac{x}{3},\frac{x}{2},\frac{x}{4},\frac{{2x}}{9},x\left( {x > 0} \right)$ is 3, then find 'x'.

Arranging the given observations in ascending order,

$\Rightarrow$

$\dfrac{2x}{9},\dfrac{x}{4},\dfrac{x}{3},\dfrac{x}{2},x$

$\Rightarrow$ Here, number of observation

$=5$

$Median=\left(\dfrac{5+1}{2}\right)^{th}observation$

$=3^{rd}observation$

$\therefore$

$Median=\dfrac{x}{3}$

$\therefore$

$3=\dfrac{x}{3}$

$\therefore$

$x=9$

Find the mean of the following frequency distribution using steps deviation method.
C.I 0-100 100-200 200-300 30-400 400-500 500-600 600-700
F 10 15 25 25 10 10 5

Class Interval	Frequency $f_i$	Mid values $x_i$	$d_i=x_i-A$	$u_i=\frac{d_i}{h}$	$f_iu_i$
0-100	10	50	-300	-3	-30
100-200	15	150	-200	-2	-30
200-300	25	250	-100	-1	-25
300-400	25	350	0	0	0
400-500	10	450	100	1	10
500-600	10	550	200	2	20
600-700	5	650	300	3	15

$A=350,h=100,N= \sum f_i=100$

$f_iu_i=40$

$Mean (\overline x)=A+h[\frac{1}{N} \sum f_iu_i]$

Put the values

$Mean =390$

Median of the data $\dfrac{x}{3}, \dfrac{x}{6}, \dfrac{x}{9}. \dfrac{x}{2}, \dfrac{x}{8}, \dfrac{x}{4}$ is $5$ , then find the value of 'x'.

$\Rightarrow$ The given numbers are

$\dfrac{x}{3},\,\dfrac{x}{6},\,\dfrac{x}{9},\,\dfrac{x}{2},\,\dfrac{x}{8},\,\dfrac{x}{4}$

$\Rightarrow$ Arranging given numbers in ascending order

$\dfrac{x}{9},\,\dfrac{x}{8},\,\dfrac{x}{6},\,\dfrac{x}{4},\,\dfrac{x}{3},\,\dfrac{x}{2}$

$\Rightarrow$ Number of observation

$(n)=6$

$\Rightarrow$ Number of observation is even number.

$\therefore$

$Median =\left(\dfrac{n}{2}+1\right)^{th}observation=\left(\dfrac{6}{2}+1\right)^{th}=4^{th}observation$

Here,

$4^{h}$ observation is

$\dfrac{x}{4}$

$\therefore$ According to the question,

$\Rightarrow$

$\dfrac{x}{4}=5$

$\therefore$

$x=20$

Calculate the mean of the following data, using direct method:
Class 25-35 35-45 45-55 55-65 65-75
Frequency 6 10 8 12 4

$Class$	$Frequency$ $f_i$	$Mid-value$ $x_i$	$f_ix_i$
$25-35$	$6$	$30$	$180$
$35-45$	$10$	$40$	$400$
$45-55$	$8$	$50$	$400$
$55-65$	$12$	$60$	$720$
$65-75$	$4$	$70$	$280$
	$\sum f_i=40$		$\sum f_ix_i=1980$

$\Rightarrow$

$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{1980}{40}=49.5$

Calculate the mean of the following frequency distribution.
Class $90-100$ $100-110$ $110-120$ $120-130$ $130-140$
Frequency $18$ $78$ $80$ $100$ $72$

$Class$	$Frequency$ $f_i$	$x_i$	$f_ix_i$
$90-100$	$18$	$95$	$1710$
$100-110$	$78$	$105$	$8190$
$110-120$	$80$	$115$	$9200$
$120-130$	$100$	$125$	$12500$
$130-140$	$72$	$135$	$9720$
	$\sum f_i=348$		$\sum f_ix_i=41320$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{41320}{348}=118.73$

If the mean of a data is $20.2$ . Find the value of m.
Observation $10$ $15$ $20$ $25$ $30$
Frequency $6$ $8$ m $10$ $6$

$Observation$ $x_i$	$Frequency$ $f_i$	$f_ix_i$
$10$	$6$	$60$
$15$	$8$	$120$
$20$	$m$	$20m$
$25$	$10$	$250$
$30$	$6$	$180$
	$\sum f_i=30+m$	$\sum f_ix_i=610+20m$

$\Rightarrow$

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

$\therefore$

$20.2=\dfrac{610+20m}{30+m}$

$\therefore$

$606+20.2m=610+20m$

$\therefore$

$0.2m=4$

$\therefore$

$m=20$

Find the mean of the first five whole numbers.

${\textbf{Step -1: Listing first five whole numbers.}}$

${\text{We know that, whole numbers are all the non negative numbers}}{\text{.}}$

${\text{First five whole numbers are \{ 0, 1, 2, 3, 4\} }}$

${\textbf{Step -2: Finding mean of given numbers.}}$

${\text{Mean of n given numbers = }}\dfrac{{{\text{Sum of numbers}}}}{{\text{n}}}$

$\Rightarrow {\text{ Mean of first five whole numbers = }}\dfrac{{{\text{0 + 1 + 2 + 3 + 4}}}}{{\text{5}}}$

$\Rightarrow {\text{ Mean of first five numbers = }}\dfrac{{{\text{10}}}}{{\text{5}}}$

$\Rightarrow {\text{ Mean of first five whole numbers = 2}}$

${\textbf{Hence, The mean of first five whole numbers is 2}}{\textbf{.}}$

The data given below shows the number of goals scored by a football team over $30$ games

4

3

2

1

4

0

1

0

5

1

0

0

1

4

3

2

1

0

1

3

4

2

0

1

0

3

2

1

1

0

Find the mode of the above data.

Mode is the value of the observation having maximum frequency.

So,

$1$ appears

$9$ times in the given data that is the maximum frequency so mode of the given data is

$1$ .

Find the mean of each of the data. $8,-2,9,6,13,17,12$

Given data is

$8,-2,9,6,13,17,12$ .

Formula to find mean of discrete data is

$\bar{X} =\dfrac{sum\space of \space data \space all \space data \space value}{number \space of \space data value}$

Sum of data value is

$= 8+(-2)+9+6+13+17+12=63$

Number of data value

$=7$

then,

$\bar{X}=\dfrac{63}{7}$

$\bar{X}=9$

So mean of data is

$9$ .

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 mean.

$Weekly\,wages$	$Class-mark$ $x_i$	$No.\,of\,worker$ $f_i$	$f_ix_i$
$50-55$	$52.5$	$5$	$262.5$
$55-60$	$57.5$	$20$	$1150$
$60-65$	$62.5$	$10$	$625$
$65-70$	$67.5$	$10$	$675$
$70-75$	$72.5$	$9$	$652.5$
$75-80$	$77.5$	$6$	$465$
$80-85$	$82.5$	$12$	$990$
$85-90$	$87.5$	$8$	$700$
		$\sum f_i=80$	$\sum f_ix_i=5520$

$\Rightarrow$ Required mean

$=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{5520}{80}=69$

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 $11$ $10$ $7$ $4$ $4$ $3$ $1$

$No.\,of\,days$	$Class\,mark$ $x_i$	$No.\,of\,students$ $f_i$	$f_ix_i$
$0-6$	$3$	$11$	$33$
$6-10$	$8$	$10$	$80$
$10-14$	$12$	$7$	$84$
$14-20$	$17$	$4$	$68$
$20-28$	$24$	$4$	$96$
$28-38$	$33$	$3$	$99$
$38-40$	$39$	$1$	$39$
		$\sum f_i=40$	$\sum f_ix_i=499$

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

$=\dfrac{499}{40}$

$=12.5$

Find mean of the following data by using step-deviation method
Class 10-14 15-19 20-24 25-29 30-34
Frequency $15$ $110$ $135$ $115$ $25$

Let assumed mean

$A=22$
width of given interval

$h=5$

$Class$	$x_i$	$d_i=x_i-A$	$u_i=\dfrac{d_i}{h}$	$f_i$	$f_iu_i$
$9.5-14.5$	$12$	$-10$	$-2$	$15$	$-30$
$14.5-19.5$	$17$	$-5$	$-1$	$110$	$-110$
$19.5-24.5$	$22$	$0$	$0$	$135$	$0$
$24.5-29.5$	$27$	$5$	$1$	$115$	$115$
$29.5-34.5$	$32$	$10$	$2$	$25$	$50$
				$\sum f_i=400$	$\sum f_iu_i=25$

$Mean=A+\dfrac{\sum f_iu_i}{\sum f_i}\times h$

$=22+\dfrac{25}{400}\times 5$

$=22+0.312$

$=22.312$

Find the median of the data: $3.3,3.5,3.1,3.7,3.2,3.8$

Arranging given numbers in ascending order,

$3.1,\,3.2,\,3.3,\,3.5,\,3.7,\,3.8$

Number of observations

$=6$

Here, number of observation is even number.

$\therefore$

$Median=\dfrac{1}{2}\left[\left(\dfrac{6}{2}\right)^{th}observation+\left(\dfrac{6}{2}+1\right)^{th}observation\right]$

$=\dfrac{1}{2}(3^{rd}obs+4^{th}obs)$

$=\dfrac{1}{2}(3.3+3.5)$

$=\dfrac{6.8}{2}$

$\therefore$

$Median=3.4$

Find the median of the following data: $14, 16, 14, 12, 11, 14, 16, 12, 10, 8$ .

Arranging this data, we get

$8,10,11,12,12,14,14,14,16,16$

As there are 10 observations, we have the median =

$\cfrac{12+14}{2} = 13$

The following distribution gives ages of $1000$ persons who visited a shopping centre on sunday:
Age(in years) $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$
Number of persons $105$ $222$ $220$ $138$ $102$ $113$ $100$
Use step deviation method, find the mean number of the people who visited the shopping centre on Sunday.

$\Rightarrow$

$A=35$ and

$h=10$

$C.I$	$Class\,mark$ $x_i$	$d_i=x_i-A$	$u_i=\dfrac{d_i}{h}$	$f_i$	$f_iu_i$
$0-10$	$5$	$-30$	$-3$	$105$	$-315$
$10-20$	$15$	$-20$	$-2$	$222$	$-444$
$20-30$	$25$	$-10$	$-1$	$220$	$-220$
$30-40$	$35$	$0$	$0$	$138$	$0$
$40-50$	$45$	$10$	$1$	$102$	$102$
$50-60$	$55$	$20$	$2$	$113$	$226$
$60-70$	$65$	$30$	$3$	$100$	$300$
				$\sum f_i=1000$	$\sum f_iu_i=-351$

$\Rightarrow$ Required mean

$=A+\dfrac{\sum f_iu_i}{\sum f_i}\times h$

$=35+\dfrac{(-351)}{1000}\times 10$

$=35-3.51$

$=31.49$

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 number of workers getting $Rs. 65$ or more but less than $Rs. 85$ as weekly wages.

$Weekly\,wages$ $(in\,Rs.)$	$No.\,of\,workers$
$65-70$	$10$
$70-75$	$9$
$75-80$	$6$
$80-85$	$12$
$Total$	$37$

$\Rightarrow$ The number of workers getting

$Rs.65$ or more but less than

$Rs.85=37$

Find the median of first eight $(8)$ positive composite numbers.

$4,6,8,9,10,12,14,15$ are the first $8$ composite numbers.

Then number of terns $8\,\,\left( Even \right)$

Now, median = average of $\left( 9,10 \right)$

$=\left( \dfrac{9+10}{2} \right)=\dfrac{19}{2}$

$=9.5$

The following observations have been arranged in ascending order. if the median of these observations is 58, find the value of $x$ .
$24$ , $27$ , $43$ , $48$ , $x-1$ , $x+3$ , $68$ , $73$ , $80$ , $90$ .

The given numbers are,

$24,27,43,48,x-1,x+3,68,73,80,90$

Number of observations

$(n)=10$

$\Rightarrow$

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

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

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

$5^{th}obs=x-1$ and

$6^{th}=x+3$

$\therefore$

$Median=\dfrac{x-1+x+3}{2}$

$\Rightarrow$

$58=\dfrac{2x+2}{2}$

$\Rightarrow$

$58=x+1$

$\therefore$

$x=57$

Calculate Mean, Median and Mode for the following data. Which of these three measures if minimum?
Marks s(out of 15) $7$ $4$ $10$ $9$ $15$ $12$ $7$ $9$ $9$

Arranging given marks in ascending order,

$4,7,7,9,9,9,10,12,15$

Number of observations

$(n)=9$

$\Rightarrow$

$Mean=\dfrac{4+7+7+9+9+9+10+12+15}{9}=\dfrac{82}{9}=9.11$

$\Rightarrow$ Here, number

$9$ occurs maximum number of times i.e

$3$ times.

$\therefore$

$Mode=9$

Number of observation is odd number

$\therefore$

$Median=\left(\dfrac{n+1}{2}\right)^{th}obs=\left(\dfrac{9+1}{2}\right)^{th}obs=5^{th}obs$

Here,

$5^{th}obs=9$

$\therefore$

$Median=9$

$\Rightarrow$ So, we can see

$mean$ measure is maximum and

$mode,median$ measures are minimum.

Find the median of $64$ and $225$ .

For set of two numbers the mean and median are same. Hence the median is

$\dfrac{64+225}{2}=144.5$

Find the missing frequency $f_{1}$ and $f_{2}$ in the table given below, it is being given that the mean of the given frequency distribution is $50$ .
Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ Total
Frequency $17$ $f_{1}$ $32$ $f_{2}$ $19$ $120$

$Class$	$Class-Mark$ $x_i$	$Frequency$ $f_i$	$f_ix_i$
$0-20$	$10$	$17$	$170$
$20-40$	$30$	$f_1$	$30f_1$
$40-60$	$50$	$32$	$1600$
$60-80$	$70$	$f_2$	$70f_2$
$80-100$	$90$	$19$	$1710$
		$\sum f_i=f_1+f_2+68$	$\sum f_ix_i=3480+30f_1+70f_2$

$\sum f_i=120$ [ Given ]

$\sum f_i=f_1+f_2+68$

$120=f_1+f_2+68$

$\therefore$

$f_1+f_2=52$

$\Rightarrow$

$f_1=52-f_2$ ---- ( 1 )

$\Rightarrow$

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

$\Rightarrow$

$50=\dfrac{3480+30f_1+70f_2}{120}$

$\Rightarrow$

$6000=3480+30(52-f_2)+70f_2$ [ From ( 1 ) ]

$\Rightarrow$

$2520=1560-30f_2+70f_2$

$\Rightarrow$

$40f_2=960$

$\therefore$

$f_2=24$

Substituting value of

$f_2$ in ( 1 )

$f_1=52-24=28$

$\therefore$

$f_1=28,\,f_2=24$

The lengths of $50$ leaves of a plant are measured corrcct he nearest milimetre and the data obtained is represent in the following table.
Length $(mm)$ $109-117$ $118-126$ $127-135$ $136-144$ $145-153$ $154-162$ $163-171$
No of leaves $4$ $6$ $14$ $13$ $6$ $4$ $3$

$Length$ $(mm)$	$x_i$	$f_i$	$f_ix_i$
$108.5-117.5$	$113$	$4$	$452$
$117.5-126.5$	$122$	$6$	$732$
$126.5-135.5$	$131$	$14$	$1834$
$135.5-144.5$	$140$	$13$	$1820$
$144.5-153.5$	$149$	$6$	$894$
$153.5-162.5$	$158$	$4$	$632$
$162.5-171.5$	$167$	$3$	$501$
		$\sum f_i=50$	$\sum f_ix_i=6865$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{6865}{50}=137.3$

There are 10 observations arranged in ascending order as given below.
$45,47,50,52,x,x+2,60,62,63,74.$
The median of these observation is $53$ . Find the value of $x$

1053125_1176310_ans_bd9334491a6a4106b1e7e03516895522.jpg

(a)The class interval having highest frequency is-
(b)The class-mark of class 50-60 is-
(c)The size of each class intervals is-

$Class-Interval$	$Frequency$
$0-10$	$6$
$10-20$	$2$
$20-30$	$13$
$30-40$	$17$
$40-50$	$11$
$50-60$	$4$
$60-70$	$8$
$70-80$	$7$

$(i)$ The class interval having highest frequency

$=30-40$

$(ii)$ The class mark of class

$50-60$

$=\dfrac{50+60}{2}=\dfrac{110}{2}=55$

$(iii)$ The size of each class interval

$=10-0=10$

Find the mode of $90,55,67,55,75,40,35,55,95$

Mode is the number which is repeated more often than any other. Here it is 55.

Find the median of the data $13,16,12,14,19,12,14,13$ .

we need median of

$13,16,12,14,19,12,14,13$

ascending order :

$12,12,13,13,14,14,16,19.$

median

$=$ middle most value

$\therefore 13,14$ are middle values

$\therefore$ median

$=\dfrac{13+14}{2}=13.5$

$\therefore$ median of the given data is

$13.5$

The table below- shows the daily expenditure on food of $25$ households in a locality.
Daily expenses (in Rs.) 100-150 150-200 200-250 250-300 300-350
No. of households 4 5 12 2 2
Find the mean daily expenses on food by a suitable method

Daily Expense	Households (fi)	Mid Point (xi)	fi xi
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
	25		5275

Mean

$= \frac{\sum fixi}{\sum fi} = \frac{5275}{25} = 211$

1118213_1190509_ans_47b4c4447e3a47389fe6b8d35ba56b34.jpg

Find the mean of the following distribution by step-deviation method:
Daily expenditure (in Rs.) $100-150$ $150-200$ $200-250$ $250-300$ $300-350$
No. of households $4$ $5$ $12$ $2$ $2$

$C.I$	$Frequency$ $f_i$	$Class-mark$ $x_i$	$x_i-A$	$u_i=\dfrac{x_i-A}{50}$	$f_iu_i$
$100-150$	$4$	$125$	$-100$	$-2$	$-8$
$150-200$	$5$	$175$	$-50$	$-1$	$-5$
$200-250$	$12$	$A=225$	$0$	$0$	$0$
$250-300$	$2$	$275$	$50$	$1$	$2$
$300-350$	$2$	$325$	$100$	$2$	$4$
	$\sum f_i=25$				$\sum f_iu_i=-7$

$\Rightarrow$

$Mean=A+\dfrac{\sum f_iu_i}{\sum f_i}\times h$

$=225+\dfrac{(-7)}{25}\times 50$

$=225-14$

$=211$

Find mean:
The marks obtained by 200 students are given below :
Marks No of students
0-10 5
10-20 10
20-30 11
30-40 20
40-50 27
50-60 38
60-70 40
70-80 09
80-90 14
90-100 6

Mean =

$\frac{\sum fixi}{\sum fi}$

Hence, fi = No. of students

xi = class mark

$= \sum fixi = 9490$

$=\sum fi = 180$

Mean =

$\frac{9490}{180} = 52.72$

1115683_1202007_ans_27e725d27ef74e188b510459d90a178d.jpg

Find the mean salary of $80$ workers of a factory from the following tables:
Salary (in Rs} Numbers of workers
$5000$ $22$
$6000$ $18$
$7000$ $15$
$8000$ $10$
$9000$ $8$
$10000$ $7$

$Mean =\bar x=\cfrac{\sum fixi}{\sum fi}$

$\bar x=\cfrac{(5000\times 22)+(6000\times 18)+(7000\times 15)+(8000\times 10)+(9000\times 8)+(10000\times 7)}{80}=6812.5\text{ }PA.$

The following table draws the income of finance of a grape season find the mean of their income.
Income (thousand Rs) $20-30$ $30-40$ $40-50$ $50-60$
Farmer $10$ $11$ $15$ $16$

Mean of income

$=\cfrac{10+11+15+16}{4}\\=\cfrac{52}{4}\\=13$

Mark obtained 25-35 35-45 45-55 55-65 65-75
Number of students 7 31 33 17 11
Find mean

$Marks\,obtained$	$No.\,of\,students$ $f_i$	$Class-mark$ $x_i$	$f_ix_i$
$25-35$	$7$	$30$	$210$
$35-45$	$31$	$40$	$1240$
$45-55$	$33$	$50$	$1650$
$55-65$	$17$	$60$	$1020$
$65-75$	$11$	$70$	$770$
$75-85$	$1$	$80$	$80$
	$\sum f_i=100$		$\sum f_ix_i=4970$

$\Rightarrow$

$Mean(\overline)=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{4970}{100}=49.7$

From one footwear shop, $12$ pairs of chappals were sold. The sizes of these chappals are given below. Find their mode: $7,8,6,7,7,5,9,7,6,7,8,7$

$5\>\longrightarrow\>1\>times\\6\>\longrightarrow\>2\>times\\7\>\longrightarrow\>6\> times\\8\>\longrightarrow\>2\>times$

As 7 appears most number of time. Hence 7 is the mode of the given data.

In the following table, the toll paid by drives and the number of vehicles is shown. Find the mean of the toll by assumed mean method.
Toll (Rupees) $300-400$ $400-500$ $500-600$ $600-700$ $700-800$
No. of vehicles $80$ $110$ $120$ $70$ $40$

$CI$	$Class-mark$ $x_i$	$d_i=x_i-A$	$f_i$	$f_id_i$
$300-400$	$350$	$-200$	$80$	$-16000$
$400-500$	$450$	$-100$	$110$	$-11000$
$500-600$	$550=A$	$0$	$120$	$0$
$600-700$	$650$	$100$	$70$	$7000$
$700-800$	$750$	$200$	$40$	$8000$
			$\sum f_i=420$	$\sum f_id_i=-12000$

$\overline{x}=A+\dfrac{\sum f_id_i}{\sum f_i}$

$=550-\dfrac{12000}{420}$

$=550-28.57$

$=521.43$

$\therefore$ Mean of toll is

$Rs.521.43$

Hence, this is the answer.

The marks obtained in an examination of $400$ marks is given in the following frequency distribution table. Find the mean of the distribution.
Marks 200-240 240-280 280-320 320-360 360-400
Frequency 10 20 30 24 16

Now,

$Mean = \dfrac{{sum\,\,of\,\,\left( {fx} \right)}}{f}$

$= \dfrac{{30640}}{{100}}$

Then,

We have,

$=306.40$

1199105_1273686_ans_b51bbfa5b1fd4af98ab7f3e6caef0105.JPG

Calculate the mean from the following data:

$\sum f_i=5+15+30+8+2=60\\\sum x_if_i=5(5)+20(15)+45(30)+70(8)+85(2)\\$

$25+300+1350+560+170=2405\\\\\bar x=\dfrac{\sum x_i f_i}{\sum f_i}=\dfrac{2405}{60}=40.08$

Find the median of :
(i) 5,7,9,11,15,17,2,23 and 19

$5,7,9,11,15,17,2,23,19$

Arranging in ascending order

$2,5,7,9,11,15,17,19,23$

Total no of terms

$n=9$

Hence middle term

$=\cfrac{n+1}{2}=5$

Hence

$5^{th}$ term

$=11$ .

In the following table age distribution of patient in the city, suffering from some diseases is given that the average age ( in years) per patient.
Age No. of patient
5-14 6
15-24 11
24-24 21
35-44 23
45-54 54
55-60 5

$Age$	$Class-mark$ $x_i$	$No.\,of\,patient$ $f_i$	$f_ix_i$
$4.5-14.5$	$9.5$	$6$	$57$
$14.5-24.5$	$19.5$	$11$	$214.5$
$24.5-34.5$	$29.5$	$21$	$619.5$
$34.5-44.5$	$39.5$	$23$	$908.5$
$44.5-54.5$	$49.5$	$54$	$2673$
$54.5-60.5$	$59.5$	$5$	$297.5$
		$\sum f_i=120$	$\sum f_ix_i=4770$

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{4770}{120}=39.75$

Find the mode of the given set of numbers $2,3,5,3,4,7,3,2,1,7,3$ .

Here, 3 is repeated the most number of times which is 4.

Therefore, Mode of the set is 3.

1133059_1210053_ans_94c0c2f6e0a5479496e0483423ac2b3e.jpg

The amount collected by $15$ students for flood relief find is given below. Answer the following.
Amount Number of students
$0-10$ $3$
$10-20$ $4$
$20-30$ $5$
$30-40$ $2$
$40-50$ $1$
(ii)Find the mean amount collected by a student.

We prove that

mean

$=\dfrac {Sum\ of\ all\ observation}{Number\ of\ observation}$

$=\dfrac {10\times 3+20\times 4+30\times 5+40\times 2+50\times 1}{15}$

$=\dfrac {30+80+150+80+50}{15}$

$=\dfrac {390}{15}$

mean

$=26$

Find the mode of the following data
i) $12,14,12,16,15,13,14,18,19,12,14,15,16,15,16,16,$
ii) $15,17,13,16,16,15,13,15,17,15,14,15,13,15,14,$

(i)

$12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16$

Mode is most appearing data in given data.

On above data mode is

$=16$ which appeared most in given data.

(ii)

$15, 17, 13, 16, 16, 15, 13, 17, 15, 14, 15, 13, 15, 14$

On above data mode is

$=15$ .

1204170_1393345_ans_3cd8f2d88584480d9f0f6d59faa503e4.JPG

Mean for the following frequency distribution table.
0-5 5-10 10-15 15-20 20-25 25-30
5 3 9 10 8 5

$Class$	$Class-mark$ $x_i$	$Frequency$ $f_i$	$f_ix_i$
$0-5$	$2.5$	$5$	$12.5$
$5-10$	$7.5$	$3$	$22.5$
$10-15$	$12.5$	$9$	$112.5$
$15-20$	$17.5$	$10$	$175$
$20-25$	$22.5$	$8$	$180$
$25-30$	$27.5$	$5$	$137.5$
		$\sum f_i=40$	$\sum f_ix_i=640$

$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{640}{40}=16$

Find the median of 72, 66, 87, 92, 63, 78.

Given data is

$72,66,87,92,63,78$

The increasing order is

$63,66,72,78,87,92$

The median will be the average of

$72,78$

$\implies$ median

$=\dfrac{72+78}{2}=75$

Find the mean of the following distribution
300-400 400-500 500-600 600-700 700-800 800-900
No. of girls 4 18 9 7 6 3

$Class-interval$	$Class-mark$ $x_i$	$f_i$	$f_ix_i$
$300-400$	$350$	$4$	$1400$
$400-500$	$450$	$18$	$8100$
$500-600$	$550$	$9$	$4950$
$600-700$	$650$	$7$	$4550$
$700-800$	$750$	$6$	$4500$
$800-900$	$850$	$3$	$2550$
		$\sum f_i=47$	$\sum f_ix_i=26050$

$Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{26050}{47}=554.26$

A sequence, $a,ax,ax^{2}, ......, ax^{n}$ , has odd number of terms. Find its median.

$a,ax,ax^{2},ax^{3},.....,ax^{n}$

As there are odd number of terms , the median is :

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

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

Median =

$a(x^{\dfrac{n+2}{2}-1})=a.x^{n/2}$

Find the mode of the data :
(i) 10,8,4,7,8,11,15,8,6,8,
(ii) 27,23,39,18,27,21,27,27,40,36,27

(i)10,8,4,7,8,11,15,8,6,8

In above data most occur number is 8

Hence mode =8

(ii)27,23,39,18,27,21,27,27,40,36,27

Here mode =27

1204182_1390342_ans_8c1cc82959e6421aa10c745ba38d45f5.JPG

Calculate the mean of the distribution given below using the shortcut method.
marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80
number of students 2 6 10 12 9 7 4

$Marks$	$Class-Mark$ $x_i$	$No.\,of\,students$ $f_i$	$f_ix_i$
$10.5-20.5$	$15.5$	$2$	$31$
$20.5-30.5$	$25.5$	$6$	$153$
$30.5-40.5$	$35.5$	$10$	$355$
$40.5-50.5$	$45.5$	$12$	$546$
$50.5-60.5$	$55.5$	$9$	$499.5$
$60.5-70.5$	$65.5$	$7$	$458.5$
$70.5-80.5$	$75.5$	$4$	$302$
		$\sum f_i=50$	$\sum f_ix_i=2345$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{2345}{50}=46.9$

The arithmetic mean of following data is $17$ . find the value of $P$ .
Term(x) 10 15 20 25
Frequency(f) 7 10 P 6

$Term$ $x_i$	$Frequency$ $f_i$	$f_ix_i$
$10$	$7$	$70$
$15$	$10$	$150$
$20$	$P$	$20P$
$25$	$6$	$150$
	$\sum f_i=23+P$	$\sum f_ix_i=370+20P$

$\Rightarrow$

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

$\Rightarrow$

$17=\dfrac{370+20P}{23+P}$

$\Rightarrow$

$391+17P=370+20P$

$\Rightarrow$

$21=3P$

$\Rightarrow p=\dfrac{21}{3}$

$\therefore$

$P=7$

Following observations have been written in ascending order. If median of the data is $22$ then find the value of $x$ .
$11, 12, 14, 16, 18, x+2, x+4, 30, 32, 35, 41$

Given data is

$11,12,14,16,18,x+2,x+4,30,32,35,41$

The data is in ascending order

So the median will be mean of

$x+2,x+4$

$\implies \dfrac{x+2+x+4}{2}=22$

$\implies x+3=22\implies x=19$

If the numbers $3,6,7,10,x,15,19,20,25,28$ are in ascending order and their median is $13$ . Calculate the value of $x$ .

1310296_1534471_ans_04334e9690b74bfe90c411411a05f7a4.jpeg

The average weight of 3 students P,Q and R is 84kg.Anothe students S joins the group and the average becomes 80kg. If another man T whose weight is 3kg more than that of S, replaces P, then the average weight of Q,R,S and T becomes 79 kg then the weight of P is

Explanation : P+Q+R = 84*3 = 252 P+Q+R+S = 4*80 = 320 S = 320 252 = 68 Q+R+S+T = 79*4 = 316 Q+R+2S+3 = 316 S =68, Q+R =177 P = 252 177 = 75

The average salary of all the workers in factory isThe average salary of 9 mechanic is 5000 and for the rest of the workers itFind the total number of workers in the factory. a

$\Rightarrow$ Let

$x$ be the total worker in the factory.

The average salary of all worker in factory is

$7000$

$\therefore$ The total salary of all worker

$=7000x$

The average salary of

$9$ mechanic is

$5000$

$\therefore$ Total salary of

$9$ worker

$=9\times 5000=45000$

$\Rightarrow$ Total salary of rest of worker

$=4000(x-9)$

$\Rightarrow$

$7000x=45000+4000(x-9)$

$\Rightarrow$

$7x=45+4(x-9)$

$\Rightarrow$

$7x=45+4x-36$

$\Rightarrow$

$3x=9$

$\therefore$

$x=3$

$\therefore$ Total worker

$=x+9=3+9=12$

$15$ student secured the following marks in a test in Statistic. Find the median marks
35, 28,13,17,20,30,19,29,11,10,29,23,18, 25,17

Given data of

$15$ students is

$35,28,13,17,20,30,19,29,11,10,29,23,18,25,17$

The ascending order of these data is

$10,11,13,17,17,18,19,20,23,25,28,29,29,30,35$

The middle element in this order is

$20$

So median marks is

$20$

Find the mode of the data:
10,8,4,7,8,11,15,8,6,8

1309944_1533452_ans_74b8e10a679c45d98605fb22e40761ca.jpeg

For the following distribution, find the mean:
Class Marks $10$ $20$
$30$
$40$
$50$
$60$
No of students $3$ $1$ $2$ $7$ $8$ $5$

$\sum f_i=3+1+2+7+8+5=26\\\sum x_if_i=30+20+60+280+400+300=1090\\\bar x=\dfrac{1090}{26}=41.9$

Write the empirical relationship between the three measures of central tendency?

3 median = mode + 2 mean

Find the median and mode of the data: $35,32,35,42,38,32,34$

Arranging the data in ascending order such that same numbers are put together, we get

$32,32,34,35,35,38,42$

Here,

$n=7$

$\therefore$ Median

$=$ Value of

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

$=$ Value of

$4^{th}$ observation

$=35$ .

Here,

$32$ and

$35$ both occur twice.

$\therefore 32$ and

$35$ are the two modes.

Find the median of $\dfrac{2}{3}, \dfrac{4}{5}, \dfrac{1}{2}, \dfrac{3}{4}, \dfrac{6}{5}$ .

The numbers in ascending order-

$\cfrac{1}{2}, \cfrac{2}{3}, \cfrac{3}{4}, \cfrac{4}{5}, \cfrac{6}{5}$

Number of observations

$= 5 \Rightarrow$ Odd,

Therefore,

Median

$=$ middle observation

$= \; {3}^{rd}$ observation

$= \cfrac{3}{4}$

Hence the median of the given data is

$\cfrac{3}{4}$ .

Write the formula to find mean for grouped data direct method.

Mean =

$\bar{x} = \dfrac{\sum f_{i}u_{i}}{\sum f_{i}}$

Where,

$f_i$ = frequency for that particular class

$u_i$ = mid point of that particular class

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

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

$3.1, 3.0,2.9,2.9, 2.8, 2.8, 2.7, 2.7,2.6, 2.5,2.5, 2.4,2.3, 2.2,2.1$

Range = maximum - minimum

$3.1-2.1$

$1$ kg.

If the variable takes the values $\displaystyle 0, 1, 2, \cdots ,n$ with frequencies proportional to $\displaystyle q^{n},^{n}C_{r}q^{n-1}p,^{n}C_{2}q^{n-2}p^{2},\cdots ,p^{n}$ where $p+q=1,$ then show that mean square deviation is $\displaystyle n^{2}p^{2}+npq$ and variance is $npq.$

$\displaystyle \mu _{1}'=E\left ( x \right )=\sum_{x=0}^{0}x\left ( \frac{n}{x} \right )p^{x}q^{n-x}$

$\displaystyle \left ( \frac{n}{x} \right )=^{n}C_{x}=\frac{n}{x}\left ( \frac{n-1}{x-1} \right )$

$\displaystyle =np \sum_{x-1}^{n}\left ( \frac{n-1}{x-1} \right )p^{x-1}q^{\left ( n-1 \right )-\left ( x-1 \right )}$

$\displaystyle =np \left ( p+q \right )^{n-1}=np$

$\displaystyle \therefore p+q=1$
mean square deviation about

$x=0$ is

$\displaystyle \mu _{2}=E\left ( x^{2} \right )=\sum_{x=0}^{n}x^{2}\left ( \frac{n}{x} \right )p^{x}q^{n-x}$

$\displaystyle \sum_{x=0}^{n}\left [ x\left ( x-1 \right )+1\right ]\left ( \frac{n}{x} \right )p^{x}q^{n-x}$

$\displaystyle \sum_{x=0}^{n}x\left ( x-1 \right )\left ( \frac{n}{x}p^{x}q^{n-x}+\sum_{x=0}^{n} \right )x\left ( \frac{n}{x} \right )p^{x}q^{n-x}$

$\displaystyle n\left ( n-1 \right )p^{2}\sum_{x=2}^{n}\left ( \frac{n-2}{n-2} \right )p^{x-2}q^{\left ( n-2 \right )-\left ( x-2 \right )}+np$

$\displaystyle n\left ( n-1 \right )p^{2}\left ( q+p \right )^{n-2}+np$

$\displaystyle n\left ( n-1 \right )p^{2}+np=n^{2}p^{2}+np\left ( 1-p \right )$

$\displaystyle=n^{2}p^{2}+npq$

$\displaystyle \therefore p+q=1$ variance =

$\displaystyle \mu _{2}=\mu \frac{1}{2}-\left ( \mu _{1}' \right )^{2}$

$\displaystyle =n^{2}p^{2}+npq-\left ( np \right )^{2}=npq$

Find' the average marks of student from the following data:
Marks
No. of students
Above 0
80
Above 10
78
Above 20
71
Above 30
65
Above 40
53
Above 50
45
Above 60
25
Above 70
14
Above 80
10
Above 90
8
Above 100
0
mean is 51.1
If true then enter $1$ and if false then enter $0$

Marks CI No.of Students

$x$

${ f }_{ x }$

$0-10$

$2$

$5$

$10$

$10-20$

$7$

$15$

$105$

$30-40$

$6$

$25$

$150$

$40-50$

$12$

$35$

$420$

$50-60$

$8$

$45$

$360$

$60-70$

$20$

$55$

$1100$

$70-80$

$11$

$65$

$715$

$80-90$

$4$

$75$

$300$

$90-100$

$2$

$85$

$170$

$180-200$

$8$

$95$

$760$

$80$

$4090$

$\bar { x }$ =

$\cfrac { \sum { fx } }{ \sum { f } } =\cfrac { 4090}{ 80 } =51.1$

The distribution below shows the number of wickets taken by bowlers in one day' cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?
No. of wickets
No. of bowlers
20-60
7
60-100
5
100-150
16
150-250
12
250-350
2
350-450
3

Mean is 152.89.
If true, then enter $1$ and if false, then enter $0$

No. of wickets	$x_i$ (Midpoint)	No. of bowlers ( $f_i$ )	$f_ix_i$
20 - 60	40	7	280
60 - 100	80	5	400
100 - 150	125	16	2000
150 - 250	200	12	2400
250 - 350	300	2	600
350 - 450	400	3	1200
		$\sum f_i = 45$	$\sum f_ix_i = 6880$

For grouped data when Class Interval is given, we find the Class Mark.

