# NCERT Solutions For Class 11 Maths Chapter 15

## NCERT Solutions Class 11 Maths Statistics Mean Median

NCERT solutions class 11 maths chapter 15 analysis of statistics is the perfect study material to learn statistics for class 11. The NCERT solutions for class 11 maths chapter 15 analysis of statistics is provided here to help students prepare in a better way. The NCERT solutions for class 11 maths chapter 15 is created according to the latest syllabus of the Central Board of Secondary Education. Check the NCERT solutions for class 11 maths chapter 15 pdf given below.

### NCERT Solutions Class 11 Maths Chapter 15 Exercises

Exercise 15.1

Q1. Calculate the mean deviation about the mean for the given data

5, 8, 9, 10, 11, 13, 14, 18

Sol:

The given data is,

5, 8, 9, 10, 11, 13, 14, 18

Mean,

x¯=5+8+9+10+11+13+14+188$\bar{x} = \frac{5 + 8 + 9 + 10 + 11 + 13 + 14 + 18}{8}$ x¯=888$\bar{x} = \frac{88}{8}$ x¯=11$\bar{x} = 11$

The deviations of the respective observations from the mean x¯$\bar{x}$,

xix¯$x_{i} – \bar{x}$ are -6, -3, -2, -1, 0, 2, 3, 7

The absolute values of the deviations,

|xix¯|$\left |x_{i} – \bar{x} \right |$ are 6, 3, 2, 1, 0 , 2, 3, 7

The required mean deviation about the mean is,

M.D.(x¯)=8i=1|xix¯|8$M.D. \;\left (\bar{x} \right ) = \frac{\sum_{i = 1}^{8} \left | x_{i} – \bar{x} \right |}{8}$ M.D.(x¯)=6+3+2+1+0+2+3+78$M.D. \;\left (\bar{x} \right ) = \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8}$

M.D.(x¯)=248$M.D. \;\left (\bar{x} \right ) = \frac{24}{8}$ = 3

Q2. Calculate the mean deviation about the mean for the given data

39, 71, 49, 41, 43, 56, 64, 47, 55, 45

Sol:

The given data is,

39, 71, 49, 41, 43, 56, 64, 47, 55, 45

Mean,

x¯=39+71+49+41+43+56+64+47+55+4510$\bar{x} = \frac{39 + 71 + 49 + 41 + 43 + 56 + 64 + 47 + 55 + 45}{10}$ x¯=51010$\bar{x} = \frac{510}{10}$ x¯=51$\bar{x} = 51$

The deviations of the respective observations from the mean x¯$\bar{x}$,

xix¯$x_{i} – \bar{x}$ are -12, 20, -2, -10, -8, 5, 13, -4, 4, -6.

The absolute values of the deviations,

|xix¯|$\left |x_{i} – \bar{x} \right |$ are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6.

The required mean deviation about the mean is,

M.D.(x¯)=10i=1|xix¯|10$M.D. \;\left (\bar{x} \right ) = \frac{\sum_{i = 1}^{10} \left | x_{i} – \bar{x} \right |}{10}$ M.D.(x¯)=12+20+2+10+8+5+13+4+4+610$M.D. \;\left (\bar{x} \right ) = \frac{12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6}{10}$

M.D.(x¯)=8410$M.D. \;\left (\bar{x} \right ) = \frac{84}{10}$ = 8.4

Q3. Calculate the mean deviation about the median for the given data:

13, 16, 11, 10, 11, 12, 17, 14, 13, 16, 18, 17

Sol:

The given data is,

13, 16, 11, 10, 11, 12, 17, 14, 13, 16, 18, 17

There are 12 observations which is even number.

Arrange the data in ascending order:

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

Median,

M=(122)thobservation+(122+1)thobservation2$M = \frac{\left (\frac{12}{2} \right )^{th} observation + \left ( \frac{12}{2} + 