Class Mark = Mid Point of an interval =

$\dfrac{\left( \text{ Lower limit } + \text{ Upper limit }\right)}{2}$

Class Mark is taken as

$x_i$ for each class interval. Find

$x_if_i$

Mean =

$\dfrac{\sum f_ix_i}{\sum f_i} = \dfrac{6880}{45} = 152.8888889 =152.89$ (rounding off to 100th place)

So, on an average the number of wickets taken by a bowler is 152.89

Compute the mean for the following data:
Marks
No. of students
Less than 10
0
Less than 30
10
Less than 50
25
Less than 70
43
Less than 90
65
Less than 110
87
Less than 130
96
Less than 150
100
mean is 74.80
If true then enter $1$ and if false then enter $0$

We can construct the table shown above with the help of the given data.

We know that, the class-mark or mid-value

$x_i=\dfrac{\text{Upper limit}+\text{lower limit}}{2}$

Also, we know that, mean

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

From the table, we have,

$\sum x_if_i=7480$ and

$\sum f_i=100$

$\therefore \ \overline x=\dfrac{7480}{100}=74.80$

Hence, the given statement is true.

The following table gives the distribution of expenditures of different families on education. Find the mean expenditure on education of a family.
Expenditure (in. Rs.)
No. of families
1000-1500
24
1500-2000
10
2000-2500
33
2500-3000
28
3000-3500
30
3500-4000
22
4000-4500
16
4500-5000
7
mean is 2823.53
If true then enter $1$ and if false then enter $0$

$For\quad grouped\quad data\quad when\quad Class\quad Interval\quad \quad is\quad given,\quad we\quad find\quad the\quad Class\quad Mark.\\ Class\quad Mark\quad =\quad mid\quad point\quad of\quad an\quad interval\quad =\quad \frac { \left( Lower\quad limit\quad +\quad Upper\quad limit \right) }{ 2 } \\ Class\quad Mark\quad is\quad taken\quad as\quad { x }_{ i }\quad for\quad each\quad class\quad interval,\quad find\quad { x }_{ i }{ f }_{ i }.\quad \\ Now\quad Mean\quad Expenditure\quad =\quad \frac { \Sigma { x }_{ i }{ f }_{ i } }{ \Sigma { f }_{ i } } =\quad \frac { 480000 }{ 170 } =\quad 2823.529\quad =\quad Rs.2823.53\\$

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
No. of mangoes
$50-52$
$53-55$
$56-58$
$59-61$
$62-64$
No. of boxes
$15$
$110$
$135$
$115$
$25$
If mean number of mangoes kept in a pocket box. Which method of finding the mean did you choose?

$For\quad grouped\quad data\quad when\quad Class\quad Interval\quad \quad is\quad given,\quad we\quad find\quad the\quad Class\quad Mark.\\ Class\quad Mark\quad =\quad mid\quad point\quad of\quad an\quad interval\quad =\quad \dfrac { \left( Lower\quad limit\quad +\quad Upper\quad limit \right) }{ 2 } \\ Class\quad Mark\quad is\quad taken\quad as\quad { x }_{ i }\quad for\quad each\quad class\quad interval,\quad find\quad { x }_{ i }{ f }_{ i }.\quad \\ Now\quad Mean\quad =\quad \dfrac { \Sigma { x }_{ i }{ f }_{ i } }{ \Sigma { f }_{ i } } =\quad \dfrac { 22875 }{ 400 } =\quad 57.1875\quad =\quad 57.19$

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

Let A = 30, h = 20

Class	$\displaystyle x_{i}$	$\displaystyle f_{i}$	$\displaystyle u_{i}=\frac{x_{i}-A}{h}$	$\displaystyle f_{i}u_{i}$
0 - 20	10	5	-1	-5
20 - 40	30	$\displaystyle f_{i}$	0	0
40 - 60	50	10	+1	10
60 - 80	70	$\displaystyle f_{2}$	+2	$\displaystyle 2f_{2}$
80 - 100	90	7	+3	21
100 - 120	110	8	+4	32
		$\displaystyle \sum f_{i}=30+f_{1}+f_{2}$		$\displaystyle \sum f_{i}u_{i}=58+2f_{2}$

Given

$\displaystyle 30+f_{1}+f_{2}=50\Rightarrow f_{1}+f_{2}=20$ .........(i)

$\displaystyle \overline{x}=A+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h \right )\Rightarrow 62.8=30+\left ( \frac{58+2f_{2}}{50}\times 20 \right )$

$\displaystyle 32.8=(58+2f_{2})\times \frac{2}{5}\Rightarrow 32.8\times 5=116+4f_{2}\Rightarrow 164=116+4f_{2}$

$\displaystyle 4f_{2}=164-116\Rightarrow 4f_{2}=48\Rightarrow f_{2}=12$
Now

$\displaystyle f_{1}+f_{2}=20\Rightarrow f_{1}+12=20\Rightarrow f_{1}=8$
So the missing frequencies are

$\displaystyle f_{1}=8\: and\: f_{2}=12$

If the mean for the following distribution by using deviation method
$\displaystyle x_{i}$ 15 20 22 24 25 30 33 38
Frequency 5 8 11 20 23 18 13 2
is $\cfrac{m}{100}$ .Find $m$

${ x }_{ i }$	Frequency( ${ f }_{ i }$ )	${ f }_{ i }{ x }_{ i }$
$15$	$5$	$75$
$20$	$8$	$160$
$22$	$11$	$242$
$24$	$20$	$580$
$25$	$23$	$575$
$30$	$18$	$540$
$33$	$13$	$429$
$38$	$2$	$76$
	$\sum { { f }_{ i }= }$ 100	$\sum { { f }_{ i }{ x }_{ i }= }$ 2577

Arithmeric Mean $=\cfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\cfrac { 2577 }{ 100 } =25.77$

Find the mean marks from the following data(nearest integer)
Marks No. of Students
Below 10 5
Below 20 9
Below 30 17
Below 40 29
Below 50 45
Below 60 60
Below 70 70
Below 80 78
Below 90 83
Below 100 85

Changing less than type frequency distribution in general frequency distribution

Marks	$\displaystyle x_{i}$	$\displaystyle f_{i}$	$\displaystyle A=45,h=10$ $\displaystyle u_{i}=\frac{x-A}{h}$	$\displaystyle f_{i}u_{i}$
0 - 10	5	5	-4	-20
10 - 20	15	4	-3	-12
20 - 30	25	8	-2	-16
30 - 40	35	12	-1	-12
40 - 50	45	16	0	0
50 - 60	55	15	+1	15
60 - 70	65	10	+2	20
70 - 80	75	8	+3	24
80 - 90	85	5	+4	20
90 - 100	95	2	+5	10
		$\displaystyle \sum f_{i}=85$		$\displaystyle \sum f_{i}u_{i}=29$

According to step deviation formula for mean

$\displaystyle \overline{x}=A+\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h \right )\Rightarrow \overline{x}=45+\left ( \frac{29}{85}\times 10 \right )\Rightarrow \overline{x}=45+3.41$

$\displaystyle \overline{x}=48.41$

So the mean marks is 48.41

Find the mean of the following frequency distribution
Height from mean sea level 200 600 1000 1400 1800
No. of Villages 115 107 265 143 202

Let the assumed mean

$A = 1000, K = 400$

Height

$x_i$ No. of Villages

$f_i$

$d_i = x_i - A$

$u_i = \displaystyle\frac{x_i - A}{K}$

$f_iu_i$

200 115 -800 -2 -230

600 107 -400 -1 -107

1000 265 0 0 0

1400 143 400 1 143

1800 202 800 2 404

832

$\sum{f_iu_i} = 210$

$Mean = A + K\left(\displaystyle\frac{\sum{f_iu_i}}{\sum{f_i}}\right) = 1000 + \displaystyle\frac{400\times210}{832} = 1000 + 100.96 = 1100.96 \simeq 1101$

Find the mean wage of the following distributions
Category A B C D E F G
Wages per day 50 60 70 80 90 100 110
No. of Workers 2 4 8 12 10 6 8

Taking the assumed mean

$a = 80$ , the calculations will be as follows

Wages per day

$(x_i)$ No. of workers

$(f_i)$ Deviations

$d_i = x_i - a$

$d_if_i$

50 2 -30 -60

60 4 -20 -80

70 8 -10 -80

80 12 0 0

90 10 10 100

100 6 20 120

110 8 30 240

$\sum f_i = 50$

$\sum {d_if_i} = 240$

$Mean = a + \displaystyle\frac{\sum {d_if_i}}{\sum f_i} = 80 + \displaystyle\frac{240}{50} = 80+4.8 = 84.8$

Find out the mean profits from the following frequency distribution
Profit per shop
(in Rs) Number of shops
$\displaystyle (f_{i})$
100 - 200 10
200 - 300 18
300 - 400 20
400 - 500 26
500 - 600 30
600 - 700 28
700 - 800 18

Calculation of Mean

Profit (in Rs)	No. of shops $\displaystyle (f_{i})$	Mid-point $\displaystyle (x_{i})$	Deviation $\displaystyle d_{i}=x_{i}-450$	$\displaystyle f_{i}d_{i}$
100 - 200	10	150	-300	-300
200 - 300	18	250	-200	-3600
300 - 400	20	50	-100	-2000
400 - 500	26	70	0	0
500 - 600	30	90	100	3000
600 - 700	30	90	100	5600
600 - 800	18	750	300	5400
	150			5400

Hence by assumed mean method

$\displaystyle \bar{x}=a+\frac{1}{n}\sum_{i=l}^{k}f_{i}d_{i}=450+\frac{5400}{150}$
= 450 + 36 = 486
Therefore mean profit is Rs 486

The Arithmetic Mean of the following frequency distribution is $47$ Determined the value of $p$
C.I. $0-20$ $20-40$ $40-60$ $60-80$ $80-100$
Frequency $8$ $15$ $20$ $p$ $3$

Calculation of Arithmetic Mean

C.I.	Frequency $\displaystyle (f_{i})$	Class Mark $\displaystyle (x_{i})$	$\displaystyle u_{i}=f_{i}x_{i}$
0 - 20	8	10	80
20 - 40	15	30	450
40 - 60	20	50	1000
60 - 80	p	70	70p
80 - 100	30	90	450
Total		48 + p	1980 + 70 p

$\displaystyle \bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{1980+70p}{48+p}$

$\displaystyle \therefore \frac{1980+70p}{48+p}=47$

$\displaystyle \Rightarrow 1980+70p = 47(48+p)$

$\displaystyle \Rightarrow 1980+70p=(47)(48)+47p$

$\displaystyle \Rightarrow 1980+70p=2256+47p$

$\displaystyle \Rightarrow 70p-47p=2256-1980$

$\displaystyle \Rightarrow 23p=276$

$\displaystyle \Rightarrow p=\dfrac{276} {23}=12$
Hence

$p = 12$

Calculate the mean of the following distribution of marks obtained by 75 students in paper in an examination
Marks No. of
students Marks No. of students
0 - 10 2 50 - 60 12
10 - 20 3 60 - 70 14
20 - 30 4 70 - 80 4
30 - 40 8 80 - 90 4
49 - 50 16 90 - 100 3

Taking a = 55, h = 10, we have the following table for computation of the mean

Class

interval Class

Frequency,

$\displaystyle y_{i}$ Class mid-point

$\displaystyle (x_{i})$

$\displaystyle u_{i}=\frac{x_{i}-55}{10}$

$\displaystyle f_{i}u_{_{l}}$

0 - 10 2 5 -5 -10

10 - 20 3 15 -4 -12

20 - 30 4 25 -3 -12

30 - 40 8 35 -2 -16

40 - 50 16 45 -1 -16

50 - 60 12 55 0 0

60 - 70 14 65 1 14

70 - 90 4 85 3 12

90 - 90 4 85 3 12

90 - 100 3 95 4 12

Total 75

-10

$\displaystyle \bar{u}=\frac{1}{n}\sum_{i=l}^{k}f_{i}u_{i}=\frac{1}{75}(-10)=-\frac{10}{75}$

$\displaystyle \therefore \bar{x}=a+h\bar{u}=55+10\left ( \frac{-10}{75} \right )$

= 55 - 1.33

= 53.67

Hence means marks are 53.67

The weekly observation on cost of living Index in a ccrtain city for the year 2000 - 2001 are given below Compute the mean weekly cost of living index
Cost of Living Index Number of weeks
140 - 150 5
150 - 160 10
160 - 170 20
170 - 180 9
180 - 190 6
190 - 200 2
Total 52

Let a = 175, h = 20

Calculation of Mean

Class

interval No.of

weeks

$\displaystyle (f_{i})$ mid-point

$\displaystyle (x_{i})$

$\displaystyle u_{i}=\frac{x_{i}-175}{10}$ a

$\displaystyle f_{i}u_{i}$

140 - 150 5 145 -3 -15

150 - 160 10 155 -2 -20

160 - 170 20 165 -1 -20

170 - 180 9 175 0 0

180 - 190 9 175 1 6

180 - 190 6 185 1 6

190 - 200 2 195 2 4

Total 52 -45

$\displaystyle \bar{u}=\frac{1}{n}\sum_{i=l}^{6}f_{i}u_{i}=-\frac{45}{52}$

$\displaystyle \therefore \bar{x}=a+h\tilde{u}=175+10\left ( -\frac{45}{52} \right )$

$\displaystyle =175-\frac{450}{52}=175-8.65= 166.35$

Hence mean weekly cost of living index

= 166.35

= 166.3 (approx)

Find the means of the following frequency distribution by direct method
Class interval Frequency
0 - 20 8
20 - 40 10
40 - 60 15
60 - 80 9
80 - 100 8

Class interval	Frequency $\displaystyle (f_{i})$	Mid-point $\displaystyle (x_{i})$	Product $\displaystyle (f_{i}x_{i})$
0 - 20	8	10	80
20 - 40	10	30	300
40 -80	15	50	750
60 - 80	9	70	630
80 - 100	8	90	720
50			2480

Mean

$\displaystyle (\bar{x})=\frac{1}{n}\sum_{i=l}^{k}f_{i}x_{i}=\frac{2480}{50}=49.6$

Find the mean deviation about the mean for the data
$\displaystyle X_{i}$ 10 30 50 70 90
$\displaystyle f_{i}$ 4 24 28 16 8

$\displaystyle X_{i}$	$\displaystyle f_{i}$	$\displaystyle f_{i} \times X_{i}$	$\displaystyle \mid X_{i}-\bar{X} \mid$	$\displaystyle f_{1} \mid X_{i}-\bar{X} \mid$
10	4	40	40	160
30	24	720	20	480
50	28	1400	0	0
70	16	1120	20	320
90	8	720	40	320
	80	4000		1280

$\displaystyle N=\sum_{i=1}^{5}f_{1}=80 ,$

$\displaystyle \sum_{i=1}^{5}f_{1}x_{i}=4000$

$\displaystyle \therefore \bar{X}=\frac{1}{N}\sum_{i=1}^{5}f_{1}X_{1}=\frac{1}{80}\times 4000=50$

$\displaystyle \therefore MD(\bar{x})=\frac{1}{N}\sum_{i=1}^{1}f_{1}\mid x_{i}-\bar{x} \mid =\frac{1}{80}\times 1280=16$

Find the mean deviation about the mean for the data:
Income per day Numbers of persons
0-100 4
100-200 8
200-300 9
300-400 10
400-500 7
500-600 5
600-700 4
700-800 3

Income per day	Number of persons $(f_i)$	Mid-point $X_i$	$(f_i X_i)$	$\|X_i-\bar{X}\|$	$f_i \| X_i -\bar{X} \|$
0-100	4	50	200	308	1232
100-200	8	150	1200	208	1664
200-300	9	250	2250	108	972
300-400	10	350	3500	8	80
400-500	7	450	3150	92	644
500-600	5	550	2750	192	960
600-700	4	650	2600	292	1168
	50		17900		7896

Here

$\displaystyle N=\sum_{i=1}^{8}f_{i}=50$

$\sum_{i=1}^{8}f_{1}x_{i}=17900$

$\displaystyle \therefore \bar{X}=\frac{1}{N}\sum_{i=1}^{8}f_{i}x_{i}=\frac{1}{50}\times 17900=358$

$\therefore \displaystyle M.D.\left (\bar{x} \right )=\frac{1}{N}\sum_{i=1}^{8}f_{i}\mid x_{1}-\bar{x} \mid =\frac{1}{50}\times 7896=157.92$

Find the mean deviation about the mean for the data.
$\displaystyle x_{i}$ 5 10 15 20 25
$\displaystyle f_{i}$ 7 4 6 3 5

$\displaystyle X_{i}$	$\displaystyle f_{i}$	$\displaystyle f_{i} \times X_{i}$	$\displaystyle \mid X_{i}-\bar{X} \mid$	$\displaystyle f_{1} \mid X_{i}-\bar{X} \mid$
5	7	35	9	63
10	4	40	4	16
15	6	90	1	6
20	3	60	6	18
25	5	125	11	55
	25	350		158

$\displaystyle N=\sum_{i=1}^{5}f_{1}=25,$

$\displaystyle \sum_{i=1}^{5}f_{1}x_{i}=350$

$\displaystyle \therefore \bar{X}=\frac{1}{N}\sum_{i=1}^{5}f_{1}X_{1}=\frac{1}{25}\times 350=14$

$\displaystyle \therefore MD(\bar{x})=\frac{1}{N}\sum_{i=1}^{1}f_{1}\mid x_{i}-\bar{x} \mid =\frac{1}{25}\times 158=6.32$

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

To find the class mark

$\left( { x }_{ i } \right)$ for each interval, using the following relation:

$\left( { x }_{ i } \right) =\displaystyle\frac { Upper class limit + Lower class limit }{ 2 }$
Class size

$\left( h \right)$ for this data

$=10$
Taking 70 as assumed mean

$\left( a \right)$ , calculating

${ d }_{ i },{ u }_{ i },{ f }_{ i }{ u }_{ i }$ :
From the table:

$\sum { { f }_{ i } } =35$

$\sum { { f }_{ i }{ u }_{ i } } =-2$
Mean,

$\bar { x } =a+\left( \displaystyle\frac { \sum { { f }_{ i }{ u }_{ i } } }{ \sum { { f }_{ i } } } \right) \times h$

$=70+\left( \displaystyle\frac { -2 }{ 35 } \right) \times 10$

$=70-\displaystyle\frac { 20 }{ 35 }$

$=70-\displaystyle\frac { 4 }{ 7 }$

$=70-0.57$

$=69.43$
Therefore, mean literacy rate is

$69.43$ %

1028479_427004_ans_3a9059dce1a84fb0b9d9435199fe1045.bmp

The mean and variance of $7$ observation are $8$ and $16$ respectively. If five of the observation are $2,4, 10, 12$ and $14$ . Find the remaining two observations.

Let the remaining two observation be

$x$ and

$y$
The observation are

$2, 4, 10, 12, 14, x, y$
Mean

$=\displaystyle \bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$

$\displaystyle \Rightarrow 56=42+x+y$

$\displaystyle \Rightarrow x+y=14$ ......(1)
Variance

$=16= \displaystyle \frac{1}{n}\sum_{i=l}^{7}(X_{i}-\bar{X})^{2}$

$\displaystyle 16=\frac{1}{7}\left [ (-6)^{2}+(-4)^{2}+(2)^{2}+(4)^{2}+(6)^{2}+x^{2}+y^{2}-2 \times 8(x+y)+2\times (8)^{2} \right ]$

$\displaystyle 16=\frac{1}{7}\left [ 36+16+4+6+36+x^{2}+y^{2}-16(14)+2(64) \right ]$ ...........[using (1)]

$\displaystyle 16=\frac{1}{7}\left [ 108+x^{2}+y^{2}-224+128 \right ]$

$\displaystyle 16=\frac{1}{7}\left [ 12+x^{2}+y^{2} \right ]$

$\displaystyle \Rightarrow x^{2}+y^{2}=112-12=100$

$\displaystyle x^{2}+y^{2}=100$ ..........(2)
From (1), we obtain

$\displaystyle x^{2}+y^{2}+2xy=196$ ....(3)
From (2) and (3), we obtain

$2xy = 196 - 100$

$\displaystyle \Rightarrow 2xy = 96$ ...........(4)
subtracting (4) from (2), we obtain

$\displaystyle x^{2}+ y^{2}-2xy=100 - 96$

$\displaystyle \Rightarrow (x-y)^{2} = 4$

$\displaystyle \Rightarrow x-y=\pm 2$ ............(5)
Therefore, from (1) and (5) we obtain

$x = 8$ and

$y = 6$ when

$x - y = 2$

$x = 6$ and

$y = 8$ when

$x - y = -2$
Thus the remaining observation are

$6$ and

$8$

Find the mean deviation about the mean for the data
Height of cms Number of boys
95-105 9
105-115 13
115-125 26
125-135 30
135-145 12
145-155 10

The following table is formed

Height of cms	Number of boys $\displaystyle f_{i}$	mid-point $\displaystyle X_{i}$	$\displaystyle f_{i} \times X_{i}$	$\displaystyle \mid x_{i}-\bar{x} \mid$	$\displaystyle f_{i}\mid x_{i}-\bar{x} \mid$
95-105	9	100	900	25.3	227.7
105-115	13	110	1430	14.3	198.9
115-125	26	120	3120	5.3	137.8
125-135	30	130	3900	4.7	141
135-145	12	140	1680	14.7	176.4
145-155	10	150	1500	24.7	247

Here

$\displaystyle N=\sum_{i=1}^{6}f_{i}=100, \sum_{i=1}^{6}f_{1}x_{i}=12530$

$\displaystyle \therefore \bar{X}=\frac{1}{N}\sum_{i=1}^{6}f_{i}x_{i}=\frac{1}{100}\times 12530=125.3$

$\therefore \displaystyle M.D.\left (\bar{x} \right )=\frac{1}{N}\sum_{i=1}^{6}f_{i}\mid x_{1}-\bar{x} \mid =\frac{1}{100}\times 1128.8=11.28$

Consider the above distribution of daily wages of $50$ worker of a factory.
Find the mean daily wages.

To find the class mark

${ x }_{ i }$ for each interval, using the following relation:

Class mark

${ x }_{ i } =\displaystyle\frac { \text{Upper class limit} + \text{Lower class limit} }{ 2 }$

Class size

$h =20$

Taking

$150$ as assumed mean

$a.$

Values of

${ d }_{ i },{ u }_{ i }$ and

${ f }_{ i }{ u }_{ i }$ has been calculated in the table.

And,

$\displaystyle \sum { { f }_{ i } } =50$

$\displaystyle \sum { { f }_{ i }{ u }_{ i } } =-12$

So, mean will be equal to,

$\overline{ x } =a+\left( \displaystyle\frac { \sum { { f }_{ i }{ u }_{ i } } }{ \sum { { f }_{ i } } } \right) \times h$

$=150+\left( \displaystyle\frac { -12 }{ 50 } \right)\times 20$

$=150-\displaystyle\frac { 24 }{ 5 }$

$=150-4.8$

$=145.2$

$\therefore \;$ The mean daily wage of the workers of the factory is equal to

$\text{Rs. }145.20$ .

1028502_426936_ans_31889ac7977e42d096e2c7fea0cd1db3.bmp

An analysis of monthly wages paid to workers in two firms $A$ and $B$ belonging to the same industry gives the following results :
Firm A Firm B
No.of wage earners 586 648
Mean of monthly wages Rs 5253 Rs 5253
Variance of the distribution of wages 100 121
(i) Which firm $A$ or $B$ pays larger amount as monthly wages ?
(ii) Which firm $A$ or $B$ shows greater variability in individual wages ?

(i) Number of wage earner in firm

$A=586$

Mean of monthly wage of firm

$A =Rs 5253$

$\text { Mean of monthly age of firm A }=\cfrac { \text { Total amount paid } }{\text { No. of wage earners in firm A} }$

$5253=\cfrac{\text{Total amount paid}}{586}$

Total amount paid by Firm

$A$

$=5253 \times 586$

Number of wage earner in firm

$B=648$

Mean of monthly wage of firm

$B =Rs 5253$

$\text { Mean of monthly age of firm B }=\cfrac { \text { Total amount paid } }{\text { No. of wage earners in firm B} }$

$5253=\cfrac{\text{Total amount paid}}{648}$

Total amount paid by Firm

$B$

$=5253 \times 648$

Clearly, firm

$B$ paid larger amount as monthly wage.

(ii) Variance of the distribution of wages in firm

$A$

$\displaystyle (\sigma _{1}^{2}) = 100$

$\displaystyle \therefore$ Standard deviation of the distribution of wages in firm

$\displaystyle A (\sigma _{1})=\sqrt {100}= 10$

Variance of the distribution of wages in firm

$B \displaystyle (\sigma _{1}^{2}) = 121$

$\displaystyle \therefore$ Standard deviation of the distribution of wages in firm

$\displaystyle B (\sigma _{2})= \sqrt {121}= 11$

The mean of monthly wages of both the firms is same i.e.

$5253$ . So, the firm with greater standard deviation will have more variability.
Thus firm

$B$ has greater variability in the individual wages

The table below shows the daily expenditure on food of $25$ households in a locality.
Find the mean daily expenditure on food by a suitable method.

By direct method :

$Class$	$Frequency$ $f_i$	$Mid-value$ $x_i$	$f_ix_i$
$100-150$	$4$	$125$	$500$
$150-200$	$5$	$175$	$875$
$200-250$	$12$	$225$	$2700$
$250-300$	$2$	$275$	$550$
$300-350$	$2$	$325$	$650$
	$\sum f_i=25$		$\sum f_ix_i=5275$

$Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{5275}{25}=211$

$\therefore$ The mean of daily expenditure on food is

$211$ .

Using Step-Deviatopn Method, find the mean (rounded off to nearest integer) of following frequency distribution:

$A = 25, h = 10, N = 50$ and

$\sum { fi ui } = 4$

$\bar { x } =A+h\left[ \displaystyle\frac { 1 }{ N } \sum { fi ui } \right]$

$\Rightarrow$ Mean

$= 25 + 10 \times \left({4}/{50}\right)$

$\Rightarrow$ Mean

$= 25 + 0.8$

$\therefore$ Mean

$= 25.8$

To find out the concentration of $S{O}_{2}$ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

From the table:

$\sum { { f }_{ i } } =30$

$\sum { { f }_{ i }{ u }_{ i } } =-31$

And, class size of each interval

$= 0.04$

Let

${ x }_{ i }$ be the class mark for each interval

$\Rightarrow { x }_{ i } =\displaystyle\dfrac { Upper\ class\ limit + Lower\ class\ limit }{ 2 }$

Let us take

$0.14$ as the assumed mean

$\left( a \right)$

Mean,

$\overline { x } =a+\left( \displaystyle\dfrac { \sum { { f }_{ i }{ u }_{ i } } }{ \sum { { f }_{ i } } } \right) \times h$

$=0.14+\left( \displaystyle\dfrac { -31 }{ 30 } \right) \times 0.04$

$=0.14-0.04133$

$=0.09867$

$\approx 0.099\ ppm$

Therefore, mean concentration of

$S{O}_{2}$ in the air

$= 0.099\ ppm$

1028494_426987_ans_3c7ae0c2c1c34c77a0dfa81077e39379.png

In a mathematics test given to $15$ students, the following marks (out of $100$ ) are recorded:
$41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60$
Find the mean, median and mode of this data.

The marks of

$15$ students is

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

Mean

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

$= \dfrac{822}{15} =54.8$ ;

Number of observations

$= 15$ (odd)

Median score

$= 8$ th number arranging in ascending order which is

$52.$

Maximum frequency

$=3$ of

$52$

$\therefore$ Mode

$=52.$ .

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in $20$ houses in a locality.Find the mean number of plants per house.

Number of plants
$0-2$
$2-4$
$4-6$
$6-8$
$8-10$
$10-12$
$12-14$
Number of houses
$1$
$2$
$1$
$5$
$6$
$2$
$3$

Which method did you use for finding the mean, and why?

No. of plants	No. of houses $(f_i)$	Mid point $(X_i)$	$f_ix_i$
$0-2$	$1$	$1$	$1$
$2-4$	$2$	$3$	$6$
$4-6$	$1$	$5$	$5$
$6-8$	$5$	$7$	$35$
$8-10$	$6$	$9$	$54$
$10-12$	$2$	$11$	$22$
$12-14$	$3$	$13$	$39$
Total	$20$		$162$

Calculating mean, we get

Mean,

$\bar x = \dfrac1n\sum f_ix_i$

Here,

$n = 20, \sum f_ix_i = 162$

Therefore, Mean,

$\bar x = \dfrac{162}{20} = 8.1$ plants

We have used direct method because numerical values of

$f$ and

$x$ are small.

What is the measures of central tendency for the data set $5, 5, 10, 10, 5, 20, 25$ ?

Mean, median and mode of a data set are the measures of central tendency.
Step 1: Mean

$=$

$\dfrac{\sum X}{n}$

$\sum X$

$=$ sum of the data values.

$n =$ number of the data values
Mean

$=$

$\dfrac{5+5+10+10+5+20+25}{7}=11.4$
Step 2: The data set in the ascending order is

$5, 5, 5, 10, 10, 20, 25$
So, Median of the set is

$10$ . [Median is the middle data value of the ordered set.]
Step 3: Mode is the data value(s) that appear most often in the data set. So, the modes of the data set are

$5$ .
So, the measures of central tendency of the given set of data are mean

$= 11.4$ , median

$= 10$ and mode

$= 5$ .

Consider the following distribution of daily wages of $50$ workers of a factory.

Daily wages (in C)
$100-120$
$120-140$
$140-160$
$160-180$
$180-200$
Number of workers
$12$
$14$
$8$
$6$
$10$

Find the mean daily wages of the workers of the factory by using an appropriate method.

Class interval	Mid-value $(X_i)$	Frequency $(f_i)$	$u_1=\dfrac{x_i-150}{20}$	$f_iu_i$
$100-120$	$110$	$12$	$-2$	$-24$
$120-140$	$130$	$14$	$-1$	$-14$
$140-160$	$150$	$8$	$0$	$0$
$160-180$	$170$	$6$	$1$	$6$
$180-200$	$190$	$10$	$2$	$20$
Total		$50$		$-12$

Lets take assumed mean,

$A$ as

$150$

Class interval

$=20$

$\therefore u_i = \dfrac{x_i-A}h = \dfrac{x_i - 150}{20}$

$\bar x = A+h\dfrac{\sum f_iu_i}{\sum f_i} = 150 + 20\times \dfrac{-12}{50} = 150 - 4.8 = 145.2$

Hence, mean is Rs

$145.20$

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is $Rs.\ 18$ . Find the missing frequency $f$ .
Daily pocket allowance (in $Rs.$ )
$11-13$
$13-15$
$15-17$
$17-19$
$19-21$
$21-23$
$23-25$
Number of children
$7$
$6$
$9$
$13$
$f$
$5$
$4$

Lets take assumed mean,

$A$ as

$16$

Class interval

$=2$

$\therefore u_i = \dfrac{x_i-A}h = \dfrac{x_i - 16}{2}$

Now, we have

$A = 16, \bar x = 18$ and

$h =2$

We know that,

Mean,

$\bar x = A + h \left(\dfrac1N\sum f_iu_i\right)$

$\Rightarrow 18 = 16 + 2 \left(\dfrac{2f+24}{f+44}\right)$

$\Rightarrow \dfrac{2f+24}{f+44} = 1$

$\Rightarrow f+44 = 2f + 24$

$\Rightarrow f = 20$

Hence, the value of missing frequency,

$f = 20$

The following table gives the literacy rate (in percentage) of $35$ cities. Find the mean literacy rate.
Literacy rate (in %)
45-55
55-65
65-75
75-85
85-95
Number of cities
3
10
11
8
3

Here, we use the step deviation method to find meaning.

Let's take assumed mean,

$A$ as

$70$

Class interval

$=10$

$\therefore u_i = \dfrac{x_i-A}h = \dfrac{x_i - 70}{10}$

$mean (\bar x) = A+h\dfrac{\sum f_iu_i}{\sum f_i} = 70 + 10\times \dfrac{-2}{35} = 70 - 0.57 = 69.43$

Hence, the mean literacy rate is

$69.43\%$

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 a 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.

Concentration of $SO_2$ (In ppm)	Mid-value $(x_i)$	Frequency $(f_i)$	$f_ix_i$
$0.00-0.04$	$0.02$	$4$	$0.08$
$0.04-0.08$	$0.06$	$9$	$0.54$
$0.08-0.12$	$0.10$	$9$	$0.90$
$0.12-0.16$	$0.14$	$2$	$0.28$
$0.16-0.20$	$0.18$	$4$	$0.72$
$0.20-0.24$	$0.22$	$2$	$0.44$
Total		$30$	$2.96$

Let

$\bar x$ be the required mean.

Mean,

$\bar x = \dfrac1n\sum f_ix_i$

Here,

$n = 30, \sum f_ix_i = 2.96$

Therefore, Mean,

$\bar x = \dfrac{2.96}{30} = 0.099$ ppm

We have used direct method because numerical values of

$f$ and

$x$ are small.

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes
$50-5$ 2
$53-5$ 5
56-58
$59-6$ 1
$62-64$
Number of boxes
$15$
$110$
$135$
$115$
$25$
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

No. of mangoes	Mid - value $(x_i)$	Frequency $(f_i)$	$u_i = \dfrac{x_i-60}{3}$	$f_iu_i$
50 - 52	51	15	-3	-45
53 - 55	54	110	-2	-220
56 - 58	57	135	-1	-135
59 - 61	60	115	0	0
62 - 64	63	25	3	25
Total		400		-375

Here, we have used exclusive method in class interval formation.

If we make the series an inclusive one the mid-values remain same.

So there is no need to convert the series.

Lets take assumed mean,

$A$ as

$60$

Class interval

$=3$

$\therefore u_i = \dfrac{x_i-A}h = \dfrac{x_i - 60}{3}$

$\bar x = A+h\dfrac{\sum f_iu_i}{\sum f_i} = 60 + 3\times \dfrac{-375}{400} = 57.1875$

Hence, the mean number of mangoes kept in a packing box is

$57.19$

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heart beats per minute
$65-68$
$68-71$
$71-74$
$74-77$
$77-8$ 0
$80-83$
$83-86$
Number of women
$2$
$4$
$3$
$8$
$7$
$4$
$2$

Class interval	Mid value $(x_i)$	$(f_i)$	$u_i=\dfrac{x_i-75.5}{3}$	$f_iu_i$
$65-68$	$66.5$	$2$	$-3$	$-6$
$68-71$	$69.5$	$4$	$-2$	$-8$
$71-74$	$72.5$	$3$	$-1$	$-3$
$74-77$	$75.5$	$8$	$0$	$0$
$77-80$	$78.5$	$7$	$1$	$7$
$80-83$	$81.5$	$4$	$2$	$8$
$83-86$	$84.5$	$2$	$3$	$6$
Total		$30$		$4$

Lets take assumed mean,

$A$ as

$75.5$

Class interval

$=3$

$\therefore u_i = \dfrac{x_i-A}h = \dfrac{x_i - 75.5}{3}$

$\bar x= A+h\dfrac{\sum f_iu_i}{\sum f_i} = 75.5 + 3\times \dfrac{4}{30} = 75.5 +0.4 = 75.9$

Hence, the mean heart beats per minute for these women is

$75.9$

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

Lets take assumed mean,

$A$ as

$225$

Class interval

$=50$

$\therefore u_i = \dfrac{x_i-A}h = \dfrac{x_i - 225}{50}$

$\bar x = A+h\dfrac{\sum f_iu_i}{\sum f_i} = 225 + 50\times \dfrac{-7}{25} = 225 -14 = 211$

Hence, the mean daily expenditure on food is Rs.

$211$

Calculate the mean for the following frequency distribution:
Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Frequency $3$ $7$ $10$ $6$ $8$ $2$ $4$

Marks $(I)$	10-20	20-30	30-40	40-50	50-60	60-70	70-80
$x_i$	15	25	35	45	55	65	75
$f_i$	3	7	10	6	8	2	4
$f_ix_i$	45	175	350	270	440	130	300

$\bar x=\cfrac{\sum f_ix_i}{\sum f_i}=\cfrac{1710}{40}=42.75$

Find the median of $9$ observations $14, 36, 25, 28, 35, 32, 56, 42, 50$

$n=9.$

$n$ is odd, median is the mid value.

$\therefore 35$ is the median.

1171546_564767_ans_115af18fbd544cc6918fbcee8a5d6f11.png

The blood group of $50$ donors, participated in blood donation camp are $A, AB, B, A, O, AB, O, O, A, AB, B, A, O, AB, O, O, A, B, A, O, AB, O, O, A, AB,$ $B, O, AB, O, B, A, O, AB, O, O, A, AB, B, A, O, AB, O, A, AB, B, A, O, AB, O, O$ .
Find the mode of the above verbal data:

Blood group of

$50$ donors,

$A,A,A,A,A,A,A,A,A,A,A,A$

$B,B,B,B,B,B,B$

$AB,AB,AB,AB,AB,AB,AB,AB,AB,AB,AB,AB$

$O,O,O,O,O,O,O,O,O,O,O,O,O,O,O,O,O,O,O$

from the above data we can confer that the mode of the above data is Blood group

$O.$

Calculate the mean for the following frequency distribution:
Marks $15-19$ $20-24$ $25-29$ $30-34$ $35-39$ $40-44$
Frequency $6$ $5$ $9$ $12$ $6$ $2$

Marks $(I)$	15-19	20-24	25-29	30-34	35-39	40-44
$x_i$	17	22	27	32	37	42
$f_i$	6	5	9	12	6	2
$f_ix_i$	102	110	243	384	222	84

$\bar x=\cfrac{\sum f_ix_i}{\sum f_i}=\cfrac{1145}{40}=57.25$

Find the mean weight from the following table.
Weight ( $kg$ ) $29$ $30$ $31$ $32$ $33$
No. of children $02$ $01$ $04$ $03$ $05$

Weighted Mean

$=\cfrac{(2\times 29)+(1 \times 30)+(4 \times 31)+ (3\times 32)+(5 \times 33)}{2+1+4+3+5}$

Weighted Mean

$=31.53$

Find the median of the data: $22, 28, 34, 49, 44, 57, 18, 10, 33, 41, 66, 59$

Data in ascending Order

$10,18,22,28,33,34,41,44,49,57,59,66$

So, Median

$=\cfrac{34+41}{2}=37.50$

Find the mode of $10, 12, 11, 10, 15, 20, 19, 21, 11, 9, 10$

On arrangement of the numbers in ascending order

$9$

$10,10,10$

$11,11$

$12$

$15$

$19$

$20$

$2$

$\therefore$ Since

$10$ is used often,

$mode=10.$

There are four unknown numbers. The mean of the first two numbers is $4$ and the mean of the first three is $9$ . The mean of all four number is $15$ , if one of the four number is $2$ find the other numbers

$\textbf{Step - 1 : Formulate the equations from the means of 2,3,4 numbers }$

$\text{Let the four numbers be a, b, c, d}$

$\dfrac{a+b}{2}=4$

$\Rightarrow a+b=8$

$\bf{..........eqn (1)}$

$\dfrac{a+b+c}{3}=9$

$\Rightarrow a+b+c=27$

$\bf{......... eqn (2)}$

$\dfrac{a+b+c+d}{4}=15$

$\Rightarrow a+b+c+d=60$

$\bf{......... eqn (3)}$

$\textbf{Step - 2 : Try to solve the above 3 equations }$

$\text{eqn 2 - eqn 1}$

$\implies$

$\text{(a+b+c)-(a+b) = 27-8}$

$\Rightarrow\text{ c = 19}$

$\text{eqn 3 - eqn 2}$

$\implies$

$\text{(a+b+c+d)-(a+b+c) = 60-27}$

$\Rightarrow \text{ d = 33}$

$\text{Already given one number is 2, so let b = 2}$

$\text{Substituting b = 2 in equation 1 we get}$

$a+b = 8$

$\Rightarrow a = 8-2$

$\Rightarrow a = 6$

$\textbf{Hence, The 4 numbers are 6, 2, 19 and 33.}$

Calculate the mean of the following distribution:
Class Interval $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency $8$ $5$ $12$ $35$ $24$ $16$

Class	Mid value $\left(x\right)$	$f$	$fx$
0-10	5	8	40
10-20	15	5	75
20-30	25	12	300
30-40	35	35	1225
40-50	45	24	1080
50-60	55	16	880
Total		100	3600

Mean

$(\bar { x }) =\dfrac { \sum { fx } }{ \sum { f } }$

$=\dfrac { 3600 }{ 100 }$

$=36$ .

The sum of deviations of $n$ observations from $25$ is $25$ and sum of deviations of the same $n$ observations from $35$ is $-25$ . Find the mean of the observations

${\textbf{Step -1: Simplifying the given details.}}$

${\text{According to question, }}\dfrac{{\sum {\left( {x - 25} \right)} }}{n} = 25{\text{ and }}\dfrac{{\sum {\left( {x - 35} \right)} }}{n} = - 25.$

$\Rightarrow \dfrac{{\sum {x - \sum {25} } }}{n} = 25{\text{ and }}\dfrac{{\sum {x - \sum {35} } }}{n} = - 25.$

$\Rightarrow \dfrac{{\sum {x - 25n} }}{n} = 25{\text{ and }}\dfrac{{\sum {x - 35n} }}{n} = - 25.$

${\textbf{Step -2: Adding these two equations. }}$

$\therefore 2\sum {x = 60n} .$

$\therefore \dfrac{{\sum x }}{n} = 30.$

${\textbf{Hene, mean of observations = 30 .}}$

The histogram below represents the scores obtained by $25$ students in a Mathematics mental test. Use the data to to calculate mean of the distribution.

C.I.	$f$	$x$	$f\cdot x$
$0 - 10$	$2$	$5$	$10$
$10 - 20$	$5$	$15$	$75$
$20 - 30$	$8$	$25$	$200$
$30 - 40$	$4$	$35$	$140$
$40 - 50$	$6$	$45$	$270$
	$\sum f = 25$		$\sum f\cdot x = 695$

$Mean = \dfrac {\sum fx}{\sum f}$

$= \dfrac {695}{25} = 27.8$

Find the median of the data; $3.3, 3.5, 3.1, 3.7, 3.2, 3.8$

Median

$n=6$ [If

$n$ is even, median is average of two middle numbers.]

$\cfrac{3.3+3.5}{2}=\cfrac{6.8}{2}=3.4$

$\therefore$ Median

$3.4$ .

1018987_564806_ans_e2ef43933d394f23a7782344732179dd.png

Find the mean of the following distribution by step deviation method.
Class Interval $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Frequency $10$ $6$ $8$ $12$ $5$ $9$

C.I.	Frequency $(f_i)$	Mid-value(x)	$d_i=\displaystyle\frac{x-a}{h}$	$f_id_i$
$20-30$	$10$	$25$	$-2$	$-20$
$30-40$	$6$	$35$	$-1$	$-6$
$40-50$	$8$	$45$	$0$	$0$
$50-60$	$12$	$55$	$1$	$12$
$60-70$	$5$	$65$	$2$	$10$
$70-80$	$9$	$75$	$3$	$27$
	$\sum f_i=50$			$\sum f_id_i=23$

Here,

$a=45$ and

$h=10$

$\therefore$ Mean

$=a+\displaystyle\frac{\sum f_id_i}{\sum f_i}\times h$

$=45+\displaystyle\frac{23}{50}\times 10$

$=45+4.6=49.6$

Below is the given frequency distribution of words in an essay:
Number of Words Number of Candidates
$600 - 800$ $14$
$800 - 1000$ $22$
$1000 - 1200$ $30$
$1200 - 1400$ $18$
$1400 - 1600$ $16$
Find the mean number of words written

Number of Words	Class Mark $(x_{i})$	No. of Candidates $(f_{i})$	$f_{i}x_{i}$
$600 - 800$	$700$	$14$	$9,800$
$800 - 1000$	$900$	$22$	$19,800$
$1000 - 1200$	$1100$	$30$	$33,000$
$1200 - 1400$	$1300$	$18$	$23,400$
$1400 - 1600$	$1500$	$16$	$23,400$
		$\sum f_{i} = 100$	$\sum f_{i}x_{i} = 1,10,000$

$\overline {X} = \dfrac {\sum f_{i}x_{i}}{\sum f_{i}}$

$= \dfrac {1,10,000}{100}$

$= 1,100$

$\therefore$ Mean number of words in an essay is

$1100$ .

Per day expenses of 25 families of the frequency distribution of a Dhani of a village is given as follows.
Per day expense (In Rs.) 25-35 35-45 45-55 55-65 65-75
Number of families 3 7 6 6 3
Find the mean expense of families by Direct Method.

Mean expense of families can be calculated as:

$\overline x = \cfrac{\sum x_i n_i}{\sum n_i}$

$\overline x = \cfrac{30 \times 3 + 40 \times 7 + 50 \times 6 + 60 \times 6 + 70 \times 3}{25}$

$\overline x = \cfrac{1240}{25} =Rs.\ 49.6$

Find the mean for the following distribution table, using Deviation method.
Class Interval $0 - 9$ $10 - 19$ $20 - 29$ $30 - 39$ $40 - 49$ $50 - 50$ $60 - 60$
Frequency $8$ $10$ $22$ $26$ $18$ $9$ $7$

Let us find the class marks

$x_{i}$ of each class by taking the average of upper class limit and lower class limit and put them in a table.

Let us assume

$a=35$ and

$h=10$

Class Interval	Frequency	Class marks $(x_{i})$	$d_{i}=x_i-a$	$u_i=\dfrac{x_i-a}{h}$	$f_i u_i$
$0.5-9.5$	$8$	$5$	$-30$	$-3$	$-24$
$9.5-19.5$	$10$	$15$	$-20$	$-2$	$-20$
$19.5-29.5$	$22$	$25$	$-10$	$-1$	$-22$
$29.5-39.5$	$26$	$35$	$0$	$0$	$0$
$39.5-49.5$	$18$	$45$	$10$	$1$	$18$
$49.5-59.5$	$9$	$55$	$20$	$2$	$18$
$59.5-69.5$	$7$	$65$	$30$	$3$	$21$
Total	$\sum {f_i}=100$				$\sum{f_i u_i}=-9$

Using step deviation method,

$\overline x=a+\dfrac{\sum{f_i u_i}}{\sum{f_i}}\times h$

$=35+\dfrac{-9}{100}\times 9$

$=35-0.81$

$=34.19$

Calculate the Arthimetic mean of the following data.
x $15$ $25$ $35$ $45$ $55$ $65$ $75$ $85$
f $12$ $20$ $15$ $14$ $16$ $11$ $7$ $8$

Table for calculating mean

$x$

$f$

$fx$

15 12 180

25 20 500

35 15 525

45 14 630

55 16 880

65 11 715

75 7 525

85 8 680

$\sum f =103$ and

$\sum fx =4635$

Then, Arithmetic mean

$=\dfrac{\Sigma fx}{\Sigma f}=\dfrac{4635}{103}=45$

The measurements (in $mm$ ) of the diameters of the head of the screws are given below:
Diameter (in $mm$ ) No. of Screws
$30 - 35$ $9$
$36 - 38$ $21$
$39 - 41$ $30$
$42 - 44$ $20$
$45 - 47$ $18$
Calculate mean diameter of head of a screw by 'Assumed Mean Method'

Diameter (in $mm$ ) C.I.	Class Mark $(x_{i})$	$x_{i} - A = x_{i} - 40 = d_{i}$	No. of Screws $(f_{i})$	$f_{i}d_{i}$
$33 - 35$	$34$	$-6$	$9$	$-54$
$36 - 38$	$37$	$-3$	$21$	$-63$
$39 - 41$	$40\rightarrow A$	$0$	$30$	$0$
$42 - 44$	$43$	$3$	$20$	$60$
$45 - 47$	$46$	$6$	$18$	$108$
			$\sum f_{i} = 98$	$\sum f_{i}d_{i} = 51$

$\overline {d} = \dfrac {\sum f_{i}d_{i}}{\sum f_{i}}$

$= \dfrac {51}{98}$

$= 0.52$

$\overline {x} = A + \overline {d}$

$= 40 + 0.52$

$\overline {x} = 40.52$

$\therefore$ Mean diameter of head of a screw is

$40.52$ .

From the given information prepare the frequency table showing the value of $x_i f_i$ .
IQ
$(x_i)$ No. of students
$(f_i)$
$70-80$ $7$
$80-90$ $16$
$90-100$ $20$
$100-110$ $17$

Class Interval (IQ)	Frequency $(f_i)$ No. of students	Class Mark $(x_i)$	$(x_i f_i)$
$70-80$	$7$	$75$	$525$
$80-90$	$16$	$85$	$1360$
$90-100$	$20$	$95$	$1900$
$100-110$	$17$	$105$	$1785$

The marks distribution of $30$ students in a mathematics examination are as follows:

Class-interval of marks $10 - 25$ $25 - 40$ $40 - 55$ $55 - 70$ $70 - 85$ $85 - 100$
Number of students $2$ $3$ $7$ $6$ $6$ $6$

Find the mean by direct method and find also the mode of given data.

Let the mean be

$\overline x$

Using the direct method

$:\ \overline x = \cfrac{\sum x_i n_i}{\sum n_i}$

$= \cfrac{2 \times 17.5 + 3 \times 32.5 + 7 \times 47.5 + 6 \times 62.5 + 6 \times 77.5 + 6 \times 92.5}{30}$

$= \cfrac{1860}{30} = 62$

Mode is the data point that occurs the most frequently in the data set.

Hence, the mode is class-interval

$40-55.$

The following is the frequency distribution of time (in minutes) a worker takes to complete the work. Find mean time taken by a worker to complete the work by using 'Assumed Mean Method'.
Time (in minutes) No. of Workers
$20 - 24$ $2$
$25 - 29$ $10$
$30 - 34$ $15$
$35 - 39$ $20$
$40 - 44$ $28$
$45 - 49$ $25$

Interval	No. of Workers $(f_{i})$	Class Mark $(x_{i})$	Deviation $d_{i} = \dfrac {x_{i} - A}{h}$	$f_{i}d_{i}$
$20 - 24$	$2$	$22$	$-2$	$-4$
$25 - 29$	$10$	$27$	$-1$	$-10$
$30 - 34$	$15$	$32\rightarrow A$	$0$	$0$
$35 - 39$	$20$	$37$	$1$	$20$
$40 - 44$	$28$	$42$	$2$	$56$
$45 - 49$	$25$	$47$	$3$	$75$
Total	$\sum f_{i} = 100$			$\sum f_{i}d_{i} = 137$

Assumed Mean

$(A) = 32$

and class size

$h = 5$

$\therefore$ Mean

$(\overline {x}) = A + \dfrac {\sum f_{i}d_{i}}{\sum f_{i}}\times h$

$\overline {x} = 32 + \dfrac { 137}{100}\times 5$

$= 32 + 6.85$

$= 38.85$

Hence, the mean time is

$38.85$ min.

Below is given the distribution of money (in Rs.) collected by students for flood, relief fund
Money (in Rs) $0-10$ $10-20$ $20-30$ $30-40$ $$ 40-50$$
No. of Students $5$ $7$ $5$ $2$ $6$
Find mean of money (in Rs) collected by a student using 'Direct Method'.

Money (in Rs)	Class Mark, $x_i$	No. of Students, $f_i$	$f_ix_i$
$0-10$	$5$	$5$	$25$
$10-20$	$15$	$7$	$105$
$20-30$	$25$	$5$	$125$
$30-40$	$35$	$2$	$70$
$40-50$	$45$	$6$	$270$
Total		$\sum f_i = 25$	$\sum f_ix_i = 595$

Mean

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

$= \dfrac{595}{25}$

$= 23.8$
Mean of money collected by student is Rs.

$23.8$

Find the weighted A.M. of the price for the following data.
Food stuff Quantity(in kg) $w_i$ Price per kg (in Rs.) $x_i$
Rice $25$ $30$
Sugar $12$ $30$
Oil $8$ $70$

Here the x-values are the price of the given food stuff and the weights associated are the quantities (in Kg)
Then, the W.A.M

$=\displaystyle\frac{w_1x_1+w_2x_2+w_3x_3+....+w_nx_n}{w_1+w_2+w_3+......+w_n}$

$=\displaystyle\frac{25\times 30+12\times 30+8\times 70}{25+12+8}=\frac{1670}{45}$

$=Rs. 37.11$ .

For a month, a family requires the commodities listed in the table below. The weights to each commodity is given. Find the Weighted Arithmetic Mean.
Commodity Weight (in kg) Price per kg(in Rupees)
Rice $25$ $30$
Wheat $5$ $20$
Pulses $4$ $60$
Vegetables $8$ $25$
Oil $3$ $65$

Commodity	Weight $(w)$ (in kg)	Price per Kg $(x)$ (in rupees)	$xw$
Rice	25	30	750
Wheat	5	20	100
Pulses	4	60	240
Vegetables	8	25	200
Oil	3	65	195

Now,

$\sum w=45$ and

$\sum xw=1485$

Then, weighted arithmetic mean $=\dfrac{\Sigma xw}{\Sigma w}=\dfrac{1485}{45}=33$

The following table shows the weights of $12$ workers in a factory. Find the mean weight of the workers.
Weight (in kg) $60$ $64$ $68$ $70$ $72$
Number of workers $3$ $4$ $2$ $2$ $1$

Weight $(x)$ (kg)	Number of workers $(f)$	$fx$
60	3	180
64	4	256
68	2	136
70	2	140
72	1	72

Total 12 784

Then arithmetic mean = $\dfrac{\Sigma fx}{\Sigma f}=\dfrac{784}{12}=65.33$ Kg

Calculate the Arithmetic mean of the following data by direct method.
x $5$ $10$ $15$ $20$ $25$ $30$
f $4$ $5$ $7$ $4$ $3$ $2$

x	f	fx
$5$	$4$	$20$
$10$	$5$	$50$
$15$	$7$	$105$
$20$	$4$	$80$
$25$	$3$	$75$
$30$	$2$	$60$
Total	$N=25$	$\sum fx=390$

Arithmetic Mean,

$\bar{x}=\displaystyle\frac{\sum fx}{N}$

$=\displaystyle\frac{390}{25}=15.6$

Obtain the A.M of the following data.
Marks $65$ $70$ $75$ $80$ $85$ $90$ $95$ $100$
Number of students $6$ $11$ $3$ $5$ $4$ $7$ $10$ $4$

Marks $(x)$	Number of students $(f)$	$fx$
65	6	390
70	11	770
75	3	225
80	5	400
85	4	340
90	7	630
95	10	950
100	4	400
Total	50	4105

Then, arithmetic mean= $\dfrac{\Sigma fx}{\Sigma f}=\dfrac{4105}{50}=82.1$

Find the Weighted Arithmetic Mean for the following data.
Item Number of Item Cost of Item
Powder $2$ Rs. $45$
Soap $4$ Rs. $12$
Pen $5$ Rs. $15$
Instruments box $4$ Rs. $25.50$

Items	Number of items (w)	Cost of items (x)	xw
Powder	2	45	90
Soap	4	12	48
Pen	5	15	75
Instrument box	4	25.50	102
Total	15		312

Then, weighted arithmetic mean $=\dfrac{\Sigma xw}{\Sigma w}=\dfrac{312}{15}=20.8$

The following data give the number of boys of a particular age in a class of $40$ students. Calculate the mean age of the students.
Age(in years) $13$ $14$ $15$ $16$ $17$ $18$
Number of students $3$ $8$ $9$ $11$ $6$ $3$

Age (in years) x	Number of students $(f)$	$fx$
13	3	39
14	8	112
15	9	135
16	11	176
17	6	102
18	3	54
Total	40	618

Then, average mean

$=\dfrac{\sum fx}{\sum f}=\dfrac{618}{40}=15.45$

Find the median of $17, 15, 9, 13, 21, 7, 32$ .

Arrange the values in the ascending order as

$7, 9, 13, 15, 17, 21, 32,$
Here,

$n=7$ (odd number)
Therefore, Median

$=$ Middle value

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

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

$=4$ th value.
Hence, the median is

$15$ .

A cricket player has taken the runs $13, 28, 61, 70, 4, 11, 33, 0, 71, 92$ . Find the median.

Arrange the runs in ascending order as

$0, 4, 11, 13, 28, 33, 61, 70, 71, 92$ .
Here

$n=10$ (even number).
There are two middle values

$28$ and

$33$

$\therefore$ Median

$=$ Average of the two middle values

$=\displaystyle\frac{28+33}{2}=\frac{61}{2}=30.5$ .

Find the median of the following set of values.
1) $83, 66, 86, 30, 82$
2) $45, 49, 46, 44, 38, 37, 55, 51$
3) $70, 71, 70, 68, 67, 69, 70$
4) $51, 55, 46, 47, 53, 55, 51, 46$ .

The median is also the number that is halfway into the set. To find the median, the data should be arranged in order from least to greatest. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers. And if there are odd number of terms then middle number is the median.

(1)

$83,66,86,30,82$

Arranging in ascending order.

$30,66,82,83,86$

Number of terms is odd, then median is the middle number i.e.,

$82$

(2)

$45,49,46,44,38,37,55,51$

Arranging in ascending order

$37,38,44,45,46,49,51,55$

The number of terms is even then median is average of two middle numbers

$\therefore$ Median

$=\dfrac{45+46}{2}=45.5$

(3)

$70,71,70,68,67,69,70$

Arranging in ascending order

$67,68,69,70,70,70,71$

Then terms are odd then median is middle number i.e.,

$70$

(4)

$51,55,46,47,53,55,51,46$

Arranging in ascending order

$46,46,47,51,51,53,55,55$

The number of terms is even then median is average of two middle numbers

$\therefore$ Median

$=\dfrac{51+51}{2}=51$

In a study on patients, the following data were obtained. Find the arithmetic mean.
Age (in yrs) $10-19$ $20-29$ $30-39$ $40-49$ $50-59$
No. of patients $1$ $0$ $1$ $10$ $13$

Age	Midpoints	No.of patients
10-19	14.5	1
20-29	24.5	0
30-39	34.5	1
40-49	44.5	10
50-59	54.5	13
		N=25

Mean

$=\cfrac { 14.5\times 1+24.5\times 0+34.5\times 1+44.5\times 10+54.5\times13 }{ 25 }$

Mean

$=\cfrac { 14.5+0+34.5+445+708.5 }{ 25 } =\dfrac{1202.5}{25}=48.1$

Hence,average age=

$48.1$ years

Find the mode of $2, 4, 5, 2, 1, 2, 3, 4, 4, 6, 2$ .

2174065_618331_ans_2838d939aacb4969bc4c030be1771001.jpeg

The marks of ten students in a mathematics talent examinations are $75, 72, 59, 62, 72, 75, 71, 70, 70, 70$ . Obtain the mode.

Arranging the marks in ascending order

$59,62,70,70,70,71,72,72,75,75$

$70$ has the highest frequency i.e.

$3$

Therefore, mode of the given data

$=70$

Find the mode of $15, 25, 35, 45, 55, 65$ .

Each value occurs exactly one time in the series. Hence there is no mode for this data.

Compute the arithmetic mean of the following distribution.
Class Interval $0-19$ $20-39$ $40-59$ $60-79$ $80-99$
Frequency $3$ $4$ $15$ $14$ $4$

Class Interval	MidPoint ( ${ x }_{ i }$ )	Frequency ( ${ f }_{ i }$ )
0-19	9.5	3
20-39	29.5	4
40-59	49.5	15
60-79	69.5	14
80-99	89.5	4
		$\sum { { f }_{ i } } =40$

Mean

$=\cfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\cfrac { 3\times9.5 +4\times29.5+15\times49.5+14\times69.5+4\times89.5 }{ 40 } =55.5$

Find the median of the following numbers
(i) $24, 22, 23, 14, 15, 7, 21$
(ii) $17, 15, 9, 13, 21, 32, 42, 7, 12, 10$ .

(i) Let us arrange the numbers in ascending order as below.

$7, 14, 15, 21, 22, 23, 24$
Number of items

$n=7$
Median

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

$\because n$ is odd)

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

$={ 4 }^{ th }$ observation

$=21$
(ii) Let us arrange the numbers in ascending order

$7, 9, 10, 12, 13, 15, 17, 21, 32, 42$ .
Number of items

$n=10$
Median is the mean of

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

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

$\because n$ is even)

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

$={ \left( \dfrac { 10 }{ 2 } \right) }^{ th }$ observation

$={ 5 }^{ th }$ observation

$=13$

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

$={ 6 }^{ th }$ observation

$=15$ .

$\therefore$ Median

$=\dfrac { 13+15 }{ 2 } =14$

The mean of the following frequency distribution is $350$ , find the missing frequency
Class $100-200$ $200-300$ $300-400$ $400-500$ $500-600$ $600-700$
Frequency $5$ $3$ $3$ - $2$ $1$

Class-Interval	Frequency( $f_i$ )	Midpoint( $x_i$ )	$f_i x_i$
$100-200$	$5$	$150$	$750$
$200-300$	$3$	$250$	$750$
$300-400$	$3$	$350$	$1050$
$400-500$	$x$	$450$	$450x$
$500-600$	$2$	$550$	$1100$
$600-700$	$1$	$650$	$650$

From the above table,

$\sum f_i = 14 + x$

and

$\sum f_i x_i = 4300 + 450 x$

Let the missing frequency be

$'x'$ .
Mean,

$\overline x = 350$
We know that,

$\therefore \overline x = \dfrac{\sum f_i x_i}{\sum f_i}$

$\therefore 350 = \dfrac{4300 + 450 x}{14 + x}$

$\therefore 4900 + 350 x = 4300 + 450 x$

$\therefore 600 = 100 x$

$\therefore x = \dfrac{600}{100} = 6$
Hence, the missing frequency

$= 6$

The table below shows the classification of students who participated in a camp, according to their height.
Height (cm) Number of Students
130 - 135 8
135 - 140 12
140 - 145 20
145 - 150 28
150 - 155 32
155 - 160 22
160 - 165 16
165 - 170 12
Calculate the mean height of the students.

Table shows the classification of students who participated in a camp.

Height (cm)	Number of Students	Class Average	Total Height
130 - 135	8	132.5	1060
135 - 140	12	137.5	1650
140 - 145	20	142.5	2850
145 - 150	28	147.5	4130
150 - 155	32	152.5	4880
155 - 160	22	157.5	3465
160 - 165	16	162.5	2600
165 - 170	12	167.5	2010
Total	150		22645

Meanheight

$=\dfrac{22645}{150}=150.97 cm$

$\therefore$ Mean height of the students is

$150.97 cm$ .

Find the median of the following values of a variable:
$5,10,3,7,2,9,6,2,11$

The values arranged in increasing order are:

$2,2,3,5,6,7,9,10,11$
No. of values of

$n=9$ (odd)

$\therefore$ Median

$=\cfrac { (n+1)th }{ 2 } term$
Median

$=\cfrac { 9+1th\quad term }{ 2 }$
Median

$=\cfrac { 10th }{ 2 } term=5th term$
Median

$=6$

The marks obtained by $15$ students of a class are given below. Find the modal marks.
$42, 45, 47, 49, 52, 65, 65, 71, 71, 72, 75, 82, 72, 47, 72$

Arranging the given marks in ascending order

$42,45,47,47,49,52,65,65,71,71,72,72,72,75,82$

In the given data, frequency of

$72$ is the highest

$=3$

Therefore, modal marks

$=72$

Find the mode for the set of values $482, 485, 483, 485, 487, 487, 489$ .

Arranging the given data in ascending order

$482,483,485,485,487,487,489$

$485,487$ have the highest frequency i.e.

$2$

It is a bimodal data

Therefore, there are

$2$ modes

$=485,487$

The monthly income (in rupees) of $7$ households in a village are $1200, 1500, 1400, 1000,1000, 1600, 10000$ .
(i) Find the median income of the house holds.
(ii) If one more household with monthly income of $1500$ is added, what will the median income be?

(i) Arrange income in ascending order

$1000$ ,

$1200$ ,

$1400$ ,

$1500$ ,

$1600$ ,

$10000$

Here

$n=7$

Median

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

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

$= {4^{th}}$

$=1400$

(ii) If one more household with monthly salary

$1500$ is added

then, arranging the income in ascending order

$1000$ ,

$1200$ ,

$1400$ ,

$1500$ ,

$1600$ ,

$10000$

Here

$n=8$

Meadian

$=$ 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

$=$

$\frac{{ 1400 + 1500}}{2}$

$= 1450$

The following distribution shows the daily pocket allowance of children of a locality
Daily pocket allowance (in Rs.) 10-20 20-30 30-40 40-50 50-60
Number of Children $3$ $5$ $4$ $7$ $6$
Find the mean daily pocket allowance by using appropriate method

Daily allowance	Number of children ${x}_{i}$	class marks ${f}_{i}$	${ f }_{ i }{ x }_{ i }$
10-20	3	15	45
20-30	5	25	125
30-40	4	35	140
40-50	7	45	315
50-60	6	55	330
	$\sum { { x }_{ i } } =25$		$\sum { { f }_{ i }{ x }_{ i } } =955$

Mean

$=\dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { x }_{ i } } } =\dfrac { 955 }{ 25 } =38.2$

Calculate the mean of the following distribution using step deviation method.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Number of Students $10$ $9$ $25$ $30$ $16$ $10$

Class interval	Mid Values x $_i$	Frequency f $_i$	d $_i$ =X $_i$ -A	u $_i$ = $\dfrac{x_i-A}{h}$	f $_i$ u $_i$
0-10	5	10	-20	-2	-20
10-20	15	9	-10	-1	-9
20-30	25	25	0	0	0
30-40	35	30	10	1	30
40-50	45	16	20	2	32
50-60	55	10	30	3	30

Let

$A=25$ and

$h=10$

$\overline { X }$

$=A+$

$\dfrac { h\times \sum { { f }_{ i }{ u }_{ i } } }{ \sum { { f }_{ i } } }$

$=25+$

$\dfrac{10\times63}{100}$

$=25+6.3=31.3$ .

Hence, the mean is

$31.3$ .

Frequency distribution of daily commission received by $100$ salesmen is given.
Daily Commission(in Rs.) No. of Salesmen
$100-120$ $20$
$120-140$ $45$
$140-160$ $22$
$160-180$ $09$
$180-200$ $04$
Find mean daily commission received by salesman, by assumed mean method.

Let us take the assumed mean to be

$A = 150$ .

$Interval$	$x_i$	$f_i$	$d_i = x_i - A$	$f_i\,d_i$
$100-120$	$110$	$20$	$-40$	$-800$
$120-140$	$130$	$45$	$-20$	$-900$
$140-160$	$150$	$22$	$0$	$0$
$160-180$	$170$	$09$	$20$	$180$
$180-200$	$190$	$04$	$40$	$160$
$Total$		$\sum f_i$ = $100$		$\sum f_i\,d_i$ $=-1360$

Now the actual mean by the assumed mean method is given by,

$\overline{x} =$

$A\, +\, \dfrac{\sum f_i\,d_i}{\sum f_i}$

where

$\overline{x}$ is the actual mean and

$A$ is the assumed mean.

Substituting the values into the formula, we get:

$\overline{x} = 150 + \dfrac{(-1360)}{100}$

That gives,

$\overline{x} = 136.4$ . The mean daily commission is

$Rs\, 136.4$

Find the mean of the following frequency distribution by a suitable method.
C.I. 50-70 70-90 90-110 110-130 130-150 150-170
Frequency 8 12 13 27 18 22

$\Rightarrow$ By using direct method we are going find mean :

$C.I$	$Frequency$ $f_i$	$Class-mark$ $x_i$	$f_ix_i$
$50-70$	$8$	$60$	$480$
$70-90$	$12$	$80$	$960$
$90-110$	$13$	$100$	$1300$
$110-130$	$27$	$120$	$3240$
$130-150$	$18$	$140$	$2520$
$150-170$	$22$	$160$	$3520$
	$\sum f_i=100$		$\sum f_ix_i=12020$

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

$=\dfrac{12020}{100}$

$=120.2$

Find the value of $n$ , if the median of the numbers $n + 3, n + 7, ....n + 31, n + 35$ is $30$ .

The numbers are

$n+3,n+7,....n+31,n+35$

$n+(4(1)-1),n+(4(2)-1),....,n+(4(8)-1),n+(4(9)-1)$

Median is

$n+\bigg(4\bigg(\dfrac{1+9}{2}\bigg)-1\bigg)=n+19=30$

$\implies n=11$

Find the number of telephone calls received at an exchange per interval for $250$ successive one-minute intervals.

No. of calls ( $x$ ) $0$ $1$ $2$ $3$ $4$ $5$ $6$
No. of intervals ( $f$ ) $15$ $24$ $29$ $46$ $54$ $43$ $39$

Answer:-

No. of calls (x)	No. of intervals (f)	fx
0	15	0
1	24	24
2	29	58
3	46	138
4	54	216
5	43	215
6	39	234

$\Sigma{f} = 15+24+29+46+54+43+39= 250$

$\Sigma{fx} =0+24+58+138+216+215+234 =885$

$(\bar{x}) = \cfrac{\Sigma{fx}}{\Sigma{f}} = \cfrac{885}{250} = 3.54$

The number of telephone calls received at an exchange per interval for

$250$ successive one-minute intervals is

$3.54$ .

The following table gives the number of boys of a particular age in a class of $40$ students. Calculate the mean age of the students.
Age (in years) $15$ $16$ $17$ $18$ $19$ $20$
No. of students $3$ $8$ $10$ $10$ $5$ $4$

Mean age

$= \dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ mean

$=\dfrac {(15\times 3)+(16\times 8)+(17 \times 10)+(18 \times 10)+(19 \times 5)+(20 \times 4)}{3+8+10+10+5+4}$

$\Rightarrow$ mean age

$=\dfrac {45+128+170+180+95+80}{40}$

$\Rightarrow$ mean age

$=17.45$

Five coins were simultaneously tossed $1000$ times and at each toss the number of heads were observed. The number of tosses during which $0,1,2,3,4$ and $5$ heads were obtained. They are shown in the table below. Find the mean number of heads per toss.
No. of heads per toss No. of tosses
$0$ $38$
$1$ $144$
$2$ $342$
$3$ $287$
$4$ $164$
$5$ $25$
Total $1000$

Answer:-

No. of heads per toss (x)	No. of tosses (f)	fx
0	38	0
1	144	144
2	342	684
3	287	861
4	164	656
5	25	125
	$\Sigma{f} = 1000$	$\Sigma{fx} = 2470$

mean number of heads per toss

$(\bar{x}) = \cfrac{\Sigma{fx}}{\Sigma{f}} = \cfrac{2470}{1000} = 2.47$

Hence, mean number of heads per toss is 2.47.

The following table gives the number of branches and number of plants in the garden of a school.
No. of branches ( $x$ ): $2$ $3$ $4$ $5$ $6$
No. of plants ( $f$ ): $49$ $43$ $57$ $38$ $13$
Calculate the average number of branches per plant

No. of branches ( $x$ )	No. of plants $(f)$	$fx$
$2$	$49$	$98$
$3$	$43$	$129$
$4$	$57$	$228$
$5$	$38$	$190$
$6$	$13$	$78$
	$\Sigma{f} = 49+43+57+38+13 \\ \ \ \ \ \ \ =200$	$\Sigma{fx} =98+129+228+190+78 \\ \ \ \ \ \ \ \ \ = 723$

average number of branches per plant

$(\bar{x}) = \cfrac{\Sigma{fx}}{\Sigma{f}} = \cfrac{723}{200} = 3.615$

Hence, average number of branches per plant is

$3.615.$

The marks obtained out of $50$ , by $102$ students in a Physics test are given in the frequency table below:
Marks ( $x$ ) $15$ $20$ $22$ $24$ $25$ $30$ $33$ $28$ $45$
Frequency ( $f$ ) $5$ $8$ $11$ $20$ $23$ $18$ $13$ $3$ $1$
Find the average mark.

Answer:-

Marks (x)	Frequency (f)	fx
15	5	75
20	8	160
22	11	242
24	20	480
25	23	575
30	18	540
33	13	429
28	3	84
45	1	45
	$\Sigma{f} = 5+8+11+20+23+18+13+3+1 \\ \ \ \ \ \ = 102$	$\Sigma{fx} = 75+160+242+480+575+540+429+84+45\\ \ \ \ \ \ \ \ \ =2630$

$Average\ mark (\bar{x}) = \cfrac{\Sigma{fx}}{\Sigma{f}} = \cfrac{2630}{102} = 25.78$

Find the value of $p$ for the following distribution whose mean is $16.6$ .
$x$ : $8$ $12$ $15$ $p$ $20$ $25$ $30$
$f$ : $12$ $16$ $20$ $24$ $16$ $8$ $4$

Mean =

$\dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ mean = $\dfrac {(8\times 12)+(12\times 16)+(15 \times 20)+(p \times 24)+(20 \times 16)+(25 \times 8)+(30 \times 4)}{12+16+20+24+16+8+4}$

$\Rightarrow$ mean =

$\dfrac {96+192+300+24p+320+200+120}{12+16+20+84+16+8+4}=16.6$

$\Rightarrow 1660=1228+24p$

$\Rightarrow 24p=432$

$\Rightarrow p=18$

Calculate the mean for the following distribution:
$x$ : $5$ $6$ $7$ $8$ $9$
$f$ : $4$ $8$ $14$ $11$ $3$

Mean =

$\dfrac {\sum x_{i} f_{i}} {\sum f_{i}}$

$\Rightarrow$ mean = $\dfrac {(5\times 4)+(6\times 8)+(7 \times 14)+(8 \times 11)+(9 \times 3)}{4+8+14+11+3}$

$\Rightarrow$ mean =

$\dfrac {20+48+98+88+27}{4+8+14+11+3}=7.025$

Find the mean of the following frequency distribution:

Class interval: 0-8 8-16 16-24 24-32 32-40
Frequency: $6$ $7$ $10$ $8$ $9$

Answer:-

Class interval	${x}_{i}$	Frequency $({f}_{i})$	${f}_{i} {x}_{i}$
0-8	4	6	24
8-16	12	7	84
16-24	20	10	200
24-32	28	8	224
32-40	36	9	324
		$\Sigma{f}_{i} = 40$	$\Sigma{{f}_{i} {x}_{i}} = 856$

$mean = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{856}{40} = 21.4$

The following distribution gives the number of accidents met by 160 workers in a factory during a month.
No. of accidents ( $x$ ): $0$ $1$ $2$ $3$ $4$
No. of workers ( $f$ ): $70$ $52$ $34$ $3$ $1$
Find the average number of accidents per worker.

No. of accidents (x)	No. of workers (f)	fx
0	70	0
1	52	52
2	34	68
3	3	9
4	1	4
	$\Sigma{f} = 160$	$\Sigma{fx} = 133$

We know that, by direct method, the average or mean

$(\bar{x}) = \cfrac{\Sigma{fx}}{\Sigma{f}}$

$\therefore\$ Average number of accidents per worker

$= \cfrac{133}{160}$

$= 0.83$

Hence, the average number of accidents per worker is

$0.83.$

In the first proofreading of a book containing $300$ pages the following distribution of misprints was obtained:
No. of misprints
per page ( $x$ ) $0$ $1$ $2$ $3$ $4$ $5$
No. of pages ( $f$ ) $154$ $95$ $36$ $9$ $5$ $1$
Find the average number of misprints per page.

Answer:-

No. of misprints per page (x)	No. of pages (f)	fx
0	154	0
1	95	95
2	36	72
3	9	27
4	5	20
5	1	5
	$\Sigma{f} = 300$	$\Sigma{f} = 219$

Average number of misprints per page

$\bar{x}$

$= \cfrac{\Sigma{fx}}{\Sigma{f}} = \cfrac{219}{300} = 0.73$

Find the mean of each of the following frequency distributions:
Class interval: 0-6 6-12 12-18 18-24 24-30
Frequency: $6$ $8$ $10$ $9$ $7$

Answer:-

Class interval	Frequency	${x}_{i}$	${f}_{i} {x}_{i}$
0-6	6	3	18
6-12	8	9	72
12-18	10	15	150
18-24	9	21	189
24-30	7	27	189
	$\Sigma{{f}_{i}} = 40$		$\Sigma{{f}_{i} {x}_{i}} = 618$

$mean = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{618}{40} = 15.45$

Find the mean of the following frequency distribution:

Class interval: 25-35 35-45 45-55 55-65 65-75
Frequency: $6$ $10$ $8$ $12$ $4$

Answer:-

Class interval	Frequency	${x}_{i}$	${f}_{i} {x}_{i}$
25-35	6	30	180
35-45	10	40	400
45-55	8	50	400
55-65	12	60	720
65-75	4	70	280
	$\Sigma{{f}_{i}} = 40$		$\Sigma{{f}_{i} {x}_{i}} = 1980$

$Mean = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{1980}{40} = 49.5$

Find the mean of the following frequency distribution:

Class interval: 0-8 8-16 16-24 24-32 32-40
Frequency: $5$ $6$ $4$ $3$ $2$

Answer:-

Class interval	Frequency	${x}_{i}$	${f}_{i} {x}_{i}$
0-8	5	4	20
8-16	6	12	72
16-24	4	20	80
24-32	3	28	84
32-40	2	36	72
	$\Sigma{{f}_{i}} = 20$		$\Sigma{{f}_{i} {x}_{i}} = 328$

$mean = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{328}{20} = 16.4$

Find the mean from the following frequency distribution of marks at a test in class.
Maks ( $x$ ): $5$ $10$ $15$ $20$ $25$ $30$ $35$ $40$ $45$ $50$
No. of students ( $f$ ) $15$ $50$ $80$ $76$ $472$ $45$ $39$ $9$ $8$ $6$

Answer:-

Marks (x)	No. of students (f)	fx
$5$	$15$	$75$
$10$	$50$	$500$
$15$	$80$	$1200$
$20$	$76$	$1520$
$25$	$472$	$11800$
$30$	$45$	$1350$
$35$	$39$	$1365$
$40$	$9$	$360$
$45$	$8$	$360$
$50$	$6$	$300$
	$\Sigma{f} = 800$	$\Sigma{fx} = 18830$

Let

$\bar{x}$ be the required mean

$(\bar{x}) = \cfrac{\Sigma{fx}}{\Sigma{f}} \\\ \ \ \ = \cfrac{18830}{800} \\\ \ \ \ = 23.53$

Hence, Mean

$=23.53$

Find the mean of the following frequency distribution:

Class interval: 0-6 6-12 12-18 18-24 24-30
Frequency: $7$ $5$ $10$ $12$ $6$

Answer:-

Class interval	Frequency	${x}_{i}$	${f}_{i} {x}_{i}$
0-6	7	3	21
6-12	5	9	45
12-18	10	15	150
18-24	12	21	252
24-30	6	27	162
	$\Sigma{{f}_{i}} = 40$		$\Sigma{{f}_{i} {x}_{i}} = 630$

$mean = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{630}{40} = 15.75$

Find the mean of the following frequency distribution:

Class interval: 0-10 10-20 20-30 30-40 40-50
Frequency: $9$ $12$ $15$ $10$ $14$

Answer:-

Class interval	Frequency	${x}_{i}$	${f}_{i} {x}_{i}$
0-10	9	5	45
10-20	12	15	180
20-30	15	25	375
30-40	10	35	350
40-50	14	45	630
	$\Sigma{{f}_{i}} = 60$		$\Sigma{{f}_{i} {x}_{i}} = 1580$

$mean = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{1580}{60} = 26.33$

The number of students absent in a class were recorded for $120$ days and the information is given in the following frequency table:
No. of students
absent ( $x$ ) $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
No. of days ( $f$ ) $1$ $4$ $10$ $50$ $34$ $15$ $4$ $2$
Find the mean number of students absent per day.

No. of students absent (x)	No. of days (f)	fx
0	1	0
1	4	4
2	10	20
3	50	150
4	34	136
5	15	75
6	4	24
7	2	14
	$\Sigma{f} = 1+4+10+50+34+15+4+2\\ \ \ \ \ =120$	$\Sigma{fx} =0+4+20+150+136+75+24+14\\ \ \ \ \ \ \ \ = 423$

Mean number of students absent per day

$(\bar{x}) = \cfrac{\Sigma{fx}}{\Sigma{f}} = \cfrac{423}{120} = 3.525$

Hence, the mean number of students absent per day is

$3.525$ .

Find the mean of the following frequency distribution:

Class interval: 10-30 30-50 50-70 70-90 90-110 110-130
Frequency: $5$ $8$ $12$ $20$ $3$ $2$

Answer:-

Class interval	${x}_{i}$	Frequency $({f}_{i})$	${f}_{i} {x}_{i}$
10-30	20	5	100
30-50	40	8	320
50-70	60	12	720
70-90	80	20	1600
90-110	100	3	300
110-130	120	2	240
		$\Sigma{f}_{i} = 50$	$\Sigma{{f}_{i} {x}_{i}} = 3280$

$mean = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{3280}{50} = 65.6$

The following table shows the marks scored by $140$ students in an examination of a certain paper:
Marks: 0-10 10-20 20-30 30-40 40-50
Number of students: $20$ $24$ $40$ $36$ $20$
Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

Direct method

size	x	f	xf
0-10	5	20	100
10-20	15	24	360
20-30	25	40	1000
30-40	35	36	1260
40-50	45	20	900

mean

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

$=\dfrac{3620}{140}=25.857$

Assumed mean method

size	x	u=x-25	f	uf
0-10	5	-20	20	-400
10-20	15	-10	24	-240
20-30	25	0	40	0
30-40	35	10	36	360
40-50	45	20	20	400

mean

$=A+\dfrac{\sum fu}{\sum f}$

$=25+\dfrac{120}{140}=25.875$

Step deviation method

size	x	d=x-25	u=(x-25) / 10	f	uf
0-10	5	-20	-2	20	-40
10-20	15	-10	-1	24	-24
20-30	25	0	0	40	0
30-40	35	10	1	36	36
40-50	45	20	2	20	40

mean

$=A+h \times \dfrac{\sum fu}{\sum f}$

$=25+10 \times \dfrac{12}{140}=25.875$

If the mean of the following distribution is $27$ , find the value of $p$ .
Class: 0-10 10-20 20-30 30-40 40-50
Frequency: $8$ $p$ $12$ $13$ $10$

Answer:-

Marks	No. of students ${f}_{i}$	${x}_{i}$	${f}_{i} {x}_{i}$
0-10	8	5	40
10-20	$p$	15	15 $p$
20-30	12	25	300
30-40	13	35	455
40-50	10	45	450
	$\Sigma{{f}_{i}} = 43 + p$		$\Sigma{{f}_{i} {x}_{i}} = 1245 + 15p$

$mean = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = 27 \Rightarrow \left( given \right)$

$\therefore \; \cfrac{1245 + 15p}{43 + p} = 27$

$\Rightarrow 1245 + 15p = 27 \left( 43 + p \right)$

$\Rightarrow 1245 + 15p = 1161 + 27p$

$\Rightarrow 1245 - 1161 = 27p - 15p$

$\Rightarrow 12p = 84$

$\Rightarrow p = 7$

Find the mean of the following frequency distribution:

Class interval: 25-29 30-34 35-39 40-44 45-49 50-54 55-59
Frequency: $14$ $22$ $16$ $6$ $5$ $3$ $4$

Answer:-

class interval	Mid-point ${x}_{i}$	Frequency ${f}_{i}$	${f}_{i} {x}_{i}$
25-29	27	14	378
30-34	32	22	704
35-39	37	16	592
40-44	42	6	252
45-49	47	5	235
50-54	52	3	156
55-59	57	4	228
		$\Sigma{f}_{i} = 70$	$\Sigma{f}_{i} {x}_{i} = 2545$

$Mean = \cfrac{\Sigma {f}_{i} {x}_{i}}{\Sigma{f}_{i}} = \cfrac{2545}{70} = 36.36$

The following distribution shows the daily pocket allowance given the children of a multistorey building. The average pocket allowance is Rs. $18.00$ . Find out the missing frequency.
Class interval: 11-13 13-15 15-17 17-19 19-21 21-23 23-25
Frequency: $7$ $6$ $9$ $13$ --- $5$ $4$

Answer:-

Let the missing frequency be

$x$ .

Class interval	Frequency	${x}_{i}$	${f}_{i} {x}_{i}$
11-13	7	12	84
13-15	6	14	84
15-17	9	16	144
17-19	13	18	234
19-21	$x$	20	20 $x$
21-23	5	22	110
23-25	4	24	96
	$\Sigma{f}_{i} = 44 + x$		$\Sigma {f}_{i} {x}_{i} = 752 + 20x$

$\text{average pocket allowance } \bar{x} = \cfrac{\Sigma{f}_{i}{x}_{i}}{\Sigma{f}_{i}} = 18 \Rightarrow \left(given\right)$

$\therefore \; \cfrac{752 + 20x}{44 + x} = 18$

$\Rightarrow 752 + 20x = 18 \left(44 + x \right)$

$\Rightarrow 752 + 20x = 792 + 18x$

$\Rightarrow 2x = 40$

$\Rightarrow x = 20$

Hence, the missing frequency is

$20$ .

The weekly observations on cost of living index in a certain city for the year 2004-2005 are given below. Compute the weekly cost of living index.
Cost of living
Index Number of
students Cost of living
Index Number of students
1400-1500 $5$ 1700-1800 $9$
1500-1600 $10$ 1800-1900 $6$
1600-1700 $20$ 1900-2000 $2$

Answer:-

Cost of living index	${x}_{i}$	Frequency $({f}_{i})$	${f}_{i} {x}_{i}$
1400-1500	1450	5	7250
1500-1600	1550	10	15500
1600-1700	1650	20	33000
1700-1800	1750	9	15750
1800-1900	1850	6	11100
1900-2000	1950	2	3900
		$\Sigma{f}_{i} = 52$	$\Sigma{{f}_{i} {x}_{i}} = 86500$

Mean

$= \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}} = \cfrac{86500}{52} = 1663.46$

The weekly cost of living index is

$1663.46$ .

For the following distribution, calculate mean using all suitable methods:
Size of item 1-4 4-9 9-16 16-27
Frequency $6$ $12$ $26$ $20$

size	x	f	xf
1-4	2.5	6	15
4-9	6.5	12	78
9-16	12.5	26	325
16-27	21.5	20	430

mean

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

$=\dfrac{848}{64}=13.25$

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

Given :

$\bar{x} = 62.8$ ,

$\Sigma{f}_{i} = 50$

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

$\Sigma{f}_{i} = 30 + {f}_{1} + {f}_{2}$

$50 = 30 + {f}_{1} + {f}_{2}$

${f}_{1} + {f}_{2} = 20 \; \longrightarrow {eq}^{n} (1)$

$\bar{x} = \cfrac{\Sigma{{f}_{i} {x}_{i}}}{\Sigma{{f}_{i}}}$

$\Rightarrow 62.8 = \cfrac{2060 + 30{f}_{1} + 70{f}_{2}}{30 + {f}_{1} + {f}_{2}}$

$\Rightarrow 1884 + 62.8{f}_{1} + 62.8{f}_{2} = 2060 + 30{f}_{1} + 70{f}_{2}$

$\Rightarrow 62.8{f}_{1} - 30{f}_{1} + 62.8{f}_{2} - 70{f}_{2} = 2060 - 1884$

$\Rightarrow 32.8{f}_{1} - 7.2{f}_{2} = 176 \; \longrightarrow {eq}^{n} (2)$

On solving

${eq}^{n} (1) \; \& \; (2)$ we get,

${f}_{1} = 8$ and

${f}_{2} = 12$ .

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes $50-52$ $53-55$ $56-58$ $59-61$ $62-64$
Number of boxes $15$ $110$ $135$ $115$ $25$
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Number of mangoes	Number of boxes	${x}_{i}$	${f}_{i} {x}_{i}$
50-52	15	51	765
53-55	110	54	5940
56-58	135	57	7695
59-61	115	60	6900
62-64	25	63	1575
	$\Sigma{f}_{i} = 400$		$\Sigma{f}_{i}{x}_{i} = 22875$

$\bar{x} = \cfrac{\Sigma{f}_{i}{x}_{i}}{\Sigma{f}_{i}} = \cfrac{22875}{400} = 57.1875$

Hence, the mean number of mangoes kept in a packing box is

$57.1875$ .

The table below gives the distribution of villages under different heights from sea level in a certain region. Compute the mean height of the region:
Height (in meters): 200 600 1000 1400 1800 2200
No. of village: 142 265 560 271 89 16

Let the assumed mean be

$A = 1400$ and

$h= 400$
Calculation of Mean

Height (in meters) $x_i$	No. of villages $f_i$	$d_i= x_i- 1400$	$u_i= \dfrac{x_i-1400}{400}$	$f_iu_i$
200	142	-1200	-3	-426
600	265	-800	-2	-530
1000	560	-400	-1	-560
1400(A)	271	0	0	0
1800	89	400	1	89
2200	16	800	2	32
	$N= \sum f_i=1343$			$\sum f_iu_i=-1395$

We have,

$A= 1400, h= 400,$

$\sum f_iu_i= - 1395$ and

$N= 1343$

$Mean = A+h[\dfrac{1}{N}\sum f_iu_i]$

$\Rightarrow Mean = 1400+ 400 \times \dfrac{-1395}{\ \ \ 1343}= 1400 -415.49 = 984.51$

Apply step-deviation method to find the AM of the following frequency distribution
Variate $(x)$ :
$5$ $10$ $15$ $20$ $25$ $30$ $35$ $40$ $45$ $50$
Frequency $(f)$ :
$20$ $43$ $75$ $67$ $72$ $45$ $39$ $9$ $8$ $6$

Let the assumed mean be

$A = 25$ and

$h = 5$

Calculation of Mean

Variable $x_{i}$	Frequency $f_{i}$	Deviations $d_{i} = x_{i} - A$	$u_{i} = \dfrac{x_{i} - 25}{25}$	$f_{i} u_{i}$
$5$	$20$	$-20$	$-4$	$-80$
$10$	$43$	$-15$	$-3$	$-129$
$15$	$75$	$-10$	$-2$	$-150$
$20$	$67$	$-5$	$-1$	$-150$
$25(A)$	$72$	$0$	$0$	$0$
$30$	$45$	$5$	$1$	$45$
$35$	$39$	$10$	$2$	$78$
$40$	$9$	$15$	$3$	$27$
$45$	$8$	$20$	$4$	$32$
$50$	$6$	$25$	$5$	$30$
	$N = \sum f_{i} = 384$			$\sum f_{i}\ u_{i} = -214$

We have,

$N= 384 , A= 25, h= 5$ and

$\sum f_{i}\ u_{i} = -214$

$\therefore$ Mean

$= \overline {X}= A+ h\Big[\dfrac{1}{N}\sum f_{i}\ u_{i}\Big]$

$\Rightarrow$ Mean

$= 25 + 5 \times \left (\dfrac{-214}{384} \right )= 25 -2.786= 22.214$

A frequency distribution of the of life times 400 T.V. picture tubes tested in a tube company is given below. Find the average life of tube.
Life time ( in hrs) Frequency
300-399 14
400-499 46
500-599 58
600-699 76
700-68 68
800-899 62
900-999 48
1000-1099 22
1100-1199 6

Here, the class- intervals are formed by exclusive method. If we make the series an inclusive one the mid-values remain same. So, there is no need to convert the series into an inclusive form.
Let the assumed mean be A= 749.5 and h= 100.
Calculation of Mean

Life time (in hrs):	Frequency $f_i$	Mid-Values $x_i$	$d_i= x_i-A=x_i- 749.5$	$u_i=\frac{x_i-A}{h}$ $u_i=\frac{x_i-749.5}{100}$	$f_iu_i$
300-399	14	349.5	-400	-4	-56
400-499	46	449.5	-300	-3	-138
500-599	58	549.5	-200	-2	-116
600-699	76	649.5	-100	-1	-76
700-799	68	749.5	0	0	0
800-899	62	849.5	100	1	62
900-999	48	949.5	200	2	96
1000-1099	22	1049.5	300	3	66
1100-1199	6	1149.5	400	4	24
	$N= \sum f_i= 400$				$\sum f_iu_i$ =-138

We have,
N= 400, A= 749.5, h= 100 and

$\sum f_iu_i= -138$

$\therefore \overline {X}= A + h [\frac{1}{N} \sum f_iu_i]$

$\Rightarrow \overline {X} = 749.5+ 100( \frac{-138}{400})=749.5 - 34.5 = 715$ .
Hence, the average life time of a tube is 715 hours.

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) 118-126 127-135 136-144 145-153 154-162 163-171 172-180
No. of leaves: $3$ $5$ $9$ $12$ $5$ $4$ $2$
Find the mean length of leaf.

$Length$	$f_i$	$x_i$	$f_i x_i$
$117.5-126.5$	$3$	$122$	$366$
$126.5-135.5$	$5$	$131$	$655$
$135.5-144.5$	$9$	$140$	$1260$
$144.5-153.5$	$12$	$149$	$1788$
$153.5-162.5$	$5$	$158$	$790$
$162.5-171.5$	$4$	$167$	$668$
$171.5-180.5$	$2$	$176$	$352$
	$\sum f_i=40$		$\sum f_i x_i=5879$

$\Rightarrow$

$Mean=\dfrac{\sum f_i x_i}{\sum f_i}=\dfrac{5849}{40}=146.975$

Find the mean wage from the following data:
Wage( in Rs.) 800 820 860 900 920 980 1000
No. of workers: 7 14 19 25 20 10 5

Wage(inRs) $x_i$	No. of workers $f_i$	$d_i= x_i- A$ $= x_i-900$	$u_i=\dfrac{x_i-900}{20}$	$f_iu_i$
$800$	$7$	$-100$	$-5$	$-35$
$820$	$14$	$-80$	$-4$	$-56$
$860$	$19$	$-40$	$-2$	$-38$
$900$	$25$	$0$	$0$	$0$
$920$	$20$	$20$	$1$	$20$
$980$	$10$	$80$	$4$	$40$
$1000$	$5$	$100$	$5$	$25$
	$N=\sum f_i= 100$			$\sum f_iu_i= -44$

We have
N= 100,

$\displaystyle \sum fiui$ = -44, A = 900 and h= 20.

$\therefore Mean =\overline {X} = A+ h\Big(\dfrac{1}{N} \sum f_iu_i\Big)$

$\therefore \overline{X} = 900 + 20 \times \dfrac{-44}{100}= 900-8.8= 891.2$

$\text{Hence, mean wage = Rs. 891.2.}$

The following table gives the distribution of total household expenditure ( in rupees) of manual workers in a city.
Expenditure:
(in Rs.) 100-150 150-200 200-250 250-300 300-350 350-400 400-450 450-500
Frequency: 24 40 33 28 30 22 16 7

Find the average expenditure (in Rs.) per household.

Let the assumed mean be A= 325 and h= 50.
Calculation of Mean

Expenditure (in Rs.) $x_i$	Frequency $f_i$	Mid values $x_i$	$d_i= x_i-A= x_i- 325$	$u_i=\frac{x - A}{h}= \frac{x_i-325}{50}$	$f_iu_i$
100-150	24	125	-200	-4	-96
150-200	40	175	-150	-3	-120
200-250	33	225	-100	-2	-66
250-300	28	275	-50	-1	-66
300-350	30	325	0	0	0
350-400	22	375	50	1	22
400-450	16	425	100	2	32
450-500	7	475	150	3	21
	$N=\sum f_i=200$				$\sum f_iu_i= -235$

We have,
N= 200, A= 325, h= 50, and

$\sum f_iu_i= -235$

$\therefore \overline {X} = A+ h\frac{1}{N} \sum f_iu_i$

$\Rightarrow \overline{X}= 325 = 50 \times [ \frac{-235}{200}]$

$\Rightarrow \overline {X}= 325 -\frac{235}{4}= 325 - 58.75= 266.25$
Hence, the average expenditure is Rs. 266.25.

Find the mode of the following data:
(i) $3,5,7,4,5,3,5,6,8,9,5,3,5,3,6,9,7,4$
(ii) $3,3,7,4,5,3,5,6,8,9,3,5,3,6,9,7,4$
(iii) $15,8,26,25,24,15,18,20,24,15,19,15$

$(i)$

$3,5,7,4,5,3,5,6,8,9,5,3,5,3,6,9,7,4$

$\Rightarrow$ In the given values, only one value which appear most often is number

$5$ . Number

$5$ is appear

$5$ times.

$\therefore$ The mode is

$5$ .

$(ii)$

$3,3,7,4,5,3,5,6,8,9,3,5,3,6,9,7,4$

$\Rightarrow$ In the given values, only one value which appear most often is number

$3$ . Number

$3$ is appear

$5$ times.

$\therefore$ The mode is

$3$ .

$(ii)$

$15,8,26,25,24,15,18,20,24,15,19,15$

$\Rightarrow$ In the given values, only one value which appear most often is number

$15$ . Number

$15$ is appear

$4$ times.

$\therefore$ The mode is

$15$ .

Find the mode of the following data:
$25, 16, 19, 20, 34, 15, 19, 20, 21, 24, 19, 16, 22, 16, 18, 20, 16, 19$

The frequency table of the given data is as given below:

Value ( $x_i$ ):	15	16	18	19	20	21	22	24	25	34	48
Frequency ( $f_i$ ):	1	4	1	5	3	1	1	1	1	1	1

We observe that the value

$19$ has the maximum frequency i.e. it occurs maximum number of times. Therefore, mode of the given data is

$19$ .

The information regarding $35$ villages of a district is given below obtain an appropriate measure of average.
Population Less then $1000$ $1001-2000$ $2001-4000$ $4001-8000$ $8001-10000$ More than $10,000$
Number of villages $3$ $6$ $10$ $9$ $3$ $4$

$Population$	$Class-mark$ $x_i$	$No.\,of\,villages$ $f_i$	$f_ix_i$
$0.5-1000.5$	$500.5$	$3$	$1501.5$
$1000.5-2000.5$	$1500.5$	$6$	$9003$
$2000.5-4000.5$	$3000.5$	$10$	$30005$
$4000.5-8000.5$	$6000.5$	$9$	$54004.5$
$8000.5-10000.5$	$9000.5$	$3$	$27001.5$
$10,000.5-14000.5$	$12000.5$	$4$	$48002$
		$\sum f_i=35$	$\sum f_ix_i=169517.5$

$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{169517.5}{35}=4843.36$

Find the mean of the following distribution using short-cut method:

C.I $35-40$ $40-45$ $45-50$ $50-55$ $55-60$
F $7$ $6$ $9$ $5$ $3$

The average of each class

$=37.5,42.5,47.5,52.5,57.5$

Frequencies

$=7,6,9,5,3$

Mean of the distribution =

$\dfrac{\sum (C.I\times F_i)}{F_i}$

$Mean=\dfrac{7\times 37.5+6\times 42.5+9\times 47.5+5\times 52.5+3\times 57.5}{30}=46$

Find the average of $2,3,4,5,10,13$

Number of observations

$=6$

Average

$=\cfrac{Sum\;of\;observation}{Total\; number} \\ =\cfrac{2+3+4+5+10+13}{6}=\cfrac{37}{6}=6.16$

Find the mean marks of students from the following cumulative frequency distribution:
Marks Number of students
o and above 80
10 and above 77
20 and above 72
30 and above 65
40 and above 55
50 and above 43
60 and above 28
70 and above 16
80 and above 10
90 and above 8
100 and above 0

Here we have, the cumulative frequency distribution. So, first we convert it into an ordinary frequency distribution. we observe that are 80 students getting marks greater than or equal to 0 and 77 students have secured 10 and more marks. Therefore, the number of students getting marks between 0 and 10 is 80-77= 3.
Similarly, the number of students getting marks between 10 and 20 is 77-72= 5 and so on. Thus, we obtain the following frequency distribution.

Marks	Number of students
0-10	3
10-20	5
20-30	7
30-40	10
40-50	12
50-60	15
60-70	12
70-80	6
80-90	2
90-100	8

Now, we compute mean arithmetic mean by taking 55 as the assumed mean.
Computative of Mean

Marks $(x_i)$	Mid-value $(f_i)$	Frequency	$u_i\frac{x_i-55}{10}$	$f_iu_i$
0-10	5	3	-5	-15
10-20	15	5	-4	-20
20-30	25	7	-3	-21
30-40	35	10	-2	-20
40-50	45	12	-1	-20
50-60	55	15	0	0
60-70	65	12	1	12
70-80	75	6	2	12
80-90	85	2	3	6
90-100	95	8	4	32
Total	$\sum f_i=80$			$\sum f_iu_i$ = -26

We have,
N=

$sum f_i= 80, \sum f_iu_i= - 26$ , A= 55 and h= 10

$\therefore \overline {X} = A + h[\frac{1}{N} \sum f_iu_i]$

$\Rightarrow \overline {X} = A+ h[ \frac{1}{N} \sum f_iu_i]$

$\Rightarrow \overline {X} = 55+ 10 \times \frac{-26}{80}= 55- 3.25 = 51. 75$ Marks.

Find the mean marks of the students from the following cumulative frequency distribution:
Marks: Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80 Below 90 Below 100
Number of students: 5 9 17 29 45 60 70 78 83 85

Here we have, cumulative frequency distribution less than type. First we convert it into an ordinary frequency distribution. We observe that the number of students getting marks less than

$10$ is

$5$ and

$9$ students have secured marks less than

$20$ . Therefore, number of students getting marks between

$10$ and

$20$ ( inclusive

$0$ and exclusive 10) is

$9- 5=4$ . Similarly, the number of students getting marks between

$20$ and

$30$ is

$17-9=8$ and so on. Thus, we have the following distribution frequency distribution:

Marks:	$0- 10$	$10-20$	$20- 30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80 - 90$	$90-100$
Number of students:	$5$	$4$	$8$	$12$	$16$	$15$	$10$	$8$	$5$	$2$

Let us Let us now compute arithmetic mean by taking

$55$ and the assumed mean.
Computation of Mean

Marks	Mid-value	Frequency	$u_i\dfrac{x_i-55}{10}$	$f_iu_i$
$0-10$	$5$	$5$	$-5$	$-25$
$10-20$	$10$	$4$	$-4$	$-16$
$20-30$	$25$	$8$	$-3$	$-24$
$30-40$	$35$	$12$	$-2$	$-24$
$40-50$	$45$	$16$	$-1$	$-16$
$50-60$	$55$	$15$	$0$	$0$
$60-70$	$65$	$10$	$1$	$10$
$70-80$	$75$	$8$	$2$	$16$
$80-90$	$85$	$5$	$3$	$15$
$90-100$	$95$	$2$	$4$	$8$
Total		$N= \sum f_i= 85$		$\sum f_iu_i= -56$

We have,
N=

$N= \sum f_i = 85, \sum f_iu_i= - 56, h= 10 and A= 55$

$\therefore \overline {X} = A+ h \left(\displaystyle\frac{1}{N}\sum f_iu_i\right)$

$\Rightarrow \overline{X} = 55 +10 \times \dfrac{-56}{85}= 55- 6.59 = 48. 41$ Marks.
Hence, mean marks scored by the students =

$48.41$ .

The following data gives the distribution of total household expenditure (in rupees)of manual workers in a city:
Expenditure (in Rs.) Frequency Expenditure (in Rs.) Frequency
1000-1500 24 3000-3500 30
1500-2000 40 3500-4000 22
2000-2500 33 4000-4500 16
2500-3000 28 4500-5000 7
Find the average expenditure which is being done by the maximum number of manual workers.

We observe that the class

$1500-2000$ has the maximum frequency

$40$ . So, it is the modal class such that

$l= 1500, h= 500, f= 40, f_1= 24,$ and

$f_2= 33$

$\therefore Mode = l+\frac{f- f_1}{2f- f_1- f_2}\times h$

$\Rightarrow Mode = 1500+ \frac{40- 24}{80-24-33}\times 500$

$\Rightarrow Mode = 1500 + \frac{16}{23}\times 500= 1847.826$

Find the median of the following data.
$14, 92, 62, 53, 42, 32, 22, 41, 83, 55$ .

$\Rightarrow$ The given numbers are

$14,92,62,53,42,32,22,41,83,55.$

Arranging given numbers in ascending order,

$14,22,32,41,42,53,55,62,83,92$

Here, number of observation

$(N)=10(even)$

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

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

$Median=\dfrac{42+53}{2}$

$Median=\dfrac{95}{2}$

$\therefore$

$Median=47.5$

Construct the frequency distribution table for the following data:
Height(in cm) less than 140 less than 145 less than 150 less than 155 less than 160 less than 165
No of girls 4 11 29 40 46 51

$\begin{array} { *{ 20 }{ c } }\hline & { H{ { eight } }\left( { { { incm } } } \right) } & & { { { Numberofgirls } }\left( f \right) } & \\ \hline & { { { Lessthan140 } } } & & 4 & \\ & { 140-145 } & & 7 & \\ & { 145-150 } & & { 18 } & \\ & { 150-155 } & & { 11 } & \\ & { 155-160 } & & 6 & \\ & { 160-165 } & & 5 & \\ \hline & { } & & { \Sigma f=51 } & \hline \end{array}$

The median of $80, 60, 30x, 10, 5$ is $10$ then $x =$ ?

$80,60,30x,10,5$

$median = 10$

$\dfrac{80+60+30x+10+5}{5}=10$

$\dfrac{155+30x}{5}=10$

$155+30x=50$

$30x=-105$

$2x=-7$

$x=-3.5$

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
$65-85$ $4$
$85-105$ $5$
$105-125$ $13$
$125-145$ $20$
$145-165$ $14$
$165-185$ $8$
$185-205$ $4$

$C.I$	$x_i$	$f_i$	$f_ix_i$	$c.f$
$65-85$	$75$	$4$	$300$	$4$
$85-105$	$95$	$5$	$475$	$9$
$105-125$	$115$	$13$	$1495$	$22$
$125-145$	$135$	$20$	$2700$	$42$
$145-165$	$155$	$14$	$2170$	$56$
$165-185$	$175$	$8$	$1400$	$64$
$185-205$	$195$	$4$	$780$	$68$
		$\sum f_i=68$	$\sum f_ix_i=9320$

$\Rightarrow$

$Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{9320}{68}=137.05$

Now,

$n=68$ and

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

$\therefore$ Median class

$=125-145$

We have,

$l=125,c.f=22,f=20$ and

$h=20$

$\Rightarrow$

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

$=125+\dfrac{34-22}{20}\times 20$

$=125+12$

$\therefore$

$Median=137$

Now, modal class

$=125-145$

Then we have

$l=125,f_0=13,f_1=20,f_2=14$ and

$h=20$

$\Rightarrow$

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

$=125+\dfrac{20-13}{2\times 20-13-14}\times 20$

$=125+10.77$

$\therefore$

$Mode=135.77$

A number of people were asked how many coins they had in their pockets; the results are shown below. Calculate the mean number of coins per person.
Number of coins $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
frequency $3$ $6$ $4$ $7$ $5$ $8$ $5$ $2$

Mean

$\bar{x} = \dfrac{\sum f_ix_i}{\sum f_i} = \dfrac{139}{40} = 3.475$ Coins
1004363_1060688_ans_6f89a086280e4fd3aa6f9b89a970ead5.jpg

The weight of 50 apples were recorded as given below. Calculate the mean weight to the nearest gross, by the step deviation method
weight 80-85 85-90 90-95 95-100 100-105 105-110 110-115
No. of apples 5 8 10 12 8 4 3

For the given data, class difference

$h=85-80=5$

$CI$	$f_i$	$Class\,mark$ $x_i$	$d_i=\dfrac{x_i-A}{h}$	$f_id_i$
$80-85$	$5$	$82.5$	$-3$	$-15$
$85-90$	$8$	$87.5$	$-2$	$-16$
$90-95$	$10$	$92.5$	$-1$	$-10$
$95-100$	$12$	$A=97.5$	$0$	$0$
$100-105$	$8$	$102.5$	$1$	$8$
$105-110$	$4$	$107.5$	$2$	$8$
$110-115$	$3$	$112.5$	$3$	$9$
	$\sum f_i=50$			$\sum f_id_i=-16$

$\Rightarrow$

$\overline{x}=A+\dfrac{\sum f_id_i}{\sum f_i}\times h$

$=97.5+\dfrac{(-16)}{50}\times 5$

$=97.5-1.6$

$=95.9$

What is the median of $10, 14, 86, 2, 6, 8, 99, 1$ .

To find the median we must first arrange the set of given numbers in increasing order.

The increasing order is as follows-

1,2,6,8,10,14,86,99

When the number of terms are even the median is the average of the

$(\cfrac{n}{2})^{th}$ term

and

$(\cfrac{n+2}{2})^{th}$ term

$(\cfrac{n}{2})^{th} term=(\cfrac{8}{2})^{th} term= 8\\$

$(\cfrac{n+2}{2})^{th} term=(\cfrac{8+2}{2})^{th}=(\cfrac{10}{2})^{th} term=10\\$

$Median = \frac{8+10}{2}= 9$

Find the arithmetic mean of the following frequency distribution using the step deviation method.
Class 0-10 10-20 20-30 30-40 40-50
Frequency 9 12 15 10 14

Class	Frequency $(f_{i})$	$x_{i}$	$u_{i} = \dfrac{x_{i} - A}{h}$	$F_{i}\ u_{i}$
$0 - 10$	$9$	$5$	$-2$	$-18$
$10 - 20$	$12$	$15$	$-1$	$-12$
$20 - 30$	$15$	$\boxed{25} = A$	$0$	$0$
$30 - 40$	$10$	$35$	$1$	$10$
$40 - 50$	$14$	$45$	$2$	$28$
	$60$			$8$

Here,

$A= 25, \sum f_{i}= 60, \sum f_{i}u_{i} = 8, h = 10$

We know that, Arithmetic Mean

$= A + \dfrac{\sum f_{i}u_{i}}{\sum f_{i}}\times h$

$= 25 + \dfrac{8}{60} \times 10$

$= 25 + 1.33$

$= 26.33$

Hence, the required arithmetic mean is

$26.33$ .

The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method.
No. of Wickets $20-60$ $60-100$ $100-150$ $150-250$ $250-350$ $350-450$
No. of Bowlers $7$ $5$ $16$ $12$ $2$ $3$

2153683_1055324_ans_c2329bb15f4046f9abc96ca306f8a386.jpg

Find the mode of $90, 55, 67, 55, 75, 75, 40 , 35, 55, 95$ .

Mode of

$90, 55, 67, 55, 75, 75, 40, 35, 55, 95$

Increasing order:

$35, 40, 55, 55, 55, 67, 75, 75, 90, 95$

Frequency:

$35 - 1$

$40 - 1$

$55 - 3$

$67 - 1$

$75 - 2$

$90 - 1$

$95 - 1$ .

$\therefore$ Mode is

$55$ since it has high frequency.

The following distribution shows the daily pocket allowance of children of a locality . The mean pocket allowance is $Rs. 18.$ Find the missing frequency $f$ ?
Dally pocket allowance (in Rs) $11-13$ $13-15$ $15-17$ $17-19$ $19-21$ $21-23$ $23-25$
Number of children $7$ $6$ $9$ $13$ $f$ $5$
$4$

$Daily\,pocket\,allowance(in\,Rs.)$	$Mid-value$ $x_i$	$No.\,of\,children$ $f_i$	$f_ix_i$
$11-13$	$12$	$7$	$84$
$13-15$	$14$	$6$	$84$
$15-17$	$16$	$9$	$144$
$17-19$	$18$	$13$	$234$
$19-21$	$20$	$f$	$20f$
$21-23$	$22$	$5$	$110$
$23-25$	$24$	$4$	$96$
		$\sum f_i=44+f$	$=752+20f$

$\Rightarrow$

$\overline{x}=18$ [ Given ]

$\Rightarrow$

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

$\Rightarrow$

$18=\dfrac{752+20f}{44+f}$

$\Rightarrow$

$18=\dfrac{2(376+10f)}{44+f}$

$\Rightarrow$

$9=\dfrac{376+10f}{44+f}$

$\Rightarrow$

$396+9f=376+10f$

$\therefore$

$f=20$

If a variable takes values 1,2,3,....n with frequencies ${1^2},{2^2},...{n^2}....$
Then the mean is

$\displaystyle Mean=\dfrac{\sum\,f_ix_i}{\sum f_i}$

$=\dfrac{1\times 1^2+2\times 2^2+3\times 3^2+-------n\times n^2}{1^2+2^2+3^2+---------n^2}$

$\displaystyle =\dfrac{\sum n^3}{\sum n^2}=\dfrac{n^2(n+1)^2\times 6}{4n(n+1)(2n+1)}$

$=\dfrac{3n(n+1)}{2(2n+1)}$

Frequency distribution of daily commission received by salesman is given.
Find mean daily commission received by a salesman.

$Daily\,commission$ $(in\,Rs.)$	$Mid-value$ $x_i$	$No.\,of\,salesman$ $f_i$	$f_ix_i$
$100-120$	$110$	$20$	$2200$
$120-140$	$130$	$45$	$5850$
$140-160$	$150$	$22$	$3300$
$160-180$	$170$	$9$	$1530$
$180-200$	$190$	$4$	$760$
		$\sum f_i=100$	$\sum f_ix_i=13640$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{13640}{100}=136.4$

$\Rightarrow$ Mean daily commission received by a salesman is

$Rs.136.4$

The mean of $9, 11, 13, x, 18, 19$ is x. Find x.

$\Rightarrow$ The given numbers are

$9,11,13,x,18,19$

$\Rightarrow$ Number of observation

$=6$

$\Rightarrow$ Mean

$=x$

$\therefore$

$\overline{x}=\dfrac{9+11+13+x+18+19}{6}$

$\therefore$

$x=\dfrac{70+x}{6}$

$\therefore$

$6x=70+x$

$\therefore$

$5x=70$

$\therefore$

$x=14$

The batting scores of two cricket players A and B in 10 innings are as follows :-
Batsman A 15 17 19 27 30 36 40 90 95 110
Batsman B 10 16 21 28 37 41 36 80 82 65
Find which of the player is more consistent.

the player with less standard deviation i more consistant ,

Mean runs of A =

$\bar{x}_{A}$ =47.9.

Mean runs of B =

$\bar{x}_{B}$ =41.6

we know,

Mean deviation (A)=

$\sqrt{\sum_{t=1}^{10}}\frac{fi-\bar{x}_{A}}{10}$

$\therefore$ Mean deviation

$(A)=34.177$

Mean deviation

$(B) =7.727$

$\therefore$ player B is more consistent.

On the sports day of a school, $300$ students participated. Their ages are given in the following distribution.
Age (in years) $5-7$ $7-9$ $9-11$ $11-13$ $13-15$ $15-17$ $17-19$
Number of students $67$ $33$ $41$ $95$ $36$ $13$ $15$
Find the mean and mode of the date.

$C.I$	$Frequency$ $f_i$	$Mid-value$ $x_i$	$f_ix_i$
$5-7$	$67$	$6$	$402$
$7-9$	$33$	$8$	$264$
$9-11$	$41$	$10$	$410$
$11-13$	$95$	$12$	$1140$
$13-15$	$36$	$14$	$504$
$15-17$	$13$	$16$	$208$
$17-19$	$15$	$18$	$270$
	$\sum f_i=300$		$\sum f_ix_i=3198$

$\Rightarrow$

$Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{3198}{300}=10.66$

$\Rightarrow$ Here, Modal class

$=11-13$

So,

$l=11,\,f_0=41,\,f_1=95,\,f_2=36$ and

$h=2$

$\Rightarrow$

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

$=11+\dfrac{95-41}{2\times 85-41-36}\times 2$

$=11+\dfrac{54}{93}\times 2$

$=11+1.16$

$\therefore$

$Mode=12.16$

(a) Calculate the arithmetic mean of $5.7, 6.6, 7.2, 9.3, 6.2$
(b) The weight (in kg) of $8$ new born babies are $3,3.2,3.4,3.5, 4,3.6,4.1,3.2$ . Find the mean weight of the babies.

$(a)$

Mean

$=\dfrac{5.7+6.6+7.2+9.3+6.2}{5}$

$=\dfrac{35.1}{5}$

$=7.02$

$(b)$

The weight of

$8$ new born babies are

$3,3.2,3.4,3.5, 4,3.6,4.1,3.2$

Mean

$=\dfrac{3+3.2+3.4+3.5+4+3.6+4.1+3.2}{8}$

$=\dfrac{28}{8}$

$=3.5$

Above the chart given below and answer the question.
match no Player 1 Player 2
1 35 46
2 80 238
3 15 51
4 47 42
5 18 25
6 54
What should be the score of first player in $6^{th}$ match such that average of both player 1 and player 2 is equal ?

Let the score of first player in

$6^{th}$ match be

$x$

Average of player 1 is

$\dfrac{35+80+15+47+18+x}{6}=\dfrac{195+x}{6}$

Average of player 2 is

$\dfrac{46+238+51+42+25+54}{6}=\dfrac{456}{6}$

Given that average of player 1 is equal to average of player 2.

Therefore,

$\dfrac{195+x}{6}=\dfrac{456}{6}$

$\implies 195+x=456$

$\implies x=261$

Therefore, the score of first player in

$6^{th}$ match is

$261$

Class 20-60 60-100 100-150 150-250 250-350 350-450
Fi 7 5 18 12 2 3
Find mean by step deviation method ?

$\Rightarrow$ Let

$a=200$ and

$h=20$

$Class$	$Mid-value$ $x_i$	$d_i=x_i-200$	$u_i=\dfrac{d_i}{20}$	$f_i$	$f_iu_i$
$20-60$	$40$	$-160$	$-8$	$7$	$-56$
$60-100$	$80$	$-120$	$-6$	$5$	$-30$
$100-150$	$125$	$-75$	$-3.75$	$16$	$-60$
$150-250$	$200$	$0$	$0$	$12$	$0$
$250-350$	$300$	$100$	$5$	$2$	$10$
$350-450$	$400$	$200$	$10$	$3$	$30$
$Total$				$45$	$-106$

$\Rightarrow$

$\sum f_iu_i=-106$ and

$\sum f_i=45$

$\Rightarrow$

$Mean=a+h\times \dfrac{\sum f_iu_i}{\sum f_i}$

$=200+20\times \left(\dfrac{-106}{45}\right)$

$=200-\dfrac{424}{9}$

$=200-47.11$

$=152.89$

Mean of Eight observation is 25 . If one observation is 11 excluded . find mean of the remaining .
no. of observations =8

Mean of

$8$ observations is

$25$ .

$\therefore$ Total number of

$8$ observations

$=25\times 8=200$

If the observation

$11$ is excluded, then the total of

$7$ observations

$=200-11=189$

$\therefore$ The mean of

$7$ observations

$=\dfrac{189}{7}=27$

If the mean of the following frequency distribution is $8$ , then find the value of $'k'$ ?
$x$ $2$ $4$ $6$ $8$ $10$ $12$
$frequency = f$ $4$ $7$ $10$ $16$ $k$ $3$

From the given frequency distribution table we can find,

$\Sigma f_ix_i=8 + 28 + 60 + 128 + 10k + 36\\ \therefore \sum { f } =40+k\\ \sum { f_i x_i } =260+10k$

So,

$\therefore f=40+k$

and

$\\ f\cdot x=260+10k$

Also, mean =

$8$

Using the formula,

$mean=\dfrac { \sum { f_i x_i } }{ \sum { f } } \\ \Rightarrow 8=\dfrac { 260+10k }{ 40+k } \\ \Rightarrow 320+8k=260+10k\\ \Rightarrow 2k=60\\ \Rightarrow k=30$

Hence, the value of

$k$ is

$30.$

Consider the following distribution of daily wages of 50 workers of a factory:
Find the mean daily wages of the workers of the factory by using appropriate method.
Daily wages(in rs) Number of workers
500-200 12
520-540 14
540-560 8
560-580 6
580-600 10

Daily wages	$(fi)$	$(xi)$	$(fi xi)$
$600-20$	$12$	$510$	$6120$
$520-540$	$14$	$630$	$7420$
$540-560$	$8$	$550$	$4400$
$560-580$	$6$	$570$	$3420$
$580-600$	$10$	$590$	$$5900
	$\displaystyle \sum fi=50$		$\displaystyle \sum fi\ xi\Rightarrow 27260$

mean

$\dfrac {\displaystyle \sum fi\ xi}{\displaystyle \sum fi}=\dfrac {27260}{50}=\dfrac {2726}{5}=545.2$

$\therefore \$ meam\n wage of

$50$ employees is

$545.2 rs$

In a retail market fruit vendor were selling mangoes in boxes. These boxes contained varying number of mangoes. The following was the distribution.
No.of mangoes $50-52$ $53-55$
$56-58$ $59-61$ $62-64$
No. of boxes $15$ $110$ $135$ $115$ $25$
Find the mean and median number of mangoes kept in a packing box.

1154997_1144462_ans_1e7f07b6052a4a4bb5aaee871d415d0e.jpg

Find the mean for the following distribution
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 6 8 13 7 3 2 1

Marks	Frequency (fi)	(xi)	(fi)(xi)
10-20	6	15	90
20-30	8	25	200
30-40	13	35	455
40-50	7	45	315
50-60	3	55	165
60-70	2	65	130
70-80	1	75	75
	Total=40		Total=1430

$Mean=\dfrac {\sum (fi)(xi)} {\sum fi}$

$=\dfrac{1430}{40}$

$=35.75$

Find the arithmetic mean of the sales per day in a fair price shop in a week
rs10000, rs10250, rs10790, rsrs15350, rs10110

$\Rightarrow$ Total number of days

$=6$

$\Rightarrow$ Total amount

$=Rs.(10000+10250+10790+9865+15350+10110)=Rs.66365$

$\Rightarrow$ Required arithmetic mean

$=\dfrac{Rs.66365}{6}=Rs.11060.83$

Consider the following distribution of daily wages of $50$ workers of a factory.
Daily wages (in Rs) $100 - 120$ $120 - 140$ $140 - 160$ $160 - 180$ $180 -200$
Number of workers $12$ $14$ $8$ $6$ $10$
Find the mean daily wages of the workers of the factory by using an appropriate method.

Daily wages	No.of workers (f )	x	xf
100-120	12	110	1320
120-140	14	130	1820
140-160	8	150	1200
160-180	6	170	1020
180-200	10	190	1900
	50		7260

mean

$= \bar{x} = \dfrac{\sum xf}{\sum f} = \dfrac{7260}{50} = 145.2$

So, the mean daily ways of the workers

are

$145.2.$ [Ans]

The weighted mean of first $n$ natural numbers, whose weights are proportional to the corresponding numbers is $\dfrac {2n+1}{3}$ .Is it true? If true enter 1 else 0.

$\\Let\>\, weight \>of\> n\> be\> \lambda\>n, where \> \lambda=n,n\in N \\then \>\,\, weighted \> \,\,mean\\$

$=(\dfrac{(1\times\lambda)+(2\times\lambda)+(3\times\lambda)+.......+(n\times\lambda)}{1\lambda+2\lambda+3\lambda+......+n\lambda})\\$

$=(\dfrac{1^2+2^2+3^2+.....+n^2}{1+2+3+....+n})\\$

$=(\dfrac{(\dfrac{n(n+1)(2n+1)}{6})}{(\dfrac{n(n+1)}{2})})\\$

$=(\dfrac{2n+1}{3})$

Find the mean and median of the data 35, 48, 92, 76, 64, 52, 51, 63 andIf 51 is replaced by 66, what will be the new median ?

First we need to arrange given data in ascending order

$35,48,51,52,63,64,71,76,92$

Mean

$=\cfrac { \sum { { x }_{ i } } }{ n } =\cfrac{552}{9}=61.33$

Median

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

$=\cfrac{10}{2}=5$ th term i.e.,

$63$

If we replace

$51$ by

$66$ new series will become

$35,48,52,63,64,66,71,76,92$

new median is again

$5$ th term i.e.,

$64$

The following observations have been arranged in the ascending order. If the median of the data is 78, find the value of $x$ .
$44,47,63,65 , x + 13,87,93,99,110$

The given observations are,

$44,47,63,65,x+13,87,93,99,110$

$\Rightarrow$ Number of observations

$(N)=9$ (odd)

$\Rightarrow$

$Median=\left(\dfrac{9+1}{2}\right)^{th}obs=5^{th}obs$

$5^{th}obs=x+13$

$\therefore$

$Median=x+13$

$\Rightarrow$

$78=x+13$

$\therefore$

$x=65$

Find the modes of the following data:
$5,6,3,5,4,9,5,6,4,9,5$

Making a frequency table,

Number Frequency

$3$

$1$

$4$

$2$

$5$

$3$

$6$

$2$

$9$

$2$

$\Rightarrow$ Mode

$=5$ .

The following data shows the number of visitors to a zoological park every hour. Calculate the average number of visitors to the park during the whole day.
Time 10-11 a.m 11-12 noon 12-1 p.m 1-2 p.m 2-3 p.m 3-4 p.m 4-5 p.m
Numbers of visitors 30 40 34 26 35 34 32

Average no. of visitors

$=\cfrac { 30+40+34+26+35+34+32 }{ 7 }$

$=\cfrac { 231 }{ 7 } =33$

Mean of the numbers $1,2,3 , \ldots ,$ n with respective weights $1 ^ { 2 } + 1,2 ^ { 2 } + 2,3 ^ { 2 } + 3 , \ldots , n ^ { 2 } + n$ is

$\\weighted\>mean\\=(\dfrac{1(1^2+1)+2(2^2+2)+.......+n(n^2+n)}{(1^2+1)+(2^2+2)+......+(n^2+n)})\\=(\dfrac{(1^3+1^2)+(2^2+2^2)+.......+(n^3+n)}{(1^2+2^2+.......+n^2)+(1+2+......+n)})\\=(\dfrac{(1^3+2^3+.....+n^3)+(1^2+2^2+......+n^2)}{(\dfrac{n(n+1)(2n+1)}{6})+(\dfrac{n(n+1)}{2})})\\=(\dfrac{(\dfrac{n^2(n+1)^2}{4})+(\dfrac{n(n+1)(2n+1)}{6})}{n(n+1)\left[(\dfrac{2n+1}{6})+(\dfrac{1}{2})\right]})\\=(\dfrac{(\dfrac{n(n+1)}{2})\left[(\dfrac{n(n+1)}{2})+(\dfrac{2n+1}{3})\right]}{(\dfrac{n(n+1)}{2})\left[(\dfrac{2n+1}{3})+1\right]})\\=(\dfrac{(\dfrac{n^2+n}{2})+(\dfrac{2n+1}{3})}{(\dfrac{2n+1+3}{3})})\\=(\dfrac{(\dfrac{3n^2+3n+4n+2}{6})}{(\dfrac{2n+4}{3})})\\=(\dfrac{3n^2+7n+2}{6})\times(\dfrac{3}{2(n+1)})\\=(\dfrac{(3n+1)(n+2)}{4(n+1)})$

The following observations have been arranged in ascending order:
$11,12,14,18,x+2,x+4,30,32,35,41$ . If the median of the data is $24$ , find $x$ .

Here, the number of observations

$n=10$ , since n is even, therefore

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

$\Rightarrow 24=\dfrac{5^{th}\ observation+6^{th}\ observation}{2}$

$\Rightarrow 24=\dfrac{(x+2)+(x+4)}{2}$

$\Rightarrow 24=\dfrac{2x+6}{2}$

$\Rightarrow 24=x+3$

$\Rightarrow x=21$

Hence,

$x=21$

The mean of the following frequency distribution is $25.2$ . Find the missing frequency $x$
$C.I.$ $0-10$ $10-20$
$20-30$ $30-40$ $40-50$
frequency $8$ $X$ $10$ $11$ $9$

Solution -

C.I	0-10	10-20	20-30	30-40	40-50
Frequency	8	X	10	11	9.
Midvalue	5	15	25	35	45

Mean = 25.2

$\frac{8\times 5+15x+10\times 25+11\times 35+9\times 45}{8+x+10+11+9} = 25.2$

$1080+15x = 957.6+25.2x$

$x= \frac{122.4}{10.2}$

$x = 12.$

1103967_1188659_ans_54ab351b20fa4164be1099fb9c4072db.jpg

The scores in mathematics test of $15$ student are as follow:
$9,\ 25,\ 10,\ 20,\ 9,\ 20,\ 15,\ 10,\ 5,\ 15,\ 25,\ 20,\ 20,\ 15,\ 20$
Find the mode of this data.

Scores in mathematics are as follows :

$9,25,10,20,9,20,15,10,5,15,25,20,20,15,20$

Most number of frequency for a score

$=20=5$ times

So, mode

$=20$

Hence, the answer is

$20.$

The following table gives the information of frequency distribution of weekly wages of $150$ workers of a company. Find the mean of the weekly wages by 'step deviation' method.
Weekly wages (Rupees)
$1000-2000$ $2000-3000$ $3000-4000$ $4000-5000$
No. of workers $25$ $45$ $50$ $30$

$Class$	$Class-mark$ $x_i$	$d_i=x_i-A$	$u_i=\dfrac{d_i}{h}$	$f_i$	$f_i\times u_i$
$1000-2000$	$1500$	$-2000$	$-2$	$25$	$-50$
$2000-3000$	$2500$	$-1000$	$-1$	$45$	$-45$
$3000-4000$	$3500=A$	$0$	$0$	$50$	$0$
$4000-5000$	$4500$	$1000$	$1$	$30$	$30$
				$\displaystyle\sum f_i=150$	$\displaystyle \sum f_iu_i=-65$

$\Rightarrow$ Required mean

$=A+h\dfrac{\sum f_iu_i}{\sum f_i}$

$=3500-\dfrac{65}{150}\times 1000$

$=3500-433.33$

$=Rs.3066.67$

Hence, the mean of the weekly wages is

$Rs.3066.67$ .

The weekly wages of $120$ workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.
Weekly wages (Rupees) $0-2000$ $2000-4000$ $4000-6000$ $6000-8000$
No. of workers $15$ $35$ $50$ $20$

Class (Weekly wages in thousand rupees)	Class Mark $x_i$	Frequency(Number of workers) $f_i$	Class Mark $\times$ Frequency $x_i\times f_i$
0-2000	1000	15	15000
2000-4000	3000	35	105000
4000-6000	5000	50	250000
6000-8000	7000	20	140000
		$\sum f_i=120$	$\sum f_ix_i=510000$

Mean

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

$=\dfrac{510000}{120}$

$=Rs.4250$

$\therefore$ Mean of weekly wages is

$Rs.4250$

A frequency distribution of the life times of $400$ T.V. picture tubes tested in a tube company is given. Find the average life time of a tube.
Life time (in hrs) Frequency Life time (in hrs) Frequency
$300-399$ $14$ $800-899$ $62$
$400-499$ $46$ $900-999$ $48$
$500-599$ $58$ $1000-1099$ $22$
$600-699$ $76$ $1100-1199$ $6$
$700-799$ $68$

$Life\,time$ $(in\,hrs)$	$Class-mark$ $x_i$	$Frequency$ $f_i$	$f_ix_i$
$299.5-399.5$	$349.5$	$14$	$4893$
$399.5-499.5$	$449.5$	$46$	$20677$
$499.5-599.5$	$549.5$	$58$	$31871$
$599.5-699.5$	$649.5$	$76$	$49362$
$699.5-799.5$	$749.5$	$68$	$50966$
$799.5-899.5$	$849.5$	$62$	$52669$
$899.5-999.5$	$949.5$	$48$	$45576$
$999.5-1099.5$	$1049.5$	$22$	$23089$
$1099.5-1199.5$	$1149.5$	$6$	$6897$
		$\sum f_i=400$	$\sum f_ix_i=286000$

$\Rightarrow$

$Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{286000}{400}=715$

One situation where mean would be appropriate representative value.

Grading system uses average of grades a student obtained in individual subjects to compare performances of two students.

Th following distribution shows the number of outdoor patients in $64$ hospitals. Ifthe mean is $18 .$ find the missing frequencies $f _ { 1 }$ and $f _ { 2 }.$

Total no .of patients

$64=7+6+f_1+13+f_2+5+4\\f_1+f_2=35\\f_2=35-f_1\cdots (1)$

Mean of the data

$18=\dfrac{12(7)+14(6)+16(f_1)+18(13)+20f_2+22(5)+24(4)}{64}\\84+84+16f_1+234+20f_2+110+96=1152\\4f_1+5f_2=136\\4f_1+5(35-f_1)=136\\-f_1=136-175\\f_1=39\implies f_2=-4$

Frequencies cannot be negitive

Find the mean of the following frequency distribution by the assumed mean method :
No. of apples $50-53$ $53-56$ $56-59$ $59-62$ $62-65$
No of boxes $20$ $150$ $115$ $95$ $20$

$No.\,of\,apples$	$x_i$	$f_i$	$d_i=x_i-A$	$f_id_i$
$50-53$	$51.5$	$20$	$-6$	$-120$
$53-56$	$54.5$	$150$	$-3$	$-450$
$56-59$	$57.5=A$	$115$	$0$	$0$
$59-62$	$60.5$	$95$	$3$	$285$
$62-65$	$63.5$	$20$	$6$	$120$
		$\sum f_i=400$		$\sum f_id_i=-165$

$Mean(\overline{x})=A+\dfrac{\sum f_id_i}{\sum f_i}$

$=57.5-\dfrac{165}{400}$

$=57.5-0.4$

$=57.1$

Consider the following distribution of daily wages of $50$ workers of a factory.
Daily wages (in Rs) 100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10
Find the mean daily wages of the workers of the factory by using an appropriate method.

Daily wages	Class-mark $x_i$	No.of workers $f_i$	$f_ix_i$
$100-120$	$110$	$12$	$1320$
$120-140$	$130$	$14$	$1820$
$140-160$	$150$	$8$	$1200$
$160-180$	$170$	$6$	$1020$
$180-200$	$190$	$10$	$1900$
		$\sum f_i=50$	$\sum f_ix_i=7260$

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

$=\dfrac{7260}{50}$

$=145.2$

The mean of 6 numbers isIf one number is excluded, the mean of remaining number becomes 22 find the excluded number.

Let excluded number be x

As Mean =

$\frac{sum}{Number of terms}\rightarrow sum=6\times 24=144$

and

$144-x=(22\times 5) \rightarrow x=144-110$

Ramaining numbers

$\rightarrow x=34$

1181974_1294734_ans_04aaa793bc8741d3ac0d08f14cc2d920.jpg

Find the mean of the following frequency distribution using step-deviation method.
Marks obtained number of students
Less than 10 14
Less than 20 22
Less than 30 37
Less than 40 55
Less than 50 67
Less than 60 75

Step deviation method

class interval	number of students (f)	mid point of class interval	Deviation $d_i' = \dfrac{x_i - A}{h}$	$f_i d_i'$
0-10 10-20 20-30 30-40 40-50 50-60	14 22 37 55 67 75	5 15 25 35 45 55	-2 -1 0 1 2 3	-28 -22 0 55 134 225
total	$\sum f = 270$			$\sum f_id_i' = 364$

By step deviation method

mean

$= A + \dfrac{\sum f_i d_i'}{\sum f} \times h$

$= 25 + \dfrac{364}{270} \times 10$

$= 25 + \dfrac{364}{27}$

$= 25 + 13.48$

mean

$= 38.48$

where h is class width

$h = 10$

A is assumed mean

$A = 25$

The following data gives the information on the life - time (in hours) of 75 electrical instruments.
Lifetime
(in hours) 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 10 15 12 21 8 9
find the mean lifetime of the instruments.

$Lifetime$ $(in\,hours)$	$Class-mark$ $x_i$	$Frequency$ $f_i$	$f_ix_i$
$0-20$	$10$	$10$	$100$
$20-40$	$30$	$15$	$450$
$40-60$	$50$	$12$	$600$
$60-80$	$70$	$21$	$1470$
$80-100$	$90$	$8$	$720$
$100-120$	$110$	$9$	$990$
		$\sum f_i=75$	$\sum f_ix_i=4330$

$Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{4330}{75}=57.73$

Give below is the frequency distribution of marks scored by the students:
Calculate mean marks scored by a student by using $'Direct Method'$

$\begin{array} { *{ 20 }{ l } }\begin{array}{l} class\, \, mark=\left( { upper\, \, \lim it-lower\, \, \lim it } \right) /2 \\ \sum { fi } =40 \end{array} \\ { \sum { fixi=920 } } \\ { mean=\dfrac { { \sum { fixi } } }{ { \sum { fi } } } =\dfrac { { 920 } }{ { 40 } } =23 } \\ { } \end{array}$
1208942_1311273_ans_091f75bc01f14d6185121133e982ed9d.PNG

1208942_1311273_ans_091f75bc01f14d6185121133e982ed9d.PNG

A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by 'assumed mean' method.
Production
(in thousand rupees) $25-30$ $30-35$ $35-40$ $40-45$ $45-50$
No. of farm owners $20$ $25$ $15$ $10$ $10$

A frequency distribution table for the production of oranges of some farm owners is given below.

Production (in thousand rupees)	$25 - 30$	$30-35$	$35-40$	$40-45$	$45-50$
No. of farm owners	$20$	$25$	$15$	$10$	$10$

Now we can prepare a table for calculating mean

Production	No. of farm owners $(f_i)$	Midpoint $(x_i)$	$d_i = x_i - A$	$f_i d_i$
$25-30$	$20$	$27.5$	$27.5 - 37.5 = -10$	$-200$
$30-35$	$25$	$32.5$	$32.5 - 37.5 = -5$	$-125$
$35-40$	$15$	$37.5$	$37.5 - 37.5 = 0$	$0$
$40-45$	$10$	$42.5$	$42.5 - 37.5 = +5$	$50$
$45-50$	$10$	$47.5$	$47.5 - 37.5 = +10$	$100$
	$\sum f_i = 80$			$\sum f_i d_i = -175$

Let, the assumed mean be

$A=37.5$ .

Mean

$(\bar{x}) = A + \dfrac{\sum f_i d_i}{\sum f_i}$

$= 37.5 + \dfrac{-175}{80}$

$= 37.5 - \dfrac{175}{80}$

$= 37.5 - 2.1875$

$= 35.3125$

Hence, the mean production is

$35.31\ thousand$ rupees.

Calculate the mean of the following distribution.
Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 5 8 10 12 7 8

Given that

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$
Frequency	$5$	$8$	$10$	$12$	$7$	$8$

Now we can prepare the table for calculating mean

Class	Frequency (f)	Mid point of class (x)	fx
$0-20$	$5$	$10$	$50$
$20-40$	$8$	$30$	$240$
$40-60$	$10$	$50$	$500$
$60-80$	$12$	$70$	$840$
$80-100$	$7$	$90$	$630$
$100-120$	$8$	$110$	$880$
	$\sum f = 50$		$\sum fx = 3140$

we know that mean

$(\bar{x}) = \dfrac{\sum fx}{\sum f}$

$= \dfrac{3140}{50}$

$= \dfrac{314}{5}$

$= 62.8$

Mean

$= 62.8$

What is the mean height of a students of $1400,147,145,155,160,150,144,148,152$ .

Mean

$=\dfrac{sum\,\, of\,\, the \,\, observations}{Number\,\, of \,\, observations}$

$=\dfrac{1400+147+145+155+160+150+144+148+152}{9}$

$=\dfrac{2601}{9}=289$

$\therefore$ Mean

$=289$

Find the mean of the following data
Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 3 6 8 10 8 5 3

Mean of grouped data by direct method

mid point

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

Class	Frequency (f)	Mid-point (x)	fx
$10 - 20$	$3$	$\dfrac{10 + 20}{2} = \dfrac{30}{2} = 15$	$3 \times 15 = 45$
$20 - 30$	$6$	$\dfrac{20 + 30}{2} = \dfrac{50}{2} = 25$	$6 \times 25 = 150$
$30 - 40$	$8$	$\dfrac{30 + 40}{2} = \dfrac{70}{2} = 35$	$8 \times 35 = 280$
$40 - 50$	$10$	$\dfrac{40 - 50}{2} = \dfrac{90}{2} = 45$	$10 \times 45 = 450$
$50 - 60$	$8$	$\dfrac{50 + 60}{2} = \dfrac{110}{2} = 55$	$8 \times 55 = 440$
$60 - 70$	$5$	$\dfrac{60 + 70}{2} = \dfrac{130}{2} = 65$	$5 \times 65 = 325$
$70 - 80$	$3$	$\dfrac{70 + 80}{2} = \dfrac{150}{2} = 75$	$3 \times 75 = 225$
	$N = \sum f = 43$		$\sum fx = 1915$

Arithmetic mean of grouped Data by Direct method

Mean

$= \dfrac{\sum fx}{\sum f} = \dfrac{\sum fx}{N} = \dfrac{1915}{43}$

$= 44.5348$

$\cong 44.535$ ramded 3 decimal

The following table shows the heights $(cm)$ of $50$ girls of class $X$ of a school.
Height $(cm)$
Number of girls
$120-130$ $2$
$130-140$ $8$
$140-150$ $12$
$150-160$ $20$
$160-170$ $8$
$Total$ $50$
Find the mean of the above data by step deviation method.

$\Rightarrow$

$A=145$ and

$h=10$

$Class-Interval$	$Class-mark$ $x_i$	$Frequency$ $f_i$	$u_i=\dfrac{x_i-A}{h}$	$f_iu_i$
$120-130$	$125$	$2$	$-2$	$-4$
$130-140$	$135$	$8$	$-1$	$-8$
$140-150$	$145=A$	$12$	$0$	$0$
$150-160$	$155$	$20$	$1$	$20$
$160-170$	$165$	$8$	$2$	$16$
		$\sum f_i=50$		$\sum f_iu_i=24$

$Mean =A+\dfrac{\sum f_iu_i}{\sum f_i}\times h$

$=145+\dfrac{24}{50}\times 10$

$=145+4.8$

$=149.8$

If the arithmetic mean of following data is 17, find the value of P.
Term(x) 10 15 20 25
Frequency(f) 7 10 P 6

$Term(x_i)$	$Frequency$ $(f_i)$	$f_ix_i$
$10$	$7$	$70$
$15$	$10$	$150$
$20$	$P$	$20P$
$25$	$6$	$150$
	$\sum f_i=23+P$	$\sum f_ix_i=370+20P$

We know that,

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

$\Rightarrow$

$17=\dfrac{370+20P}{23+P}$

$\Rightarrow$

$391+17P=370+20P$

$\Rightarrow$

$3P=21$

$\therefore\$

$P=7$

Hence, the value of

$P$ is

$7$ .

The median of $34,83,25,74,56$ is

given data

$34,83,25,74,56$

Ascending order is

$25,34,56,74,83$

Median is the middle most value

$56$ is the median of data.

Find mode of the following distribution

$x_i=3,9,15,21,28\\f_i=5,4,1,6,4\\\sum x_if_i=15+36+15+126+112=304\\\sum f_i=20\bar x=\dfrac{304}{20}=15.2$

Find mean of the following for example distribution.
Marks 0-20 20-40 40-60 60-80 80-100 100-120 120-140
No.of students 6 8 10 12 6 5 3

Now we can prepare table for calculating mean

marks	No of students $f$	midpoint of marks $x$	deviation $d_i = x_i - A$	$fd$
$0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$ $120-140$	$6$ $8$ $10$ $12$ $6$ $5$ $3$	$10$ $30$ $50$ $70$ $90$ $110$ $130$	$-60$ $-40$ $-20$ $0$ $20$ $40$ $60$	$-360$ $-320$ $-200$ $0$ $120$ $200$ $180$
	$\sum f = 50$			$\sum fd = -380$

Short cut method

mean

$= A + \dfrac{\sum fd}{\sum f}$ where A is assumed mean

$= 70 + \dfrac{-380}{50}$

$= 70 - \dfrac{38}{5}$

$= 70-7.6$

$=62.4$

$\therefore mean (\bar{x}) = 62.4$

The Median of data $\dfrac{k}{6}, \dfrac{k}{4}, k, \dfrac{k}{3}, \dfrac{k}{2}$ is 14 . Find $k$ .

Given data is

$\frac{k}{6},\frac{k}{4},k,\frac{k}{3},\frac{k}{2}$

The ascending order of data is

$\frac{k}{6},\frac{k}{4},\frac{k}{3},\frac{k}{2},k$

Given median is

$14$

$\implies \dfrac{k}{3}=14$

$\implies k=42$

If the mean of $8,12,21,42 , x$ is 20 then the value of $x$

Given data is

$8,12,21,42,x$

Their mean is

$20$

$\implies \dfrac{8+12+21+42+x}{5}=20$

$\implies 83+x=100\implies x=17$

The table shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method

$x_i=125,175,225,275,325$

$\sum f_i=4+5+12+2+2=25$

$\sum x_if_i=500+875+2700+550+650=5725\\\bar x=\dfrac{5725}{25}=211$

Find the average age from the following distribution :

$\sum f_1=6+11+21+23+14+5=80\\\sum x_if_i=10(6)+20(11)+30(21)+40(23)+50(14)+60(5)\\60+220+630+920+700+300=2820\\\bar x=\dfrac{\sum x_if_i}{\sum f_i}=\dfrac{2830}{80}=35.375$

Find median of data
$2,6,4,3,5,1,9,8,7$

Given data

$2,6,4,3,5,1,9,8,7$

Ascending order

$1,2,3,4,5,6,7,8,9$

Since

$n$ is odd Middle most term is Median

$5$ is median.

If the average number of 8 terms is given to be 40 and the average of first 6 terms is given to be 35.What is the average of the remaining 2 terms?

Sum of all the

$8$ terms

$= 320.$
The sum of first

$6$ terms

$= 210$
Now , the sum of remaining terms

$= 320 -210 = 110$
So , the average of

$2$ terms would be

$= 110/2 = 55$

The average of 5 terms isIf the first 4 terms are 45, 42, 119, and 84, what will be the last term?

Sum of all the terms

$= 250$
Sum of first four terms

$= 45+42+119+84 = 290$
So, the 5th term should be

$250 -290 = - 40.$

Calculate the mean, median and mode
$23, 25, 28, 25, 16, 23, 17, 22, 25, 25$ .

$3, 25, 28, 25, 16, 23, 17, 22, 25, 25$

Total no. of terms

$= 10$

Sum of data

$= 23 + 25 + 28 + 25 + 16 + 23 + 17 + 22 + 25 + 25 = 229$

Therefore,

Mean of given data

$= \cfrac{229}{10} = 22.9$

Now, since no. of terms is even.

Therefore,

Median of given data

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

$= \cfrac{16 + 23}{2} \\ = \cfrac{39}{2} = 19.5$

Since

$25$ has occured most frequently in the given set of data.

Therefore,

Mode of data

$= 25$

The average of 7 consecutive numbers is n. If the next two numbers are included, the average will be.....

The average of

$7$ consecutive numbers is

$n$ implies that the

$4th$ term is equal to n.
Now if we include next two terms then the average of

$9$ terms will be the 5th term. Now as the terms are consecutive, so the

$5th$ term will be

$n + 1.$

In a family of 6 members, the average age of the family at present is 25 while the age of the youngest member in the family is 5yrs, so what will be the average age of the family at the time of his birth ?

Average age of

$6$ family member is

$25$ .

$\therefore$ Total age of

$6$ family member

$=6\times 25=150\,years$

The age of youngest member is

$5\,years$

$\Rightarrow$ Total age of

$6$ family member at the times of youngest birth

$=150-(5\times 6)$

$=150-30$

$=120\,years$

$\therefore$ Average age of family at time of youngest birth

$=\dfrac{120}{6}=20\,years$

The measurements (in mm) of the diameters of the heads of the screws are given below. Calculate the mean diameter of the head of the screws.
Diameter (in mm) 33-35 36-38 39-41 42-44
Number of screws $\left( { f }_{ 1 } \right)$ 17 19 23 21

Given Data

The measurements (in mm) of the diameters of the heads of the screws.

Diameter (in mm)	$33 - 35$	$36- 38$	$39-41$	$42-44$
Number of screws	$17$	$19$	$23$	$21$

Diameter	(continuous) Diameter	Mid point $(x_i)$	No of screws $(f_i)$	$f_i x_i$
$33-35$	$32.5-35.5$	$34$	$17$	$578$
$36-38$	$35.5-38.5$	$37$	$19$	$703$
$39-41$	$38.5-41.5$	$40$	$23$	$920$
$42-44$	$41.5-44.5$	$43$	$21$	$903$
			$\displaystyle \sum f_i = 80$	$\displaystyle \sum f_i x_i = 310$

mean

$(\bar{x}) = \dfrac{\displaystyle\sum f_i x_i}{\displaystyle\sum f_i}$

$= 38.8$

$\therefore$ mean diameter of the head of the screws

$= 38.8$

Calculate the mean of the following distribution
CI 5-15 15-25 25-35 35-45 45-55 55-65
f 6 11 21 23 14 5

Given

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

Now can prepare a table for calculating mean

$CI$	$f$	mid point $x$	$fx$
$5-15$	$6$	$10$	$60$
$15-25$	$11$	$20$	$220$
$25-35$	$21$	$30$	$630$
$35-45$	$23$	$40$	$920$
$45-55$	$14$	$50$	$700$
$55-65$	$5$	$60$	$300$

$\displaystyle\sum fx = 2830$

$\displaystyle\sum f = 80$

$\therefore$ mean

$= \dfrac{\displaystyle\sum fx}{\displaystyle\sum f}$

$= \dfrac{2830}{80}$

$= 35.375$

$\therefore$ mean

$= 35.375$

Find the mean of the following data, by using suitable method.
Class interval Frequency
0-4 4
5-9 3
10-14 2
15-19 1

Given data

Class interval	$0-4$	$5-9$	$10-14$	$15-19$
frequency	$4$	$3$	$2$	$1$

Now we can prepare table for calculating mean

Direct method

class interval	Classinterval	Midpoint $(x_i)$	frequency $(f_i)$	$f_ix_i$
$0-4$	$-0.5 - 4.5$	$2.5$	$4$	$10$
$5-9$	$4.5-9.5$	$7.5$	$3$	$22.5$
$10-14$	$9.5-14.5$	$12.5$	$2$	$25$
$15-19$	$14.5-19.5$	$17.5$	$1$	$17.5$
		Total	$10$	$75$

We know that mean

$\bar{x} = \dfrac{\displaystyle\sum f_ix_i}{\displaystyle\sum f_i} = \dfrac{75}{10} = 7.5$

$\therefore$ mean

$= 7.5$

The heights (in cm) of 50 students of a class are given below:
Height (in cm)
156
154
155
151
157
152
153
Number of students
8
4
10
6
7
3
12
Find the median height.

2042789_1534235_ans_14a0bc3481384461a8a8cb02817a59f2.jpg

The weights of $11$ children in school cricket club are $35, 39, 32, 36, 40, 30, 34, 37, 38, 33, 31$ (kgs). Find the median weight.

Given, the weight of the children

$35, 39, 32, 36,40, 30, 34, 37, 38, 33, 31$

Arrange the data in assending order

$30, 31, 32, 33, 34,35,36, 37, 38, 39, 40$
Median

$= 35$ . (the middle number)

Consider the following distribution of daily wages of 50 workers of a factory :

Daily wages (in Rs)

500-520

520-540

540-560

560-580

580-600

Number of workers

12

14

8

6

10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Daily wages	No.of workers	Mean
500-520	12	510
520-540	14	530
540-560	8	550
560-580	6	570
580-600	10	590

Mean wages

$=\cfrac { 510\times 12+530\times 14+550\times 3+510\times 6+590\times 10 }{ 12+14+8+6+10 } \\ \quad \quad =Rs545.2$

The marks obtained by the students of class $6$ are shown:
$0-10$ $10-20$ $20-30$ $30-40$
$15$ $32$ $55$ $35$
Find the mean of the data.

Class

$fi$

$xi$

$fixi$

$0-10\\ 10-20\\ 20-30\\ 30-40$

$15\\ 32\\ 55\\ 35$

$5\\ 15\\ 25\\ 35$

$75\\ 480\\ 1375\\ 1225$

${ \sum { { f }_{ i } } }=137$

$\sum { { f }_{ i }{ x }_{ i } } =3155$

mean

$\overline { x } =\dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =23.029$

If following statement is true enter '1' else enter '0'.
The monthly pocket money of six friends is given below:
Rs. 45, Rs. 30, Rs. 40, Rs. 50, Rs. 25, Rs. 45
The amounts of pocket money in ascending order are
Rs. 25, Rs. 30, Rs. 40, Rs. 45, Rs.45, Rs. 50

1481812_1670798_ans_27c6b8212e28445c99465ccd09453216.jpeg

Find the missing frequencies if the mean of the given data is $53$ .
Age in years $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ Total
Number of People $15$ ${F}_{1}$ $21$ ${F}_{2}$ $17$ $100$

$15+F_1+21+F_2+17 = 100$

$15 + F_1 + 21 + F_2 + 17 \Rightarrow F_1 + F_2 = 47 ...(i)$

$\dfrac{15 \times 10 + 30\times F_1 + 50 \times 21 + 70 \times F_2 + 90 \times 17}{250} = 53$

$150 + 30 F_1 + 1050 + 70F_2 + 1530 = 131250 ...(ii)$

on solving (1) & (2)

$F_1 = 180.75 \quad F_2=-227.75$

Find the mean of the following frequency by a suitable method.
C.I. 50-70 70-90 90-110 110-130 130-150 150-170
Frequency 8 12 13 27 18 22

Where

$h = 20$

C.I	frequency $(f_i)$	midpoint of C.I $(x_i)$	$d_i = x_i - A$	$d'_i = (x_i - A)/h$	$f_i d'_i$
$50-70$	$8$	$60$	$60-120=-60$	$\dfrac{-60}{20} = -3$	$-24$
$70-90$	$12$	$80$	$80-120=-40$	$\dfrac{-40}{20}=-2$	$-24$
$90-110$	$13$	$100$	$100-120=-20$	$\dfrac{-20}{20} = -1$	$-13$
$110-130$	$27$	$120$	$120-120=0$	$0$	$0$
$130-150$	$18$	$140$	$140-120=20$	$\dfrac{20}{20} = 1$	$18$
$150-170$	$22$	$160$	$160-120=40$	$\dfrac{40}{20} = 2$	$44$
	$\sum f_i = 100$				$\sum f_i d_i'=1$

By step deviation method mean

$(\bar{x}) = A + \dfrac{\sum f_i d'_i}{\sum f_i} \times h$

$= 120 + \dfrac{1}{100} \times 20$

$= 120 + \dfrac{1}{5}$

$=120 + 0.2$

mean

$= 120.2$

Find the median of the following data
41, 43,127, 99,71, 92, 71, 58, 57

Given that

$41,43,127,99,71,92,71,58,57$

first of all arrange the data in ascending order

$41,43,57,58,71,71,92,99,127$

Here observe that

No' of observation

$n=9$ i.e odd

So, median that

no' of observations

$n=9$ i.e odd

So median is

$\dfrac{n+1}{2}$ the term i.e middle term

Median =

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

$\dfrac{9+1}{2}$ th term

$5$ th term

median =

$71$

The demand of different shirt sizes, as obtained by a survey, is given below:
Size: 38 39 40 41 42 43 44 Total
Number of persons(wearing it): 26 39 20 15 13 7
5

125

Find the modal shirt sizes, as observed from the survey.

$X$	$f$
$38$	$26$
$39$	$39$
$20$	$40$
$41$	$15$
$42$	$13$
$43$	$7$
$44$	$5$
Total	$125$

Here,

$x$ denotes the size of the shirts and

$f$ denotes the number of peoples ( frequency ) wearing it.

The value

$39$ occur maximum number of times, that is

$39.$

$\therefore$ The modal shirt size is

$39$

Find the median of the following data
31, 38, 27, 28, 36, 25, 35, 40

$31,38,27,28,36,25,35,40$

No' of observations (n) =

$8$

first of all we arrange the observations in an ascending order

$25,27,28,31,35,36,38,40$

Here observe that No' of observations (n) is even so

Median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{\dfrac{8}{2}\text{th term}+(\dfrac{8}{2}+1)\text{th term}}{2}$

$=\dfrac{4 \text{th term}+5\text{th term}}{2}$

$=\dfrac{31+35}{2}$

$=\dfrac{66}{2}$

Median =

$33$

Find the median of the following data
12, 17, 3, 14, 5, 8, 7, 15

$12,17,3,14,5,8,7,15$

No' of observations (n) =

$8$

first of all we arrange the observations in an ascending order

$3,5,7,8,12,14,15,17$

Here observe that No' of observations (n) is even so

Median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{\dfrac{8}{2}\text{th term}+(\dfrac{8}{2}+1)\text{th term}}{2}$

$=\dfrac{4 \text{th term}+5\text{th term}}{2}$

$=\dfrac{8+12}{2}$

$=\dfrac{20}{2}$

Median =

$10$

Find the median of the following data
25, 34, 31, 23, 22, 26, 35, 29, 20, 32

$25,34,31,23,22,26,35,29,20,32$

No' of observations (n) =

$10$

first of all we arrange the observations in an ascending order

$20,22,23,25,26,29,31,32,34,35$

Here observe that No' of observations (n) is even so

Median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{\dfrac{10}{2}\text{th term}+(\dfrac{10}{2}+1)\text{th term}}{2}$

$=\dfrac{5 \text{th term}+6\text{th term}}{2}$

$=\dfrac{26+29}{2}$

$=\dfrac{55}{2}$

Median =

$27.5$

Find the median of the following data
15, 6, 16, 8, 22, 21, 9, 18, 25

Given that

$15,6,16,8,22,21,9,18,25$

first of all arrange the data in ascending order

$6,8,9,15,16,18,21,22,25$

Here observe that

No' of observation

$n=9$ i.e odd

So median is (

$\dfrac{n+1}{2}$ )the term i.e middle term

Median =

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

$\dfrac{9+1}{2}$ th term

$\dfrac{10}{2}$ th term =

$5$ th term

median =

$16$

Heights of 25 children (in cm) in a school are as given below:
$168,165,163,160,163,161,162,164,163,162,164,163,160,163,163,165,163,162,163,164,$
$163,160,165,163,162$
What is the mode of heights?

Arranging the data in tabular form, we get

Height of Children $\mathrm{(cm)}$	Tally Bars	Frequency
$160$	III	$3$
$161$	I	$1$
$162$	IIII	$4$
$163$	IIII IIII II	$10$
$164$	III	$3$
$165$	III	$3$
$168$	I	$1$
Total		$25$

Here, Clearly

$163\space\mathrm{cm}$ occurs the most number of times.

$\therefore$ the mode of the given data is

$163\space\mathrm{cm}$ .

Find mode of the data: $2,6,5,3,0,3,4,3,2,4,5,2,4$

Arranging the data in ascending order such that same values are put together, we get:

$0,2,2,2,3,3,3,4,4,4,5,5,6$

Here,

$2,3$ and

$4$ occur three times each.

$\therefore$

$2,3$ and

$4$ are the three modes.

Find the median of the following observations : 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33.If 92 is replaced by 99 and 41 by 43 in the above data, find the new median.

given that

the observation are

$46,64,87,41,58,77,35,90,55,92,33$

No' of observations (n) =

$11$

arrange the data is ascending order

$33,35,41,55,58,64,77,87,90,92$

Here observe that No' of observations (n) are odd i.e

$n=11$

So median is

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

Median

$=\dfrac{n+1}{2}$ th term

$=\dfrac{11+1}{2}$ th term

$=6$ th term

$=58$

$\therefore median =58$

and

$92$ is replaced by

$99$

and

$41$ by

$43$ them

new data observation are

$33,35,43,46,55,58,64,77,87,90,99$ ............(1)

Here observe that no's observation are odd i.e

$n=1$

new median is

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

$=\dfrac{n+1}{2}$ th term

$=\dfrac{11+1}{2}$ th term

$=\dfrac{12}{2}$ th term

$=6$ th term (

$\therefore\,from (1)$ )

$=58$

$\therefore$ new median

$=58$

Find the mean of the following data :
x : 19 21 23 25 27 29 31
f : 13 15 16 18 16 15 13

Now we can prepare a table calculating mean

$x$	$f$	$f_x$
19	13	247
21	15	315
23	16	368
25	18	450
27	16	432
29	15	435
31	13	403
	$\sum f=106$	$\sum f_x=2650$

We know that

$mean\,(\bar{x})=\dfrac{\sum f_x}{\sum f}$

$=\dfrac{2650}{106}$

$=25$

mean of given data is

$25$

The weights (in kg) of 15 students are : $31,35,27,29,32,43,37,41,34,28,36,44,45,42,30$ . Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.

Given number are

$31,35,27,29,32,43,37,41,34,28,36,44,45,42,30$

Arranging given numbers in ascending order, we get

$27,28,29,30,31,32,34,35,36,37,41,42,43,44,45$

Here,

$n=15$

$\therefore$ Median

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

$=(\frac{15+1}{2})^{th}$ value

$=8^{th}$ value

$=35\space\mathrm{kg}$

If the weight

$44\space\mathrm{kg}$ is replaced by

$46\space\mathrm{kg}$ and

$27\space\mathrm{kg}$ is replaced by

$25\space\mathrm{kg}$ .

Then, new values in order be

$25,28,29,30,31,32,34,35,36,37,41,42,43,45,46$

So, Again

$n=15$

$\therefore$ New Median

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

$=(\frac{15+1}{2})^{th}$ value

$=8^{th}$ value

$=35\space\mathrm{kg}$

The mean of the following distribution is 23.Find the value of P.
Class intervals
0-8
8-16
16-24
24-32
32-40
40-48
Frequency
5
3
10
P
4
2

Class mark,

$x_i$ = (upper class limit + lower class limit)/2

Class intervals	Frequency $f_i$	Class mark $x_i$	$f_ix_i$
0-8	5	4	20
8-16	3	12	36
16-24	10	20	200
24-32	P	28	28P
32-40	4	36	144
40-48	2	44	88
Total	$\Sigma f_i = 24+P$		$\Sigma f_ix_i = 488+28P$

$Given\, mean = 23.4$

We know that, $Mean = \dfrac{\Sigma f_ix_i} {\Sigma f_i}$

$\therefore \ 23.4 = \dfrac{(488+28P)}{(24+P)}$

$\Rightarrow 23.4\times(24+P) = 488+28P$

$\Rightarrow 561.6+23.4P = 488+28P$

$\Rightarrow 561.6-488 = 28P -23.4P$

$\Rightarrow 73.6 = 4.6 P$

$\therefore \ P = \dfrac{73.6}{4.6} = 16$

Hence the value of P is $16$ .

Find the mean of the following data, using direct method:
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency $7$ $5$ $6$ $12$ $8$ $2$

Class interval	Frequency $f_i$	Mid-Value $x_i$	$f_i\ x_i$
$0-10$	$7$	$5$	$35$
$10-20$	$5$	$15$	$75$
$20-30$	$6$	$25$	$150$
$30-40$	$12$	$35$	$420$
$40-50$	$8$	$45$	$360$
$50-60$	$2$	$55$	$110$
	$\sum f_i=40$		$1150$

We know that, mean

$=\dfrac {\displaystyle \sum f_ix_i}{\displaystyle \sum f_i}$

$=\dfrac{1150}{40}$

$=28.75$

Hence, the mean of the given data is

$28.75$ .

Find the mean of the following distribution:
Class: $3-5$ $5-7$ $7-9$ $9-11$ $11-13$
Frequency: $5$ $10$ $10$ $7$ $8$

$C.I$	$F_i$	$x_i$	$x_iF_i$
$3 - 5$ $5-7$ $7-9$ $9-11$ $11-13$	$5$ $10$ $10$ $7$ $8$	$4$ $6$ $8$ $10$ $12$	$20$ $60$ $80$ $70$ $96$
	$\displaystyle\sum F_i=40$		$326$

Mean

$= \dfrac{\displaystyle\sum x_i F_i}{\displaystyle\sum F_i} = \dfrac{326}{40} = 8.15$

Mean

$= 8.15$

Using the step-deviation method, calculate the mean marks of the following distribution.
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 ( $x_i$ )	$d_i = x_i – A$	$u_1=d_1/h$	$f_iu_i$
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	$\Sigma f_i = 80$				$\Sigma f_iu_i = 24$

By step deviation method, $Mean = \overset{-}{x} = A+h∑\Sigma f_iu_i /\Sigma f_i$

$= 67.5+5(24/80)$

$= 67.5+5\times0.3$

$= 67.5+1.5$

$= 69$

Hence the mean of the distribution is $69$ .

Using an appropriate method, find the mean of following frequency distribution :
Class $84-90$ $90-96$ $96-102$ $102-108$ $108-114$ $114-120$
Frequency $8$ $10$ $16$ $23$ $12$ $11$
Which method did you use, and why?

Class Interval	Frequency $f_i$	Mid-Valiue $x_i$	$f_i\ x_i$
$84-90$	$8$	$87$	$696$
$90-96$	$10$	$93$	$930$
$96-102$	$16$	$99$	$1584$
$102-108$	$23$	$105$	$2415$
$108-114$	$12$	$111$	$1332$
$144-120$	$11$	$117$	$1287$
	$80$		$8244$

Mean

$=\dfrac {\displaystyle \sum f_ix_i}{\displaystyle \sum f_i}$

$=\dfrac{8244}{80}$

$=103.05$

Find the mean of the following data, using direct method:
Class $25-35$ $35-45$ $45-55$ $55-65$ $65-75$
Frequency $6$ $10$ $8$ $12$ $4$

Class Interval	Frequency $f_i$	Mid-Value $x_i$	$f_i\ x_i$
$25-35$	$6$	$30$	$180$
$35-45$	$10$	$40$	$400$
$45-55$	$8$	$50$	$400$
$55-65$	$12$	$60$	$720$
$65-75$	$4$	$70$	$280$
	$\sum f_i=40$		$\sum f_ix_i=1980$

Mean

$=\dfrac {\displaystyle \sum f_ix_i}{\displaystyle \sum f_i}$

$=\dfrac{1980}{40}$

Mean

$=49.5$

Compute the mean of the following data, using direct method:
Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$
Frequency $6$ $9$ $15$ $12$ $8$

Class Interval	Frequency $f_i$	Mid-Value $x_i$	$f_i\ x_i$
$0-100$	$6$	$50$	$300$
$100-200$	$9$	$150$	$1350$
$200-300$	$15$	$250$	$3750$
$300-400$	$12$	$350$	$4200$
$400-500$	$8$	$450$	$3600$
	$\sum f_i=50$		$\sum f_ix_i=13200$

Mean

$=\dfrac {\displaystyle \sum f_ix_i}{\displaystyle \sum f_i}$

$=\dfrac{13200}{40}$

$=264$

The following table gives the daily wages of a worker in a factory:
Calculate their mean by the short cut method.

By short cut method,

$Mean=\overset{-}{x}=A+\Sigma f_id_i/\Sigma f_i$

$=62.5+(-25/100)$

$=62.5-0.25$

$=62.25$
Hence mean of the distribution is

$62.25$

1712977_1784155_ans_71d2b3a60bb849ddb3b5163d27db70fe.jpg

Calculate the mean of the following distribution using step deviation method

By step deviation method,

$Mean=\overset{-}{x}=A+h\Sigma f_iu_i/\Sigma f_i$

$=25+10(63/100)$

$=25+10\times0.63$

$=25+6.3$

$=31.3$
Hence mean of the distribution is

$31.3$

1712953_1784151_ans_66f3abd547c24712bb5aa3f551a84063.jpg

The following are the marks obtained by $70$ boys in a class test:
Calculate the mean by: Short cut method

Here,

$A = 65$

$\bar{x}=A+\dfrac{\sum f_id_i}{\sum f_i}=65 + \left ( \dfrac{-270}{70} \right )=65-3.86=61.14$
1799815_1834235_ans_62a84119ab6c4ef196da5e4ef4620358.PNG

The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.

In order to calculate the median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformly distributed or equally spaced. Hence, we cannot say that the statement "the median of an ungrouped data and the median calculated when the same data is grouped are always the same" is always correct.

The following table gives the weekly wages of workers in a factory.
Calculate the mean by using: Short Cut Method

Here,

$A = 72.5$

$\bar{x} = A + \dfrac{\sum f_id_i}{\sum f_i}=72.5 + \left ( \dfrac{-280}{80} \right ) = 72.5 - 3.5 = 69$
1799794_1834234_ans_6ea7763d30dd406e9cebcf4ff81fd3ff.PNG

The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.

Calculate the median of the following frequency distribution table:
$\begin{array}{|l|l|}\hline Class Interval & Frequency \\\hline 1-4 & 6 \\4-7 & 30 \\7-10 & 40 \\10-13 & 16 \\13-16 & 4 \\16-19 & 4 \\\hline\end{array}$

$\begin{array}{|l|r|r|}\hline Class \ Interval & Frequency & Cumulative \ Frequency \\\hline 1-4 & 6 & 6 \\4-7 & 30 & 36 \\7-10 & 40 & 76 \\10-13 & 16 & 92 \\13-16 & 4 & 96 \\16-19 & 4 & 100 \\\hline\end{array}$

Median Class = Size of

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

$=$ Size of 50 th class

$=7-10$

Lowest limit(l)

$=7$

Class width(h)

$=3$

Frequency(f)

$=40$

and

$c.f=36$ preceding cumulative frequency of median class

$Median =L+\left(\dfrac{\dfrac{N}{2}-c f}{f} \times h\right)=7+\left(\dfrac{50-36}{40} \times 3\right)=7+\dfrac{21}{20}=\dfrac{161}{20}=8.05$

The following observations have been arranged in ascending order. If the median the data is $13$ , find the value of $x$
$3,6,7,10,x,x+4,19,20,25,28$

Given observations in ascending order

$3,6,7,10,x,x+4,19,20,25,28$

Here,

$n=10$ which is even and median

$=13$

So,

Median

$\cfrac { 1 }{ 2 } \left[ (10/2)th\quad term+(10/2+1)th\quad term \right] =\cfrac { 1 }{ 2 } (5th\quad term+6th\quad term)=\cfrac { 1 }{ 2 } (x+x+4)=(2x+4)/2=x+2\quad$

$\Rightarrow x+2=13$

$x=13-2=11$

Hence the value of

$x$ is

$11$

The following table gives the weekly wages of workers in a factory.
Calculate the mean by using: Direct Method

Mean

$= \sum f_i x_i/ \sum f_i = 5520/80 = 69$

A frequency distribution of funds collected by 120 workers in a company for the drought-affected people are given in the following table. Find the mean of the funds by 'step deviation' method.
Funds (Rupees) 0 - 500 500 - 1000 1000 - 1500 1500 - 2000 2000 - 2500
No. of workers 35 28 32 15 10

Class

(Fund in Rupees) Class Mark

$x_{i}$

$d_{i} = x_{i} - A$

$u_{i} = \dfrac{d_{i}}{h}$ Frequency

(No. of workers)

$f_{i}$

Frequency * deviation

$f_{i} * u_{i}$

0 - 500 250 -1000 -2 35 -70

500 - 1000 750 -500 -1 28 -28

1000 - 1500 1250 = A 0 0 32 0

1500 - 2000 1750 500 1 15 15

2000 - 2500 2250 1000 2 10 20

$\sum f_{i} = 20$

$\sum f_{i} u_{i} = -63$

Required Mean

$= A + h\dfrac{\sum f_{i} u_{i}}{\sum h}$

$= 1250 - \left ( \dfrac{63}{120}\right ) 500$

$= 1250 - 265.5$

$= Rs. 987.5$

Hence, the mean of the funds is Rs 987.5.

The following are the marks obtained by $70$ boys in a class test:
Calculate the mean by: Step deviation method

Here,

$A = 65$ and

$h = 10$

$\bar{x}=A+\dfrac{\sum f_iu_i}{\sum f_i}=65 + 10 \times \left ( \dfrac{-27}{70} \right )=65-3.86=61.14$
1799822_1834236_ans_263f6684b6544292baa16f552cb1043e.PNG

The mean of the following distribution is 132 and sum of frequencies isFind the values of $f_1$ and $f_2$
Class interval
0-40
40-80
80-120
120-160
160-200
200-240
Frequency
4
7
$f_1$
12
$f_2$
9

$\textbf{Step 1: Find the mid point of all the class intervals and assume 100 as assumed mean to use }$

$\textbf{short formula to calculate mean.}$

$\text{Given, }$

$\Sigma f_i = 50$

$\bar x = 132$

$\text{and }h = 40$

$\text{Let, }A \ \text{(assumed mean)}$

$= 100$

$\Sigma f_iu_i = 2f_2 + 24$

$\text{Now, }$

$132 = 100 + \dfrac{24 + 2f_2}{50} \times 40$

$\left[\because \boldsymbol{\bar x = A + \dfrac{\Sigma f_iu_i}{\Sigma f_i} \times h}\right]$

$\Rightarrow f_2 = 8$

$......(1)$

$\textbf{Step 2: Get an equation using sum of frequencies and solve it with equation (1).}$

$\text{It is given that }\Sigma f_i = 50$

$\Rightarrow f_1 + f_2 + 32 = 50$

$\Rightarrow f_1 + f_2 = 18$

$.....(2)$

$\text{Substitute }f_2=8 \text{ in equation (2), we get}$

$f_1+8=18$

$\Rightarrow f_1=10$

$\textbf{Hence, }\boldsymbol{f_1=10 \ \&\ f_2=8.}$

A milk center sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by the direct method.
Milk Sold (Litre)
1 - 2 2 - 3 3 - 4 4 - 5 5 - 6
No. of Customers 17 13 10 7 3

Mean =

$\dfrac{\sum x_{i}f_{i}}{\sum f_{i}}$

$=\dfrac{141}{50}$

$= 2.83 \;liters$

Hence, the mean of the milk sold is 2.82 liters.

1914577_1852344_ans_f66ec6d9f8c04d689b5b47e990872015.PNG

Find the decile range and percentile on the basis of following data:
X f
0 - 10 3
10 - 20 9
20- 30 8
30 - 40 5
40 - 50 7
50 - 60 5
60 - 70 7
70 - 80 6

Class (x)	Frequency (f)	Cumulative frequency
0-10	3	3
10-20	9	12
20 - 30	8	20
30 - 40	5	25
40 - 50	7	32
50 - 60	5	37
60 - 70	7	44
70 - 80	6	50
	N = 50

Calculate of decile range

$\dfrac{9N}{10} = \dfrac{9\times 50}{10} = 45$
Cumulative frequency equal to or exactly greater than 45 is 50 whose corresponding class is 70 - 80

The following tables gives the information of frequency distribution of weekly wages of150 workers of a company. Find the mean of the weekly wages by 'step deviation' method.
Weekly wages (Rupees) 1000 - 2000 2000 - 3000 3000 - 4000 4000 - 5000
NO. of workers 25 45 50 30

Class (Weekly wages rupees)	Class Mark $x_{i}$	$d_{i} = x_{i} - A$	$u_{i} = \dfrac{d_{i}}{h}$	Frequency (Number of workers) $f_{i}$	Frequency * deviation $f_{i} * u_{i}$
1000 - 2000	1500	-1000	-1	25	-25
2000 - 3000	2500	0	0	45	0
3000 - 4000	3500 = A	1000	1	50	50
4000 - 5000	4500	2000	2	30	60
				$\sum f_{i} = 150$	$\sum f_{i} u_{i} = - 85$

Required Mean

$= \dfrac{\sum f_{i} u_{i}}{\sum h}$

$= \left ( \dfrac{85}{150} \right ) = 0.57$

$= 2500 + 0.57 (1000)$

$= 2500 + 570$

$= Rs 3070$
Hence, the mean of the weekly wages is Rs 3070.

Write the formula to find mean for grouped data using step deviation method.

Following formula can be used to calculate mean for grouped data using step deviation method:

Mean =

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

where,

$a$ is the assumed mean (any value from

$x_i$ can be choosen as assumed mean)

$h$ is the class interval

and

$u_i$ is the magnitude of the difference between

$x_i$ and

$a$

Find the decile range and percentile range of the following frequency distribution
Marks (x) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Frequency (f) 5 8 20 14 3

Calculate of Decile Range -

$\dfrac{9N}{10} = \dfrac{9\times 502}{10} = = 9 \times 5 = 45$
Cumlative frequency exactly greater than 45 is 47, whose corresponding class is 30 - 40

$\therefore l = 30, f = F = 33, h , h = 10$
1954286_1874310_ans_168b534a5bb94b1398c11439b6658f63.png

Find mean by step deviation method:

Here,

$A = 87.50$ and

$h = 7$

$\bar{x}=A+ h \times \dfrac{\sum f_iu_i}{\sum f_i}=87.5 + 7 \times \dfrac{29}{150}=87.5 + 1.35 = 88.85$
1799823_1834237_ans_74411115ebcc45e99b9e7a8bdc8340bd.PNG

The following time shows the number of students and the time they utilized daily for their studies. Find the meantime spend by students for their studies by the direct method.

Time (hrs.) 0 - 2 2 - 4 4 - 6 6 - 8 8 - 10
No. of students 7 18 12 10 3

Class (Time in Hours)	Class Mark $x_{i}$	Frequency (Number of students) $f_{i}$	Classmark x Frequency $x_{i}f_{i}$
0 - 2	1	7	7
2 - 4	3	18	54
4 - 6	5	12	60
6 - 8	7	10	70
8 - 10	9	3	27
		$\sum f_{i} = 50$	$\sum x_{i}f_{i} = 218$

Mean -

$\dfrac{\sum x_{i}f_{i}}{\sum f_{i}}$

$= \dfrac{218}{50}$

$= 4.36$ hours
Hence, the mean time spent by students for their studies in 4.36 hours.

Two brands of tyres are to be compared for their mean life. The following data are available.
Life (thousand km.) $10 - 15$ $15 - 20$ $20 - 25$ $25 - 30$ $30 - 35$ $35 - 40$
No. of tyres of brand $A$ $4$ $7$ $10$ $5$ $3$ $1$
No. of tyres of brand $B$ $5$ $8$ $15$ $9$ $6$ $2$
On the basis of mean, which brand of tyres is better?

$\textbf{Step 1: Find the mean for each brand.}$

$\text{We know,}$

$\text{Mean} = \mathrm{\dfrac {\sum (Internal\ Midpoint \times Frequency)}{\sum \ Frequency}}$

$\text{For brand A,}$

$\text{Mean}= \dfrac {12.5\times 4 + 17.5\times 7 + 22.5\times 10 + 27.5 \times 5 + 32.5 \times 3 + 37.5\times 1}{4+7+10+5+3+1}$

$= \dfrac { 670}{30}$

$= 22.33$

$\text{For brand B,}$

$\text{Mean}= \dfrac {12.5\times 5 + 17.5\times 8 + 22.5\times 15 + 27.5 \times 9 + 32.5 \times 6 + 37.5\times 2}{5+8+15+9+6+2}$

$= \dfrac { 1057.5}{45}$

$= 23.5$

$\textbf{Step 2: Determining the better brand.}$

$\text{Mean of brand B is greater than mean of brand A.}$

$\text{Which means, on an average tyres of brand B runs more distance than brand A.}$

$\textbf{Hence, tyres of brand B are better.}$

Following data relates to daily wages in rupees of persons working in a factory. Compute the median daily wage:
Daily wage 200-250 250-300 300-350 350-400 400-450 450-500 500-550 550-600
No. of workers 14 28 33 30 20 15 13 7

Solution is as follows:
2146910_2026093_ans_27dc6ee21ff34357a136dc758a27631f.PNG

The mean age of a combined group of men and women is 30 years. If the mean age of the group of men is that of the group of women is 27, find out the percentage of men and women in the group.

2138963_2021669_ans_fdb700c0e4bb49808d8fd6f35c01ecc2.jpeg

Calculate arithmetic mean of the following series. Take 15 as assumed average and use short cut method.
Size 20 19 18 17 16 15 14 13 12
Frequency 1 2 4 8 11 10 7 4 2

Size (X)	Frequency (f)	d = X - A	fd
20	1	5	5
19	2	4	8
18	4	3	12
17	8	2	16
16	11	1	11
15	10	0	0
14	7	-1	-7
13	4	-2	-8
12	2	-3	-6
	∑f = 49		∑fd = 32

Mean = A + ∑fd/∑f = 15 + 32/48 = 15 + 0.67 = 13.67.

Find out the median from the following particulars:
Marks 0-10 10-20 20-30 30-40 40-50 50-60
No of students 3 9 18 30 18 12

2145644_2024954_ans_b19a5b14f1d2452193ca385b9e753def.png

Calculate median from the following series-
Class 1-10 11-20 21-30 31-40 41-50 51-60
Frequency 4 15 20 10 8 5

2139735_2022169_ans_b0a0d632e9ec402c980aaed43a629037.png

Calculate median from the following:

CI

f

0-10

2

10-20

4

20-30

6

30-40

2

40-50

4

50-60

2

Solution is as follows:

CI	f	Cumulative frequency (cf)
0-10	2	2
10-20	4	2 + 4 = 6
20-30	6	6 + 6 = 12
30-40	2	12 + 2 = 14
40-50	4	14 + 4 = 18
50-60	2	18 + 2 = 20
	N = ∑f = 20

N/2 = 20/2 = 10. Median lies in 20-30.
Median = L + [(N/2 - cf)/ ∑f] x h = 20 + [(10 - 6)/20] x 10 = 22.

Calculate median and mode from the following data:
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
No. of Students 2 18 30 45 35 20 6 3

2151550_2028164_ans_9aaa46ac52ba444bb5c7ff5414ad10bd.png

Calculate median from the following:

X

10

20

30

40

60

70

f

2

4

6

2

4

2

Solution is as follows:

2144416_2024582_ans_d9720cb7b8c34871b48781bbe78ac3db.PNG

Calculate median of the following data:

Marks

No of students (f)

46-50
5

41-45 11

36-40 22

31-35 35

26-30 26

21-25 13

16-20 10

11-15 7

Solution is given by:
2158203_2032316_ans_337277f77cdc4b618b4063385506553a.PNG

A candidate obtains the following percentage of marks:

Subjects

Marks

History

54

Geography

47

Sanskrit

75

Maths

84

Economics

58

English

78

Politics

57

It is agreed to give double weights to marks in English, Maths and Sanskrit. What is the weighted mean?

The solution is given by:
2158195_2032309_ans_01e07432afed41ffa85aed53bddced3a.PNG

Calculate median from the following data:

S.No.

Marks obtained (X)

1

10

2

12

3

14

4

17

5

18

6

20

7

21

8

30

9

32

2158944_2032729_ans_0b8b3840bc7a43de973d63361d992c72.PNG

Calculate mean from the following data. (use step-deviation method).

Class Interval

No. of workers

0-10

10

10-20

15

20-30

25

30-40

25

40-50

10

50-60

10

60-70

5

2159821_2033317_ans_f7d8843aa9b847988f5bd983774ac9e6.PNG

Calculate the weighted mean of the following distribution:
Items 12 29 14 41
Weight 6 4 5 2

Calculation of Weighted mean

Items (X)	Weight (W)	WX
12	6	72
29	4	116
14	5	70
41	2	82
	∑W = 17	∑WX = 340

Weighted Mean = ∑WX/∑W = 340/17 = 20.

Find out median of the following series:

Size

Frequency

20

14

25

18

30

33

35

30

40

20

45

15

50

13

55

7

2158990_2032877_ans_aaccbbaa0aaf4ac39b7a5ad4f87ad0af.PNG

Calculate mean by step-deviation method from the following data:

X

No. of Persons

20-29

10

30-39

8

40-49

6

50-59

4

60-69

2

2158991_2032878_ans_15c854592df54d4486125b41a043fb81.PNG

Find out the missing frequency of the group 20 - 30, when the median is 28.

Class Interval

Frequency

0-10

5

10-20

8

20-30

?

30-40

16

40-50

6

2159904_2033296_ans_06f83725c32c4d5c9dbae36cc5facc7b.PNG

Calculate the median.

Age

Number

15-19

4

20-24

20

25-29

38

30-34

24

35-39

10

40-44

4

2160062_2033395_ans_a73d23ef9802414c8228548763ea1153.PNG

Find out the Arithmetic mean from the following data by direct method:

Runs scored

No. of matches

0-20

2

20-40

4

40-60

6

60-80

4

80-100

2

Calculate arithmetic mean of the following series. Take 15 as assumed average and use Short-cut Method.

Size

Frequency

20

1

19

2

18

4

17

8

16

11

15

10

14

7

13

4

12

2

11

1

Calculate arithmetic mean using Direct Method.

Marks

Number of Students

4

12

5

10

6

18

7

20

8

24

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

Mean marks = Sum/Total number = (41+39+52+48+54+62+46+52+40+96+42+40+98+60+52)/15 = 822/15 = 54.8 marks.

The following table gives height of certain persons:

Height (inches)

Number of Persons

57

1

59

3

61

7

63

10

65

11

67

15

69

10

71

7

73

2

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

Arranging in ascending order: 39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98. N = 15, which is odd.
Median = (N+1)th term/2 = (15+1)th term/2 = 8th term = 52 marks.

Calculate the median from the following-
Marks Frequency
Less than 10 5
Less than 20 20
Less than 30 45
Less than 40 80
Less than 50 100
Less than 60 115
Less than 70 125
Less than 80 130

2163984_2034639_ans_e3bd56b43fba454b983bbda7f95e4343.png

Following is the number of heads of 8 coins when tossed 205 times:

Number of Heads

Frequency

0

2

1

19

2

46

3

62

4

47

5

20

6

5

7

2

8

2

Calculate arithmetic mean (mean number of heads per toss), using Direct Method.

Calculate median from the following data-
Mid-Value 20 30 40 50 60 70
Male (Cf) 12 25 42 46 48 50

2163985_2034669_ans_60d01cb5b98a4b1ca4f8e2b59ef08d5f.png

Following is the number of heads of 8 coins when tossed 205 times:

Number of Heads (X)

Frequency (f)

0

2

1

19

2

46

3

62

4

47

5

20

6

5

7

2

8

2

Calculate arithmetic mean (mean number of heads per toss), using Short-cut Method.

Calculate mean using step deviation method from the following information:

Class Interval

Frequency

0-5

15

5-10

10

10-15

11

15-20

12

20-25

6

25-30

6

30-35

10

35-40

15

40-45

10

45-50

5

If the arithmetic mean of the data given below is 52, find the missing frequency:

Class Interval

No. of Students

10-20

5

20-30

3

30-40

4

40-50

7

50-60

f

60-70

6

70-80

13

Read the following news report and answer on the basis of the same: For an examination, 60 students of a school appeared. Marks obtained by the 12 students were 25 each. The number of students who scored 20 and 5 each was three more than the students who scoredAlso, 8 students who scored 10 marks each. The students who scored 15 marks each were half the students who scored 10 each and remaining students had scored 30 each. Answer the following question: The value of median will be ____________________.

Marks	Number of students
5	15 = 12 + 3
10	8
15	4
20	15 = 12 + 3
25	12
30	6
Total	60

Calculation of median:

Marks	Number of students (f)	cf
5	15	15
10	8	23
15	4	27
20	15	42
25	12	54
30	6	60
	Total = 60

Median = (N+1)/2 = (60+1)/2 = 30.5th term = 20.

Find Arithmetic Mean of the following distribution using Step Deviation Method (Assume mean = 27.5).

Size (in cm.)

No. of Items

10-15

2

15-20

8

20-25

20

25-30

35

30-35

20

35-40

15

Find out median:
X
f
2
3
4
5
6
12
8
10
10
5

Calculate mean from the following data:

Marks

No. Of students

0-20

8

20-40

12

40-60

30

60-80

20

80-100

10

Following are the wages of 19 workers.
Wages 10 20 30 40 50
Number of workers 4 5 3 2 5
Find the arithmetic mean using asumed mean method.

2164476_2035008_ans_318710a58ab444c08a61b070ee8ebb0e.png

Calculate median from the following data:

Size

Frequency

20

14

25

18

30

33

35

30

40

20

45

15

50

13

55

7

Calculate arithmetic mean from the following data:

Marks

No. of Students

0-10

2

10-20

7

20-30

10

30-40

15

40-50

20

50-60

16

60-70

6

70-80

4

Calculate the arithmetic mean for the following data using step deviation method.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
No. of students $8$ $15$ $22$ $20$ $10$ $5$

Marks	Midpoint( ${ x }_{ i }$ )	No. of students( ${ f }_{ i }$ )	Deviation( ${ d }_{ i }$ )	${ u }_{ i }$	${ u }_{ i }{ f }_{ i }$
0-10	5	8	-30	-3	-24
10-20	15	15	-20	-2	-30
20-30	25	22	-10	-1	-22
30-40	35	20	0	0	0
40-50	45	10	10	1	10
50-60	55	5	20	2	10

$\sum { { f }_{ i }=80 }$ , h=

$10$

$35$ ,

${ d }_{ i }={ x }_{ i }-A$ ,

${ u }_{ i }=\cfrac { { d }_{ i } }{ h }$

Mean=

$A+h\left( \cfrac { \sum { { u }_{ i }{ f }_{ i } } }{ N } \right) \\ =35+10\left( \cfrac { -56 }{ 80 } \right) \\$

Mean=

$28$

The ages of children in a scout camp are $13, 13, 14, 15, 13, 15, 14, 15, 13, 15$ years. Find the mean, median and mode of the data.

Arranging the given data in ascending order

$13,13,13,13,14,14,15,15,15,15$

$n=10$

mean

$=\cfrac { 4(13)+2(14)+4(15) }{ 10 } =14$ years

$n$ is even, median

$=\cfrac { 4th+5th }{ 2 }$ value

$=14$ years

$13$ and

$15$ both have the highest frequencies of

$4$

It is a bimodal set of data.

mean

$=14$ ;median

$=14$ ;mode

$=13,15$

The total marks obtained by $40$ students in the Annual examination are given below
Marks $150-200$ $200-250$ $250-300$ $300-350$ $350-400$ $400-450$ $450-500$
Students $2$ $3$ $12$ $10$ $4$ $6$ $3$
Using step deviation method, find the mean of the above data.

Marks	Midpoint( ${ x }_{ i }$ )	Students( ${ f }_{ i }$ )	Deviation( ${ d }_{ i }$ )	${ u }_{ i }$	${ u }_{ i }{ f }_{ i }$
150-200	175	2	-150	-3	-6
200-250	225	3	-100	-2	-6
250-300	275	12	-50	-1	-12
300-350	325	10	0	0	0
350-400	375	4	50	1	4
400-450	425	6	100	2	12
450-500	475	3	150	3	9

$\sum { { f }_{ i }=40 }$ , h=

$50$

$325$ ,

${ d }_{ i }={ x }_{ i }-A$ ,

${ u }_{ i }=\cfrac { { d }_{ i } }{ h }$

Mean marks=

$A+h\left( \cfrac { \sum { { u }_{ i }{ f }_{ i } } }{ N } \right) \\ =325+50\left( \cfrac { 1 }{ 40 } \right) \\$

Mean marks=

$326.25$

Find the median of the following data.
(i) $18, 12, 51, 32, 106, 92, 58$
(ii) $28, 7, 15, 3, 14, 18, 46, 59, 1, 2, 9, 21$

$(i)$ Arrange data in ascending order

$12,18,32,51,58,92,106$

nedian=mid value

$=51$

$(ii)$ Arrange data in ascending order

$1,2,3,7,9,14,15,18,21,28,46,59$

There are two middle values

$=14,15$

Median

$=\cfrac { 14+15 }{ 2 } \\ =14.5$

In a retail market , fruit vendors were selling mangoes kept in packed boxes. These boxes contained different number of mangoes. The following was the distribution of mangoes according to the number of boxes ?
Number of mangoes $50-52$ $53-55$ $56-58$ $59-61$ $62-64$
Number of boxes $15$ $110$ $135$ $115$ $25$

Find the mean number of mangoes using step-deviation method.

No of mangoes	No of boxes $(F_i)$	Class mark $(x_i)$	$X_i-A$	$Ui=\dfrac{Xi-A}{h}$	$FiUi$
$50-52$	$15$	$51$	$-6$	$-2$	$-30$
$53-55$	$110$	$54$	$-3$	$-1$	$-110$
$56-58$	$135$	$57$	$0$	$0$	$0$
$59-61$	$115$	$60$	$3$	$1$	$115$
$62-64$	$25$	$63$	$6$	$2$	$50$
	$∑Fi=400$				$∑FiUi=25$

$mean,\ ({ \overset { - }{ x } } )=A+h\times \dfrac { \sum{F_iU_i} }{ \sum { { F }_{ i } } }$

$where, A= assumed\ mean=57\quad \\ h=class\ size=3\quad$

$\sum { { F }_{ i }={ 400 } }$

$\sum { F } _{ i }{ U }_{ i }={ 25 }$

$mean(\overset { - }{ x } )=A+h\times \dfrac { \sum{F_iU_i} }{ \sum { { F }_{ i } } }$

$\\ mean(\overset { - }{ x } )=57+3\times \dfrac { 25 }{ 400 } \\$

$=57+0.1875$

$=57.1875$

Hence, the mean number of mangoes in each box is

$57.1875$ .

The mean of the following frequency table is $57.6$ . But the frequencies $f_1$ and $f_2$ are missing. Find the value of missing frequencies $f_1$ and $f_2$ .
Class Interval $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$ Total
Frequency $7$ $f_1$ $12$ $f_2$ $8$ $5$ $50$

Class Interval	Frequency ( f)	Mid point x	fx
$0-20$	$7$	$10$	$70$
$20-40$	$f_1$	$30$	$30f_1$
$40-60$	$12$	$50$	$600$
$60-80$	$f_2$	$70$	$70f_2$
$80-100$	$8$	$90$	$720$
$100-120$	$5$	$110$	$550$
Total	$\sum f=32+f_2+f_2=50$		$\sum fx=30fx_1+70f_2+1940$

$\sum f=f_1+f_2+32=50$

$\Rightarrow f_1+f_2=18$ ..........(i)

Mean

$\bar{x}=\displaystyle\frac{\sum fx}{\sum f}$

$57.6=\displaystyle\frac{30f_1+70f_2+1940}{50}$

$30f_1+70f_2+1940=2,880$

$30f_1+70f_2=940$
or

$3f_1+7f_2=94$ .............(ii)

Solving (i) and (ii)

$f_1=8, f_2=10$

Thus, the value of

$f_1$ is

$8$ and the value of

$f_2=10$ .

Find the mean of the following frequency distribution:
Marks obtained Number of students
10-20 $6$
20-30 $8$
30-40 $13$
40-50 $7$
50-60 $3$
60-70 $2$
70-80 $1$

Let assumed mean

$A=45$
and interval

$i=10$

Marks	No. of students	Mid value	$u=\cfrac { x-A }{ i }$	$f\times u$
10-20	6	15	-3	-18	-47
20-30	8	25	-2	-16
30-40	13	35	-1	-13
40-50	7	45	0	0
50-60	3	55	1	3	10
60-70	2	65	2	4
70-08	1	75	3	3
Total	$\sum { f } =40$			$\sum { f } =-37$

Formula Mean

$=A+\cfrac { \sum { fu } }{ \sum { f } } \times i$

$=45+\cfrac { -37 }{ 40 } \times 10=45-\cfrac { 37 }{ 4 } =45-9.25=35.75$

Candidates of four schools appear in a mathematics test. The data were as follows.
Schools No. of candidates Average Score.
I $60$ $75$
II $48$ $80$
III Not available $55$
IV $40$ $50$
If average score of the candidates of all the four schools is $66$ , find the number of candidates that appeared from school II.

Solution:-

Given,

Average score of candidates of I school

$({\bar{x}}_{1}) = 75$

Average score of candidates of II school

$({\bar{x}}_{2}) = 80$

Average score of candidates of III school

$({\bar{x}}_{3}) = 55$

Average score of candidates of IV school

$({\bar{x}}_{4}) = 50$

Average score of the candidates of all the four schools

$({\bar{x}}_{mean})$ = 66

As we know that,

$\bar{x} = \cfrac{\text{Sum of score of all candidates}}{\text{Total number of students}}$

$\text{School I} \Rightarrow$

$\therefore {\bar{x}}_{1} = \cfrac{\text{Sum of scores of all candidates in school I }({S}_{1})}{\text{Total number of candidates in school I } ({n}_{1})}$

$\Rightarrow 75 = \cfrac{{S}_{1}}{60}$

$\Rightarrow {S}_{1} = 4500$

$\text{School II} \Rightarrow$

$\therefore {\bar{x}}_{2} = \cfrac{\text{Sum of scores of all candidates in school II }({S}_{2})}{\text{Total number of candidates in school II } ({n}_{2})}$

$\Rightarrow 80 = \cfrac{{S}_{2}}{48}$

$\Rightarrow {S}_{2} = 3840$

$\text{School III} \Rightarrow$

$\therefore {\bar{x}}_{3} = \cfrac{\text{Sum of scores of all candidates in school III }({S}_{3})}{\text{Total number of candidates in school III } ({n}_{3})}$

Let the number of candidates in school III be n.

$\Rightarrow 55 = \cfrac{{S}_{3}}{n}$

$\Rightarrow {S}_{3} = 55n$

$\text{School IV} \Rightarrow$

$\therefore {\bar{x}}_{IV} = \cfrac{\text{Sum of scores of all candidates in school IV }({S}_{4})}{\text{Total number of candidates in school IV } ({{n}_{4}})}$

$\Rightarrow 40 = \cfrac{{S}_{4}}{50}$

$\Rightarrow {S}_{4} = 2000$

$\because {\bar{x}}_{mean} = \cfrac{{S}_{1} + {S}_{2} + {S}_{3} + {S}_{4}}{{n}_{1} + {n}_{2} + {n}_{3} + {n}_{4}}$

$66 = \cfrac{4500 + 3840 + 55n + 2000}{60 + 48 + n + 40}$

$66 = \cfrac{10340 + 55n}{148 + n}$

$\Rightarrow 66(148 + n) = 10340 + 55n$

$\Rightarrow9768 + 66n = 10340 + 55n$

$\Rightarrow 66n - 55n = 10340 - 9768$

$\Rightarrow 11n = 572$

$\Rightarrow n = \cfrac{572}{11}$

$\Rightarrow n = 52$

Hence, the number of candidates that appeared from

$School \ III \ are \ 52.$

Below is the distribution of money collected
by students for flood relief.
Money No. of student
0-10
10-20
20-30
30-40
40-50 5
2
6
5
7
find mean of money collected by a student using "Direct Method"

$\therefore$ Mean

$=\dfrac{695}{25}=27.8.$

Hence, the answer is

$27.8.$

a) Find the mean median and mode of $20,15,25,7,25,17,5$
b) Construct a right angled triangle $PQR$ such that $\angle Q=90^{o}\ PQ=6cm$ , hypotenuse $PR=7cm$
c) The daily wages of $5$ person is given below:
Wages $300$ $150$ $200$ $250$ $350$
Persons $A$ $B$ $C$ $D$ $E$

$\begin{array}{l} a'\, \, \, 5,7,15,17,20,25,25 \\ Mode=25 \\ Now\, for\, \, Median \\ n=7\left( { odd } \right) \\ Median={ \left( { \frac { { n+1 } }{ 2 } } \right) ^{ th } }term \\ ={ \left( { \frac { { 7+1 } }{ 2 } } \right) ^{ th } }term \\ ={ 4^{ th } }term \\ =17\, \, m \\ b'\, \, Step\, \, of\, \, construction \\ \left( i \right) Draw\, \, a\, \, line\, \, QY \\ \left( { ii } \right) Make\, \, an\, \, angle\, \, 90^{ \circ }\, \, with\, \, Q \\ \left( { iii } \right) Cut\, \, PQ=6 \\ \left( { iv } \right) From\, \, P\, \, cut\, \, PR=7\, \, QR=7\, \, at\, \, QY. \\ then\, \, \Delta PQR\, \, is\, \, required\, \, \Delta . \end{array}$
1214081_1050849_ans_13ffff8d0a0246d3b1ea2abcf1162513.PNG

1214081_1050849_ans_13ffff8d0a0246d3b1ea2abcf1162513.PNG

Find the mean deviation about the median for the following continuous distribution:
Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60
No. of body 6 8 14 16 4 2

Class $(fi-hi)$	frequency $fi$	cumulative frequency $cfi$	Midpoint $(l+h)/2$	$fi\|xi-M\|$
$0-10$	6	6	5	137.16
$10-20$	8	6+8=14	15	102.88
$\rightarrow 20-30$	14	14+14=28	25	40.04
$30-40$	16	28+16=44	35	114.24
$40-50$	4	44+4=48	45	68.56
$50-60$	2	48+2=50	55	54.28

$N= \sum fi=50$

$\sum fi |xi- M|= 517.16$

Here,

$\frac{N}{2}= \frac{50}{2}= 25$ , which lies in the class 20-30

Therefore the median class is 20-30.

$\therefore$ Median,

$M= l+\frac{\frac{N}{2}-cf}{f}\times h$

where

$l$ is the lower limit of the median class.

$cf$ is the cumulative frequency of the class interval proceeding the median class.

$f$ is the frequency of median class.

$h$ is the width of the median class.

$\therefore M= 20+\frac{25-14}{14}\times 10$

$= 20+\frac{11}{14}\times 10$

$= 20+\frac{55}{7}$

$\Rightarrow M= 27.86$ (approx)

mean deviation about median =

$\frac{\sum fi\left | xi-M \right |}{\sum fi}$

$=\frac{517.16}{50}$

$= 10.343$ (approx)

The following distribution represents the number of books issued at the counter of a ecommerce college library on $200$ different days :
Calculate lower and upper quartile for above data.
No. of books issued $80$ $100$ $120$ $145$ $200$ $280$ $310$ $380$ $400$ $410$
No. of days $15$ $18$ $25$ $27$ $40$ $25$ $19$ $16$ $8$ $7$

The given distribution represents the no. of books issued at the counter of a ecommerce college library on 200 distribution

No of books issued	80	100	120	145	200	280	310	380	400	410
No. of days	15	18	25	27	40	25	19	16	8	7

Lower and upper quartile

No.mof books issued	No. of days (Freqvency)	Cumulative frequency
80	15	15
100	18	15 $+$ 18 $=$ 33
120 (lower quartile)	25	33 $+$ 25 $=$ 58
145	27	58 $+$ 27 $=$ 85
200	40	85 $+$ 40 $=$ 125
280	25	125 $+$ 25 $=$ 150
310	19	150 $+$ 19 $=$ 169
380 (upper quartile)	16	169 $+$ 16 $=$ 185
400	8	185 $+$ 8 $=$ 193
410	7	193 $+$ 7 $=$ 200

$N = 200$

Lower qualtile

$(Q_1) = \text{size of} \,\dfrac{N+1}{4}th\, \text{ item}$

$= \text{size of} \,\dfrac{200+1}{4}th\, \text{ item}$

$= \text{size of} \,\dfrac{201}{4}th\, \text{ item}$

$= \text{size of}\, 50.25^{th}\, \text{ item}$

So, Lower quartile

$(Q_1) = 120$

Upper quartile

$(Q_3) = \text {size of} \,\dfrac{3(N+1)}{4} th\, \text{item}$

$= \text {size of}\, \dfrac{3(200+1)}{4} th\, \text{item}$

$= \text {size of}\, 3(50.25)^{th}\, \text{item}$

$= \text {size of}\, 150.75^{th}\, \text{item}$

$(Q_3) = 280$

The mean of the following distribution is $50$ . Find a.
x $10$ $30$ $50$ $70$ $90$
y $17$ $5a+3$ $32$ $7a-11$ $18$

Mean

$=\dfrac { \sum _{ i=1 }^{ 5 }{ { f }_{ i }{ x }_{ i } } }{ \sum _{ i=1 }^{ 5 }{ { f }_{ i } } }$

So, first

$\sum _{ x=1 }^{ 5 }{ { f }_{ 1 }{ x }_{ 1 } } =\left( 10\times 17 \right) +\left( 30\left( 5a+3 \right) \right) +\left( 50\times 32 \right) +70\times \left( 7a-11 \right) +\left( 90\times 19 \right)$

$=170+150a+90+1600+490a-770+1710$

$=2800+640a$

Now,

$\sum _{ i=1 }^{ 5 }{ { f }_{ i } } =17+5a+3+32+7a-11+19$

$=60+12a$

Mean

$=\dfrac { \sum _{ i=1 }^{ 5 }{ { f }_{ i }{ x }_{ i } } }{ \sum _{ i=1 }^{ 5 }{ { f }_{ i } } }$

$\Rightarrow 50=\dfrac { 2800+640a }{ 60+12a }$

$\Rightarrow 50\left( 60+12a \right) =2800+640a$

$\Rightarrow 3000-2800=640a-600a$

$a=5$ .

The mean of five numbers isIf one number is excluded, their mean isFind the excluded number.

The mean of five numbers

$1$ and

$27$

Let be

$x_1,x_2,x_3,x_4,x_5$

$\text{mean}=\dfrac{\text{sum of observation}}{\text{no' of observations}}$

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

$x_1+x_2+x_3+x_4+x_5=27\times 5=135$ ..........(1)\

if one number is excluded

their mean is

$25$ (that means remove one observations)

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

$x_1+x_2+x_3+x_4=100$ .........(2)

$\therefore$ (2) and (1)

$x_1+x_2+x_3+x_4+x_5=135$

$100+x_5=135$

$x_5=135-100$

$x_5=35$

$\therefore$ excluded number is

$35$

The age (in years) of $12$ persons are given below.
$42,48,48,50,56,43,47,56,60,65,56,65$ . Find the mode.

Highest number of frequency of 56=3

so, mode =56.

Given below are the daily wages of $200$ workers in a factory :
Daily Wages (in Rs.) $240-300$ $300 - 360$ $360 - 420$ $420 - 480$ $480 - 540$
Number of Workers $20$ $30$ $20$ $40$ $90$
Calculate the mean daily wages.

Given are the wages of

$200$ workers in factory.

We can prepare a table for calculating the mean, as shown below:

Daily wages	Mid value $\left(x_i\right)$	No. of workers $\left(f_i\right)$	$f_ix_i$
240 - 300	270	20	5400
300 - 360	330	30	9900
360 - 420	390	20	7800
420 - 480	450	40	18000
480 - 540	510	90 $\sum {f_i}=200$	45900 $\sum{x_if_i}=87000$

We know that, mean

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

$= \dfrac{87000}{200}$

$= 435$

$\therefore\$ The mean daily wage is

$Rs.\ 435$ .

Find the arithmetic mean using the assumed-mean method:
Class interval 100-120 120-140 140-160 160-180 180-200
Frequency 10 20 30 15 5

Let mean

$=150$

$\Rightarrow A=150$

$\Rightarrow$ Mean

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

$=150+\dfrac{(-30\phi)}{8\phi}=150-\dfrac{15}{4}=150-3.75$

$=146.25$ .

1183557_1215452_ans_3f8b7d0523084da8acc7a1cced2fb673.jpg

Find the mean age from the following distribution:
$Age(in years)$ $25-29$ $30-34$ $35-39$ $40-44$ $45-49$ $50-54$ $55-59$
$No. of persons$ $4$ $14$ $22$ $16$ $6$ $5$ $3$

Now we can prepare table for calculating mean

Age	continuous age	midpoint $x$	No of persons $f$	$fx$
$25-29$ $30-34$ $35-39$ $40-44$ $45-49$ $50-54$ $55-59$	$24.5-29.5$ $29.5-34.5$ $34.5-39.5$ $39.5-44.5$ $44.5-49.5$ $49.5-54.5$ $54.5-59.5$	$27$ $32$ $37$ $42$ $47$ $52$ $57$	$4$ $14$ $22$ $16$ $6$ $5$ $3$	$108$ $448$ $814$ $672$ $282$ $260$ $171$
			$\sum f = 70$	$\sum f_x = 2755$

$\therefore mean(\bar{x}) = \dfrac{\sum f_x}{\sum f} = \dfrac{2755}{70}$

$= 39.357$ (rounded to 2 decimals)

Mean age

$\cong 39.36$

The mean weight of 8 numbers isIf each number is multiplied by 2, what will be the new mean?

The mean weight of

$8$ numbers is

$15$

i.e

$n=8,mean=15$

We know

$mean=\dfrac{\text{sum of numbers}}{\text{No' of numbers}}$

$15=\dfrac{\text{Sum of numbers}}{8}$

Sum of numbers =

$8\times 15=120$

and given if each number is multiplied by

$2$

After multiplication

$2$ in each number then their new sum will be

$120\times 2$

$=240$

$\therefore$ new mean =

$\dfrac{\text{sum of numbers }}{\text{No' of numbers}}$

$=\dfrac{240}{8}$

$=30$

$\therefore$ New mean =

$30$

In the following table, the information is given about the number of families and the siblings in the families less than $14$ years of age. Find the mode of the data.

No. of siblings

1

2

3

4

Families

15

25

5

5

1486649_1539309_ans_d0506c6c8b494a8989e1a15b7c0cdb33.jpg

Find the sum of the deviations of the variate values 3,4,6,7,8,14 from their mean.

In the given variate values

$3,4,6,7,8,14$

We know that

$mean=\dfrac{\text{sum of observations}}{\text{No of observations}}$

$mean=\dfrac{3+4+6+7+8+14}{6}$

$=\dfrac{42}{6}$

$(\bar{x})=7$

Variate value x $x_i$	mean $(\bar{x})$	Deviation $di=x_i-mean(\bar{x})$
3	7	3-7=-4
4	7	4-7=-3
6	7	6-7=-1
7	7	7-7=0
8	7	8-7=1
14	7	14-7=7

$\therefore$ Sum of Deviations of the variate values x from their mean is

$=-4-3-1+0+1+7$

$=-8+8$

$=0$

The following observation have been arranged in ascending order. If the median of the data is 63, find the value of x:
29, 32, 48, 50, x, x+2, 72, 78, 84, 95

Given that

Observation hae been arranged in ascending order

$29,32,48,50,x,x+2,72,78,84,95$

median of above Data =

$63$ .............(1)

No' of observations (n) =

$10$

Here observe that No' of observations (n) is even so

Median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{\dfrac{10}{2}\text{th term}+(\dfrac{10}{2}+1)\text{th term}}{2}$

$=\dfrac{5 \text{th term}+ 6\text{th term}}{2}$

from (1)

$63=\dfrac{x+x+2}{2}$

$63=\dfrac{2x+2}{2}$

$2x+2=63\times 2$

$2x+2=126$

$2x=126-2$

$2x=124$

$x=\dfrac{124}{2}$

$x=62$

Value of

$x = 62$

Find the median of the following data
$83,37,70,29,45,63,41,70,34,54$

Given data

$83, 37, 70, 29, 45, 63, 41, 70, 34, 54$

First of all arrange the Data in ascending order

$29, 34, 37, 41, 45, 54, 63, 70, 70, 83$

Here we observe that no. of observations in data

$n=10$

i.e.,

So median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{\dfrac{10}{2}^{th}term+\left(\dfrac{10}{2}+1\right)^{th}term}{2}$

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

$=\dfrac{45+54}{2}$

Median

$=\dfrac{99}{2}=49.5$ .

Find the value of p for the following distribution whose mean is 16.6.
x : 8 12 15 p 20 25 30
f : 12 16 20 24 16 8 4

Given that

mean of Distribution Data is

$16.6$

i.e mean

$=16.6$ .........(1)

$x$	8	12	15	P	20	25	30
$f$	12	16	20	24	16	8	4

We know

$\displaystyle \dfrac{\sum f_x}{\sum f}$

$=\dfrac{\text{sum of observations}}{\text{No of observations}}$

We have prepare a table.

$x$	$f$	$f_x$
8	12	96
12	16	192
15	20	300
P	24	24P
20	16	320
25	8	200
30	4	120

Here

$\displaystyle \sum f=100$

$\displaystyle \sum f_x=1228+24P$

$mean =\dfrac{\sum f_x}{\sum f}$

$16.6=\dfrac{1228+24P}{100}$

$1228+24P=16.6\times 100$

$24P=1660-1228$

$24P=432$

$P=\dfrac{432}{24}$

$P=18$

value of

$P=18$

Find the median of the following data
133, 73, 89, 108, 94, 104, 94, 85, 100, 120

$133,73,89,108,94,104,94,85,100,120$

No' of observations (n) =

$10$

first of all we arrange the observations in an ascending order

$73,85,89,94,94,100,104,108,120,133$

Here observe that No' of observations (n) is even so

Median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{\dfrac{10}{2}\text{th term}+(\dfrac{10}{2}+1)\text{th term}}{2}$

$=\dfrac{5 \text{th term}+6\text{th term}}{2}$

$=\dfrac{94+100}{2}$

$=\dfrac{194}{2}$

Median =

$97$

Find the median of the following data
83, 37, 70, 29, 45, 63, 41, 70, 34, 54

$83,37,70,29,45,63,41,70,34,54$

first of all we arrange the data in an ascending order

$29,34,37,41,45,54,63,70,70,83$

Here observe that No' of observations

$(n)=10$ is even so

Median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{\dfrac{10}{2}\text{th term}+(\dfrac{10}{2}+1)\text{th term}}{2}$

$=\dfrac{5 \text{th term}+6\text{th term}}{2}$

$=\dfrac{45+54}{2}$

$=\dfrac{99}{2}$

Median =

$49.5$

Numbers 50, 42, 35, 2x+10, 2x -8, 12, 11, 8 are written in descending order and their median is 25, find x.

$50,42,35,2x+10,2x-8,12,11,8$ are written in descending order

their median is

$25$ ..............(1)

Here observe that No' of numbers are even i.e

$n=8$

So Median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{\dfrac{8}{2}\text{th term}+(\dfrac{8}{2}+1)\text{th term}}{2}$

$=\dfrac{4 \text{th term}+5\text{th term}}{2}$

$25=\dfrac{2x+10+2x-8}{2}$

$4x+2=50$

$4x=50-2$

$4x=48$

$x=\dfrac{48}{4}$

$x=12$

value of

$x = 12$

If the mean of the following data is 15, find p.
x : 5 10 15 20 25
f : 6 p 6 10 5

Given that

mean of the following Data is

$15$ i.e mean

$=15$ .........(1)

i.e mean

$15$

$x$	5	10	15	20	25
$f$	6	P	6	10	5

We know

$mean=\dfrac{\sum f_x}{\sum f}$

we have to prepare a table

$x$	$f$	$f_x$
5	6	30
10	P	10P
15	6	90
20	10	200
25	5	125

$\displaystyle \sum f=27+P$

$\displaystyle \sum f_x=445+10P$

$\therefore mean=\dfrac{\sum f_x}{\sum f}$

$15=\dfrac{445+10P}{27+P}$ .........from(1)

$(27+P)15=445+10P$

$405+15P=445+10P$

$15P-10P=445-405$

$5P=40$

$P=\dfrac{40}{5}$

$P=8$

$\therefore$ value of

$P$ is

$8$

The weights (in kg ) of 15 students are : 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42,Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median. If the first answer is 35 then the second answer will be_________

given that

the weights (in kg) of

$15$ students are

$31,35,27,29,32,43,37,41,34,41,34,28,36,44,45,42,30$

No' of observations (n) =

$15$

arrange the data is ascending order

$27,28,29,30,31,32,34,35,36,37,41,42,43,44,45$

Here observe that No' of students are odd i.e

$n=15$

Median

$=\dfrac{n+1}{2}$ th term

$=\dfrac{15+1}{n}$ th term

$=8$ th term

$\therefore median =35$

if weight

$44kg$ is replaced by

$46kg$ and

$27kg$ by

$25kg$

then new Data is

$25,28,29,30,31,32,34,35,36,37,41,42,43,46,45$

Arrange data the ascending order

$25,28,29,30,31,32,34,35,36,37,41,42,43,45,46$

Here observe that no's of students are odd i.e

$n=15$

new median is

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

$=\dfrac{15+1}{2}$ th term

$=8$ th term

$=35$

$\therefore$ new median

$=35kg$

Find the median of the following data
92, 35, 67, 85, 72, 81, 56, 51, 42, 69

$92,35,67,85,72,81,56,51,42,69$

No' of observations (n) =

$10$

first of all we arrange the observations in an ascending order

$35,42,51,56,67,69,72,81,85,92$

Here observe that No' of observations (n) is even so

Median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{\dfrac{10}{2}\text{th term}+(\dfrac{10}{2}+1)\text{th term}}{2}$

$=\dfrac{5 \text{th term}+6\text{th term}}{2}$

$=\dfrac{67+69}{2}$

$=\dfrac{136}{2}$

Median =

$68$

Find the median of the following data : 41, 43, 127, 99, 61, 92, 71, 58,If 58 is replaced by 85, what will be the new median? If the first answer is 61, then the second answer will be______

given that

$41,43,127,99,61,92,71,58,57$

arrange the data in ascending order

$41,43,57,58,61,71,92,99,127$

Here observe that No' of observations (n) are odd i.e

$n=9$

So median is

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

Median

$=\dfrac{n+1}{2}$ th term

$=\dfrac{9+1}{n}$ th term

$=5$ th term

$=61$

$58$ is replaced by

$85$

then new data is

$41,43,57,85,61,71,92,99,127$

arrange the data in ascending order

$41,43,57,61,71,85,92,99,127$

Here observe that no's observation are odd i.e

$n=9$

median is

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

Median

$=\dfrac{9+1}{2}$ th term

$=5$ th term

$=71$

$\therefore$ new median

$=71$

Find the median of the following data
$133,73,89,108,94,104,94,85,100,120$

Given data

$133, 73, 89, 108, 94, 104, 94, 85, 100, 120$

First of all arrange the data in ascending order

$73, 85, 89, 94, 94, 100, 104, 108, 120, 133$

Here observe that no. of observations in data

$n=10$

so median is average of

$\dfrac{n}{2}^{th}$ term and

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

Median

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

$=\left(\dfrac{10}{2}^{th} term +\left(\dfrac{10}{2}+1\right)^{th} term\right)/2$

$=\left(5^{th} term +6^{th} term\right)/2$

$=\dfrac{94+100}{2}=\dfrac{194}{2}=97$

$\therefore$ Median

$=97$ .

Numbers $50,42,35,2x+10,2x-8,12,11,8,6$ are written in descending order and their median is 25, find $x.$

Given numbers

$50 , 42, 35 , 2x + 10, 2x - 8 , 12 , 11, 8, 6$ are in descending order

Median is

$25$ __(I)

Here we observe that

No. of observation is odd i.e.

$n = 9$

So median is

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

from (I)

$5^{th}$ term from data is

$2x - 8$

Median

$= \left(\dfrac{9+1}{2}\right)^{th}$ term

$25 = 5^{th}$ term

$25 = 2x - 8$

$2x = 25 + 8$

$2x = 33$

$x = \dfrac{33}{2}$

$x = 16.5$

$x$ value is

$16.5$

Find the median of the following data: $41,43,127,99,61,92,71,58,57,$ If 58 is replaced by 85, what will be the new median? If the old median was 61, then what's the new median?

Given data

$41, 43, 127, 99, 61, 92, 71, 58, 57$

Arrange the data in ascending order

$41, 43, 57, 58, 61, 71, 92, 99, 127$

Here no. of observations odd i.e.,

$n=9$ , so median is

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

Median

$=\dfrac{9+1}{2}th$ term

$=5^{th}$ term

$=61$

$58$ is replaced by

$85$

then new data is

$41, 43, 57, 85, 61, 71, 92, 99, 127$

Arrange the data in ascending order

$41, 43, 57, 61, 71, 85, 92, 99, 127$

Here no. of observations odd i.e.,

$n=9$ , so

median is

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

Median

$=\dfrac{9+1}{2}th$ term

$=5^{th}$ term

$=71$

New median

$=71$ .

Find the median of the following data
$31,38,27,28,36,25,35,40$

Given that the data

$31, 38, 27, 28, 36, 25, 35, 40$

First of all arrange the data ascending order

$25, 27, 28, 31, 35, 36, 38, 40$

Here observe that no. of observations n

$=8$ (even)

so median is average of

$\dfrac{n}{2}^{th}$ term &

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

Median

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

$=\dfrac{\dfrac{8}{2}^{th} term +\left(\dfrac{8}{2}+1\right)^{th} term}{2}$

$=\dfrac{4^{th} term +5^{th} term}{2}$

$=\dfrac{31+35}{2}$

$=\dfrac{66}{2}$

Median

$=33$ .

Find the median of the following observations: $46,64,87,41,58,77,35,90,55,92,33$ . If 92 is replaced by 99 and 41 by 43 in the above data, find the new median? If the first answer is 58, then the second answer is ______

Given that

the observations are

$46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33$

No. of observations(n)

$=11$

Arrange the data in ascending order

$33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92$

Here observe that no. of observations are odd i.e.,

$n=11$

So median is

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

Median

$=\dfrac{n+1}{2}$ th term

$=\dfrac{11+1}{2}$ th term

$=6^{th}$ term

$=58$

$\therefore$ Median

$=58$

and

$92$ is replaced by

$99$

and

$41$ by

$43$ then

new data observation are

$33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99$ ........

$(1)$

Here observe that no. of observations are odd i.e,

$n=1$

so new median is

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

$=\dfrac{n+1}{2}th$ term

$=\dfrac{11+1}{2}$ th term

$=\dfrac{12}{2}$ th term

$=6^{th}$ term (from

$(1)$ )

$=58$

$\therefore$ New median

$=58$ .

Find the median of the following data
$92,35,67,85,72,81,56,51,42,69$

Given data

$92, 35, 67, 85, 72, 81, 56, 51, 42, 69$

No. of observations

$(n)=10$

First of all we arrange the observations in an ascending order

$35, 42, 51, 56, 67, 69, 72, 81, 85, 92$

Here observe that no. of observations 'n' is even so

Median is average of

$\dfrac{n}{2}$ th term and

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

Median

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

$=\dfrac{10/2^{th} term+\left(\dfrac{10}{2}+1\right)^{th}term}{2}$

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

$=\dfrac{67+69}{2}$

$=\dfrac{136}{2}$

Median

$=68$ .

Find the median of the following data
$25,34,31,23,22,26,35,29,20,32$

Given data

$25, 34, 31, 23, 22, 26, 35, 29, 20, 32$

No. of observations(n)

$=10$

First of all we arrange the observations in an ascending order

$20, 22, 23, 25, 26, 29, 31, 32, 34, 35$

Here observe that no. of observation(n) is even so

Median is average of

$\dfrac{n}{2}^{th}$ term &

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

Median

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

$=\dfrac{\dfrac{10}{2}^{th} term +\left(\dfrac{10}{2}+1\right)^{th}term}{2}$

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

$=\dfrac{26+29}{2}$

$=\dfrac{55}{2}$

Median

$=27.5$ .

Find the median of the following data
$12,17,3,14,5,8,7,15$

Given data

$12, 17, 3, 14, 5, 8, 7, 15$

No. of observations

$(n)=8$

First of all we arrange the observations in an ascending order

$3, 5, 7, 8, 12, 14, 15, 17$

here observe that no. of observations(n) is even so

Median is average of

$\dfrac{n}{2}th$ term and

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

Median

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

$=(8/2^{th} term +\left(\dfrac{8}{2}+1\right)^{th}term)/2$

$=(4^{th} term +5^{th} term)/2$

$=(8+10)/2=\dfrac{20}{2}=10$

$\therefore$ Median

$=10$ .

Find the median of the following data
$41,43,127,99,71,92,71,58,57$

Given data

$41, 43, 127, 99, 71, 92, 71, 58, 57$

No. of observations

$n=9$

First of all we arrange the observations in an ascending orders

$41, 43, 57, 58, 71, 71, 92, 99, 127$

Here observe that no. of observation(n) is odd so

Median is the

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

Median

$=\left(\dfrac{9+1}{2}\right)^{th}$ term

$=\dfrac{10}{2}^{th}$ term

$=5^{th}$ term

Median

$=71$ .

Find the median of the following data
$15,6,16,8,22,21,9,18,25$

Given data

$15, 6, 16, 8, 22, 21, 9, 18, 25$

No. of observations

$(n)=9$

First of all we arrange the observations in an ascending order

$6, 8, 9, 15, 16, 21, 22, 25$

Here observe that no. of observations(n) is odd so

median is the

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

Median

$=\left(\dfrac{9+1}{2}\right)^{th}$ term

$=5^{th}$ term

Median

$=16$ .

The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x: $29,32,48,50,x,x+2,72,78,84,95$

Given that the observations have been arranged in ascending order

$29, 32, 48, 50, x, x+2, 72, 78, 84, 95$

and median

$=63$ .........

$(1)$

Here observe that no. of observations are

$(n)=10$

So median is average of

$\dfrac{n}{2}th$ term and

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

i.e., median

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

$=\dfrac{\dfrac{10}{2}^{th} term +\left(\dfrac{10}{2}+1\right)^{th} term}{2}$

Median

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

$63=\dfrac{x+(x+2)}{2}$

$2x+2=126$

$2x=126-2$

$2x=124$

$x=\dfrac{124}{2}$

$x=62$

$\therefore$ value of

$x=62$ .

The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are give as follows:
Speed (km/h) 85-100 100-115 115-130 130-145
Number of players 11 9 8 5
Calculate the median bowling speed.

1960888_1797627_ans_79a8ed92ecbf48498ecc2f4bc8d17c2c.jpg

Find the value of median from the data given below:

Weight (kg)

No. of Students

93-97

3

98-102

5

103-107

12

108-112

17

113-117

14

118-122

6

123-127

3

128-132

1

2160068_2033401_ans_25f2c5dc2bc04660b989845f9b79e3f4.PNG

The mean weight of 8 numbers is 15 kg. If each number is multiplied by 2, what will be the new mean weight? (in kg)

Given that

mean weight of

$8$ numbers is

$15$ kg

$\therefore n=8$ and mean

$=15$ kg

We know mean

$=\dfrac{sum \ of \ weights}{total \ no. \ of \ weights}$

$15=\dfrac{sum \ of \ weights}{8}$

sum of weights

$=15\times 8=120$

Given if each number is multiplied by

$2$

After multiplication of

$2$ in each number

then their new sum of weights

$=120\times 2=240$

$\therefore$ New mean

$=\dfrac{new \ sum \ of \ weights}{no. \ of \ weights}$

$=\dfrac{240}{8}$

New mean

$=30$

Hence

New mean weight is

$30$ kg.

The scores in mathematics test (out of 25) of 15 students are as follows:
$19,25,23,20,9,20,15,10,5,16,25,20,24,12,20$
Find the mode and median of this data. Are they same?

2039859_1677451_ans_272713e65990411e94c612fc1a21ad8a.png

Two Samples of size 6 & 6 respectively have means 350 & 138 and standard deviation 5 andFind the mean and standard deviation of the combined sample of size 250.

Solution is given by:
2158897_2032636_ans_3c3093049e3e4968836b4aa24fac50ce.jpg

No. of children ( $x$ )	$0$	$1$	$2$	$3$	$4$	$5$
No. of families ( $f$ )	$10$	$21$	$55$	$42$	$15$	$7$

Class interval:	50-70	70-90	90-110	110-130	130-150	150-170
Frequency:	$18$	$12$	$13$	$27$	$8$	$22$

Weekly wages (in Rs.)	$50-55$	$55-60$	$60-65$	$65-70$	$70-75$	$75-80$	$80-85$	$85-90$
No. of workers	$5$	$20$	$10$	$10$	$9$	$6$	$12$	$8$

Number of days	$0-6$	$6-10$	$10-14$	$14-20$	$20-28$	$28-38$	$38-40$
Number of students	$11$	$10$	$7$	$4$	$4$	$3$	$1$

Class	10-14	15-19	20-24	25-29	30-34
Frequency	$15$	$110$	$135$	$115$	$25$

Age(in years)	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$
Number of persons	$105$	$222$	$220$	$138$	$102$	$113$	$100$

Length $(mm)$	$109-117$	$118-126$	$127-135$	$136-144$	$145-153$	$154-162$	$163-171$
No of leaves	$4$	$6$	$14$	$13$	$6$	$4$	$3$

Daily expenses (in Rs.)	100-150	150-200	200-250	250-300	300-350
No. of households	4	5	12	2	2

Daily expenditure (in Rs.)	$100-150$	$150-200$	$200-250$	$250-300$	$300-350$
No. of households	$4$	$5$	$12$	$2$	$2$

Marks	No of students
0-10	5
10-20	10
20-30	11
30-40	20
40-50	27
50-60	38
60-70	40
70-80	09
80-90	14
90-100	6

Salary (in Rs}	Numbers of workers
$5000$	$22$
$6000$	$18$
$7000$	$15$
$8000$	$10$
$9000$	$8$
$10000$	$7$

Income (thousand Rs)	$20-30$	$30-40$	$40-50$	$50-60$
Farmer	$10$	$11$	$15$	$16$

Mark obtained	25-35	35-45	45-55	55-65	65-75
Number of students	7	31	33	17	11

Toll (Rupees)	$300-400$	$400-500$	$500-600$	$600-700$	$700-800$
No. of vehicles	$80$	$110$	$120$	$70$	$40$

Amount	Number of students
$0-10$	$3$
$10-20$	$4$
$20-30$	$5$
$30-40$	$2$
$40-50$	$1$

marks	11-20	21-30	31-40	41-50	51-60	61-70	71-80
number of students	2	6	10	12	9	7	4

Class Marks	$10$	$20$	$30$	$40$	$50$	$60$
No of students	$3$	$1$	$2$	$7$	$8$	$5$

Marks	No. of students
Above 0	80
Above 10	78
Above 20	71
Above 30	65
Above 40	53
Above 50	45
Above 60	25
Above 70	14
Above 80	10
Above 90	8
Above 100	0

No. of wickets	No. of bowlers
20-60	7
60-100	5
100-150	16
150-250	12
250-350	2
350-450	3

Marks	No. of students
Less than 10	0
Less than 30	10
Less than 50	25
Less than 70	43
Less than 90	65
Less than 110	87
Less than 130	96
Less than 150	100

Measures Of Central Tendency - Class 11 Commerce Economics - Extra Questions

Find the mean of following distribution with step-deviation method to the nearest integerClass10 - 1515 - 2020 - 2525 - 3030 - 3535 - 40Frequency5681263

Find the mode of the given set of numbers-1,1,2,4,3,2,1,2,2,41, 1, 2, 4, 3, 2, 1, 2, 2, 4

Following are the margins of victory in the football matches of a league.1,3,2,5,1,4,6,2,5,2,2,2,4,1,2,3,1,1,2,3,2,6,4,2,1,1,4,2,1,5,3,3,2,3,2,4,2,1,2,31, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 1, 1, 2, 3, 2, 6, 4, 2, 1, 1, 4, 2, 1, 5, 3, 3, 2, 3, 2, 4, 2, 1, 2, 3, find the mode of this data.

A class teacher has the following absentee record of 4040 students of a class for the whole term. Find the mean number of days a student was absent.

Find the median of the data: 15,22,9,20,6,18,11,25,1415, 22, 9, 20, 6, 18, 11, 25, 14

Find the mode for the following data:22,36,18,22,20,34,22,42,46,4222, 36, 18, 22, 20, 34, 22, 42, 46,42

Find the median of the data: 32,30,28,31,22,26,27,2132, 30, 28, 31, 22, 26, 27, 21

Find the mode for the following data:4,3,1,5,3,7,9,64, 3, 1, 5, 3, 7, 9, 6

Find the median of the data: 26,31,33,37,43,8,26,3326, 31, 33, 37, 43, 8, 26, 33

The following numbers are the sizes of shoes sold by a shop in a particular day. Find the mode.6,7,8,9,10,6,7,10,7,6,7,9,7,66, 7, 8, 9, 10, 6, 7, 10, 7, 6, 7, 9, 7, 6

Find the mode of the data: 5,3,3,5,7,6,3,4,3,5,8,55, 3, 3, 5, 7, 6, 3, 4, 3, 5, 8, 5

Find the mode for the data: 15,20,22,25,30,20,15,20,12,2015, 20, 22, 25, 30, 20, 15, 20, 12, 20

Calculate mean (¯X)(\overline{X}), when ∑fixi=100\sum f_ix_i=100 and ∑fi=20\sum f_i=20.

Find the mode of 5,6,9,6,12,3,6,11,6,75, 6, 9, 6, 12, 3, 6, 11, 6, 7.

Find the mode of 22,25,21,22,29,25,34,37,30,22,29,2522, 25, 21, 22, 29, 25, 34, 37, 30, 22, 29, 25.

If \sum f_{i}x_{i} = 75\sum f_{i}x_{i} = 75 and \sum f_{i} = 15\sum f_{i} = 15, then find the mean \overline {x}\overline {x}

Calculate the arithmetic mean of given frequency distribution table.Class-interval0-100-1010-2010-2020-3020-3030-4030-4040-5040-5050-6050-60Frequency77141428282626161699

Find the mode of following numbers 6, 3, 2, 6, 5, 66, 3, 2, 6, 5, 6 and 88.

The following table gives the number of children of 150150 families in a village.No. of children (xx)001122334455No. of families (ff)1010212155554242151577Find the average number of children per family.

If the mean of the following data is 1515, find pp.xx:551010151520202525ff:66pp66101055

Find the mean of the following frequency distribution:Class interval:50-7070-9090-110110-130130-150150-170Frequency:1818121213132727882222

Find the missing value of pp for the following distribution whose mean is 12.5812.58.xx:558810101212pp20202525ff:2255882222774422

Find the mean of the following data:xx1919212123232525272729293131ff1313151516161818161615151313

If the mean of the following data is 20.620.6. Find the value of pp.xx10101515pp25253535ff33101025257755

Find the missing frequency (pp) for the following distribution whose mean is 7.687.68xx:3355779911111313ff:66881515pp8844

The arithmetic mean of the following data is 1414. Find the value of kk.xx:551010151520202525ff:77kk884455

Find the mean median and mode of 20,15,25,7,25,17,520,15,25,7,25,17,5

Find a if the mean mark is 11.Mark01234No. of studentsa4321

Calculate mean with step Deviation method from the following data.Marks0-100-1010-2010-2020-3020-3030-4030-4040-5040-50No. of student4466101020201010

Find the mode of the following data:120, 110, 130, 110, 120, 140, 130, 120, 140, 120.120, 110, 130, 110, 120, 140, 130, 120, 140, 120.

Following are the lives in hours of 1515 pieces of the components of aircraft engine. Find the median.715,724,725,710,729,745,694,699,696,712,734,728,716,705,719715,724,725,710,729,745,694,699,696,712,734,728,716,705,719

XC.F.Less than 101210-201920-302830-403040-5040Mean= ?Median = ?

The arithmetic mean for the following distribution is 5353. Find the value of pp.ClassClass0-200-2020-4020-4040-6040-6060-8060-8080-10080-100FrequencyFrequency151518182121pp1717

If the median of the scores \frac{x}{3},\frac{x}{2},\frac{x}{4},\frac{{2x}}{9},x\left( {x > 0} \right)\frac{x}{3},\frac{x}{2},\frac{x}{4},\frac{{2x}}{9},x\left( {x > 0} \right) is 3, then find 'x'.

Find the mean of the following frequency distribution using steps deviation method.C.I0-100100-200200-30030-400400-500500-600600-700F1015252510105

Median of the data \dfrac{x}{3}, \dfrac{x}{6}, \dfrac{x}{9}. \dfrac{x}{2}, \dfrac{x}{8}, \dfrac{x}{4}\dfrac{x}{3}, \dfrac{x}{6}, \dfrac{x}{9}. \dfrac{x}{2}, \dfrac{x}{8}, \dfrac{x}{4} is 55, then find the value of 'x'.

Calculate the mean of the following data, using direct method:Class25-3535-4545-5555-6565-75Frequency6108124

Calculate the mean of the following frequency distribution.Class90-10090-100100-110100-110110-120110-120120-130120-130130-140130-140Frequency1818787880801001007272

If the mean of a data is 20.220.2. Find the value of m.Observation10101515202025253030Frequency6688m101066

Find the mean of the first five whole numbers.

The data given below shows the number of goals scored by a football team over 3030 games 4 3 2 1 4 0 1 0 5 1 0 0 1 4 3 2 1 0 1 3 4 2 0 1 0 3 2 1 1 0 Find the mode of the above data.

Find the mean of each of the data. 8,-2,9,6,13,17,128,-2,9,6,13,17,12

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 mean.

A class teacher has the following absentee record of 4040 students of a class for the whole term. Find the mean number of days a student was absent.Number of days0-60-66-106-1010-1410-1414-2014-2020-2820-2828-3828-3838-4038-40Number of students111110107744443311

Find mean of the following data by using step-deviation methodClass10-1415-1920-2425-2930-34Frequency15151101101351351151152525

Find the median of the data: 3.3,3.5,3.1,3.7,3.2,3.83.3,3.5,3.1,3.7,3.2,3.8

Find the median of the following data: 14, 16, 14, 12, 11, 14, 16, 12, 10, 814, 16, 14, 12, 11, 14, 16, 12, 10, 8.

Find the median of first eight (8)(8) positive composite numbers.

The following observations have been arranged in ascending order. if the median of these observations is 58, find the value of xx. 2424, 2727, 4343, 4848, x-1x-1, x+3x+3, 68 68, 7373, 8080, 9090.

Calculate Mean, Median and Mode for the following data. Which of these three measures if minimum? Marks s(out of 15)774410109915151212779999

Find the median of 6464 and 225225.

Find the missing frequency f_{1}f_{1} and f_{2}f_{2} in the table given below, it is being given that the mean of the given frequency distribution is 5050.Class0-200-2020-4020-4040-6040-6060-8060-8080-10080-100TotalFrequency1717f_{1}f_{1}3232f_{2}f_{2}1919120120

The lengths of 5050 leaves of a plant are measured corrcct he nearest milimetre and the data obtained is represent in the following table.Length (mm)(mm)109-117109-117118-126118-126127-135127-135136-144136-144145-153145-153154-162154-162163-171163-171No of leaves446614141313664433

There are 10 observations arranged in ascending order as given below.45,47,50,52,x,x+2,60,62,63,74.45,47,50,52,x,x+2,60,62,63,74.The median of these observation is 5353. Find the value of xx

(a)The class interval having highest frequency is-(b)The class-mark of class 50-60 is-(c)The size of each class intervals is-

Find the mode of 90,55,67,55,75,40,35,55,9590,55,67,55,75,40,35,55,95

Find the median of the data 13,16,12,14,19,12,14,1313,16,12,14,19,12,14,13.

The table below- shows the daily expenditure on food of 2525 households in a locality.Daily expenses (in Rs.)100-150150-200200-250250-300300-350No. of households451222Find the mean daily expenses on food by a suitable method

Find the mean of the following distribution by step-deviation method:Daily expenditure (in Rs.)100-150100-150150-200150-200200-250200-250250-300250-300300-350300-350No. of households445512122222

Find mean:The marks obtained by 200 students are given below :Marks No of students0-10510-201020-301130-402040-502750-603860-704070-800980-901490-1006

Find the mean salary of 8080 workers of a factory from the following tables:Salary (in Rs}Numbers of workers5000500022226000600018187000700015158000800010109000900088100001000077

The following table draws the income of finance of a grape season find the mean of their income.Income (thousand Rs)20-3020-3030-4030-4040-5040-5050-6050-60Farmer1010111115151616

Mark obtained 25-3535-4545-5555-6565-75Number of students731331711 Find mean

From one footwear shop, 1212 pairs of chappals were sold. The sizes of these chappals are given below. Find their mode: 7,8,6,7,7,5,9,7,6,7,8,77,8,6,7,7,5,9,7,6,7,8,7

In the following table, the toll paid by drives and the number of vehicles is shown. Find the mean of the toll by assumed mean method.Toll (Rupees)300-400300-400400-500400-500500-600500-600600-700600-700700-800700-800No. of vehicles 808011011012012070704040

The marks obtained in an examination of 400400 marks is given in the following frequency distribution table. Find the mean of the distribution.Marks200-240240-280280-320320-360360-400Frequency1020302416

Calculate the mean from the following data:

Find the median of :(i) 5,7,9,11,15,17,2,23 and 19

In the following table age distribution of patient in the city, suffering from some diseases is given that the average age ( in years) per patient.AgeNo. of patient5-14615-241124-242135-442345-545455-605

Find the mode of the given set of numbers 2,3,5,3,4,7,3,2,1,7,32,3,5,3,4,7,3,2,1,7,3.

The amount collected by 1515 students for flood relief find is given below. Answer the following. Amount Number of students0-100-103310-2010-204420-3020-305530-4030-402240-5040-5011(ii)Find the mean amount collected by a student.

Find the mode of the following datai) 12,14,12,16,15,13,14,18,19,12,14,15,16,15,16,16,12,14,12,16,15,13,14,18,19,12,14,15,16,15,16,16,ii) 15,17,13,16,16,15,13,15,17,15,14,15,13,15,14,15,17,13,16,16,15,13,15,17,15,14,15,13,15,14,

Mean for the following frequency distribution table.0-55-1010-1515-2020-2525-305391085

Find the median of 72, 66, 87, 92, 63, 78.

Find the mean of the following distribution 300-400400-500500-600600-700700-800800-900No. of girls4189763

A sequence, a,ax,ax^{2}, ......, ax^{n}a,ax,ax^{2}, ......, ax^{n}, has odd number of terms. Find its median.

Find the mode of the data : (i) 10,8,4,7,8,11,15,8,6,8,(ii) 27,23,39,18,27,21,27,27,40,36,27

Calculate the mean of the distribution given below using the shortcut method.marks11-2021-3031-4041-5051-6061-7071-80number of students261012974

The arithmetic mean of following data is 1717. find the value of PP. Term(x)10152025Frequency(f)710P6

Find the mean of following distribution with step-deviation method to the nearest integer
Class 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40
Frequency 5 6 8 12 6 3

Find the mode of the given set of numbers-
$1, 1, 2, 4, 3, 2, 1, 2, 2, 4$

Following are the margins of victory in the football matches of a league.
$1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 1, 1, 2, 3, 2, 6, 4, 2, 1, 1, 4, 2, 1, 5, 3, 3, 2, 3, 2, 4, 2, 1, 2, 3$ , find the mode of this data.

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.

Find the median of the data: $15, 22, 9, 20, 6, 18, 11, 25, 14$

Find the mode for the following data:
$22, 36, 18, 22, 20, 34, 22, 42, 46,42$

Find the median of the data: $32, 30, 28, 31, 22, 26, 27, 21$

Find the mode for the following data:
$4, 3, 1, 5, 3, 7, 9, 6$

Find the median of the data: $26, 31, 33, 37, 43, 8, 26, 33$

The following numbers are the sizes of shoes sold by a shop in a particular day. Find the mode.
$6, 7, 8, 9, 10, 6, 7, 10, 7, 6, 7, 9, 7, 6$

Find the mode of the data: $5, 3, 3, 5, 7, 6, 3, 4, 3, 5, 8, 5$

Find the mode for the data: $15, 20, 22, 25, 30, 20, 15, 20, 12, 20$

Calculate mean $(\overline{X})$ , when $\sum f_ix_i=100$ and $\sum f_i=20$ .

Find the mode of $5, 6, 9, 6, 12, 3, 6, 11, 6, 7$ .

Find the mode of $22, 25, 21, 22, 29, 25, 34, 37, 30, 22, 29, 25$ .

If $\sum f_{i}x_{i} = 75$ and $\sum f_{i} = 15$ , then find the mean $\overline {x}$

Calculate the arithmetic mean of given frequency distribution table.
Class-interval $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency $7$ $14$ $28$ $26$ $16$ $9$

Find the mode of following numbers $6, 3, 2, 6, 5, 6$ and $8$ .

The following table gives the number of children of $150$ families in a village.
No. of children ( $x$ ) $0$ $1$ $2$ $3$ $4$ $5$
No. of families ( $f$ ) $10$ $21$ $55$ $42$ $15$ $7$
Find the average number of children per family.

If the mean of the following data is $15$ , find $p$ .
$x$ : $5$ $10$ $15$ $20$ $25$
$f$ : $6$ $p$ $6$ $10$ $5$

Find the mean of the following frequency distribution:

Class interval: 50-70 70-90 90-110 110-130 130-150 150-170
Frequency: $18$ $12$ $13$ $27$ $8$ $22$

Find the missing value of $p$ for the following distribution whose mean is $12.58$ .
$x$ : $5$ $8$ $10$ $12$ $p$ $20$ $25$
$f$ : $2$ $5$ $8$ $22$ $7$ $4$ $2$

Find the mean of the following data:
$x$ $19$ $21$ $23$ $25$ $27$ $29$ $31$
$f$ $13$ $15$ $16$ $18$ $16$ $15$ $13$

If the mean of the following data is $20.6$ . Find the value of $p$ .
$x$ $10$ $15$ $p$ $25$ $35$
$f$ $3$ $10$ $25$ $7$ $5$

Find the missing frequency ( $p$ ) for the following distribution whose mean is $7.68$
$x$ : $3$ $5$ $7$ $9$ $11$ $13$
$f$ : $6$ $8$ $15$ $p$ $8$ $4$

The arithmetic mean of the following data is $14$ . Find the value of $k$ .
$x$ : $5$ $10$ $15$ $20$ $25$
$f$ : $7$ $k$ $8$ $4$ $5$

Find the mean median and mode of $20,15,25,7,25,17,5$

Find a if the mean mark is $1$ .
Mark 0 1 2 3 4
No. of students a 4 3 2 1

Calculate mean with step Deviation method from the following data.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
No. of student $4$ $6$ $10$ $20$ $10$

Find the mode of the following data:
$120, 110, 130, 110, 120, 140, 130, 120, 140, 120.$

Following are the lives in hours of $15$ pieces of the components of aircraft engine. Find the median.
$715,724,725,710,729,745,694,699,696,712,734,728,716,705,719$

X C.F.
Less than 10 12
10-20 19
20-30 28
30-40 30
40-50 40
Mean= ?
Median = ?

The arithmetic mean for the following distribution is $53$ . Find the value of $p$ .
$Class$ $0-20$ $20-40$ $40-60$ $60-80$ $80-100$
$Frequency$ $15$ $18$ $21$ $p$ $17$

If the median of the scores $\frac{x}{3},\frac{x}{2},\frac{x}{4},\frac{{2x}}{9},x\left( {x > 0} \right)$ is 3, then find 'x'.

Find the mean of the following frequency distribution using steps deviation method.
C.I 0-100 100-200 200-300 30-400 400-500 500-600 600-700
F 10 15 25 25 10 10 5

Median of the data $\dfrac{x}{3}, \dfrac{x}{6}, \dfrac{x}{9}. \dfrac{x}{2}, \dfrac{x}{8}, \dfrac{x}{4}$ is $5$ , then find the value of 'x'.

Calculate the mean of the following data, using direct method:
Class 25-35 35-45 45-55 55-65 65-75
Frequency 6 10 8 12 4

Calculate the mean of the following frequency distribution.
Class $90-100$ $100-110$ $110-120$ $120-130$ $130-140$
Frequency $18$ $78$ $80$ $100$ $72$

If the mean of a data is $20.2$ . Find the value of m.
Observation $10$ $15$ $20$ $25$ $30$
Frequency $6$ $8$ m $10$ $6$

The data given below shows the number of goals scored by a football team over $30$ games

4

3

2

1

4

0

1

0

5

1

0

0

1

4

3

2

1

0

1

3

4

2

0

1

0

3

2

1

1

0

Find the mode of the above data.

Find the mean of each of the data. $8,-2,9,6,13,17,12$

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 mean.

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 $11$ $10$ $7$ $4$ $4$ $3$ $1$

Find mean of the following data by using step-deviation method
Class 10-14 15-19 20-24 25-29 30-34
Frequency $15$ $110$ $135$ $115$ $25$

Find the median of the data: $3.3,3.5,3.1,3.7,3.2,3.8$

Find the median of the following data: $14, 16, 14, 12, 11, 14, 16, 12, 10, 8$ .

Find the median of first eight $(8)$ positive composite numbers.

The following observations have been arranged in ascending order. if the median of these observations is 58, find the value of $x$ .
$24$ , $27$ , $43$ , $48$ , $x-1$ , $x+3$ , $68$ , $73$ , $80$ , $90$ .

Calculate Mean, Median and Mode for the following data. Which of these three measures if minimum?
Marks s(out of 15) $7$ $4$ $10$ $9$ $15$ $12$ $7$ $9$ $9$

Find the median of $64$ and $225$ .

Find the missing frequency $f_{1}$ and $f_{2}$ in the table given below, it is being given that the mean of the given frequency distribution is $50$ .
Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ Total
Frequency $17$ $f_{1}$ $32$ $f_{2}$ $19$ $120$

The lengths of $50$ leaves of a plant are measured corrcct he nearest milimetre and the data obtained is represent in the following table.
Length $(mm)$ $109-117$ $118-126$ $127-135$ $136-144$ $145-153$ $154-162$ $163-171$
No of leaves $4$ $6$ $14$ $13$ $6$ $4$ $3$

There are 10 observations arranged in ascending order as given below.
$45,47,50,52,x,x+2,60,62,63,74.$
The median of these observation is $53$ . Find the value of $x$

(a)The class interval having highest frequency is-
(b)The class-mark of class 50-60 is-
(c)The size of each class intervals is-

Find the mode of $90,55,67,55,75,40,35,55,95$

Find the median of the data $13,16,12,14,19,12,14,13$ .

The table below- shows the daily expenditure on food of $25$ households in a locality.
Daily expenses (in Rs.) 100-150 150-200 200-250 250-300 300-350
No. of households 4 5 12 2 2
Find the mean daily expenses on food by a suitable method

Find the mean of the following distribution by step-deviation method:
Daily expenditure (in Rs.) $100-150$ $150-200$ $200-250$ $250-300$ $300-350$
No. of households $4$ $5$ $12$ $2$ $2$

Find mean:
The marks obtained by 200 students are given below :
Marks No of students
0-10 5
10-20 10
20-30 11
30-40 20
40-50 27
50-60 38
60-70 40
70-80 09
80-90 14
90-100 6

Find the mean salary of $80$ workers of a factory from the following tables:
Salary (in Rs} Numbers of workers
$5000$ $22$
$6000$ $18$
$7000$ $15$
$8000$ $10$
$9000$ $8$
$10000$ $7$

The following table draws the income of finance of a grape season find the mean of their income.
Income (thousand Rs) $20-30$ $30-40$ $40-50$ $50-60$
Farmer $10$ $11$ $15$ $16$

Mark obtained 25-35 35-45 45-55 55-65 65-75
Number of students 7 31 33 17 11
Find mean

From one footwear shop, $12$ pairs of chappals were sold. The sizes of these chappals are given below. Find their mode: $7,8,6,7,7,5,9,7,6,7,8,7$

In the following table, the toll paid by drives and the number of vehicles is shown. Find the mean of the toll by assumed mean method.
Toll (Rupees) $300-400$ $400-500$ $500-600$ $600-700$ $700-800$
No. of vehicles $80$ $110$ $120$ $70$ $40$

The marks obtained in an examination of $400$ marks is given in the following frequency distribution table. Find the mean of the distribution.
Marks 200-240 240-280 280-320 320-360 360-400
Frequency 10 20 30 24 16

Find the median of :
(i) 5,7,9,11,15,17,2,23 and 19

In the following table age distribution of patient in the city, suffering from some diseases is given that the average age ( in years) per patient.
Age No. of patient
5-14 6
15-24 11
24-24 21
35-44 23
45-54 54
55-60 5

Find the mode of the given set of numbers $2,3,5,3,4,7,3,2,1,7,3$ .

The amount collected by $15$ students for flood relief find is given below. Answer the following.
Amount Number of students
$0-10$ $3$
$10-20$ $4$
$20-30$ $5$
$30-40$ $2$
$40-50$ $1$
(ii)Find the mean amount collected by a student.

Find the mode of the following data
i) $12,14,12,16,15,13,14,18,19,12,14,15,16,15,16,16,$
ii) $15,17,13,16,16,15,13,15,17,15,14,15,13,15,14,$

Mean for the following frequency distribution table.
0-5 5-10 10-15 15-20 20-25 25-30
5 3 9 10 8 5

Find the mean of the following distribution
300-400 400-500 500-600 600-700 700-800 800-900
No. of girls 4 18 9 7 6 3

A sequence, $a,ax,ax^{2}, ......, ax^{n}$ , has odd number of terms. Find its median.

Find the mode of the data :
(i) 10,8,4,7,8,11,15,8,6,8,
(ii) 27,23,39,18,27,21,27,27,40,36,27

Calculate the mean of the distribution given below using the shortcut method.
marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80
number of students 2 6 10 12 9 7 4

The arithmetic mean of following data is $17$ . find the value of $P$ .
Term(x) 10 15 20 25
Frequency(f) 7 10 P 6

Following observations have been written in ascending order. If median of the data is $22$ then find the value of $x$ .
$11, 12, 14, 16, 18, x+2, x+4, 30, 32, 35, 41$

If the numbers $3,6,7,10,x,15,19,20,25,28$ are in ascending order and their median is $13$ . Calculate the value of $x$ .

$15$ student secured the following marks in a test in Statistic. Find the median marks
35, 28,13,17,20,30,19,29,11,10,29,23,18, 25,17

Find the mode of the data:
10,8,4,7,8,11,15,8,6,8

Expenditure (in. Rs.)	No. of families
1000-1500	24
1500-2000	10
2000-2500	33
2500-3000	28
3000-3500	30
3500-4000	22
4000-4500	16
4500-5000	7

No. of mangoes	$50-52$	$53-55$	$56-58$	$59-61$	$62-64$
No. of boxes	$15$	$110$	$135$	$115$	$25$

Marks	No. of Students
Below 10	5
Below 20	9
Below 30	17
Below 40	29
Below 50	45
Below 60	60
Below 70	70
Below 80	78
Below 90	83
Below 100	85

Height from mean sea level	200	600	1000	1400	1800
No. of Villages	115	107	265	143	202

Category	A	B	C	D	E	F	G
Wages per day	50	60	70	80	90	100	110
No. of Workers	2	4	8	12	10	6	8

Profit per shop (in Rs)	Number of shops $\displaystyle (f_{i})$
100 - 200	10
200 - 300	18
300 - 400	20
400 - 500	26
500 - 600	30
600 - 700	28
700 - 800	18

Marks	No. of students	Marks	No. of students
0 - 10	2	50 - 60	12
10 - 20	3	60 - 70	14
20 - 30	4	70 - 80	4
30 - 40	8	80 - 90	4
49 - 50	16	90 - 100	3

Cost of Living Index	Number of weeks
140 - 150	5
150 - 160	10
160 - 170	20
170 - 180	9
180 - 190	6
190 - 200	2
Total	52

Class interval	Frequency
0 - 20	8
20 - 40	10
40 - 60	15
60 - 80	9
80 - 100	8

Income per day	Numbers of persons
0-100	4
100-200	8
200-300	9
300-400	10
400-500	7
500-600	5
600-700	4
700-800	3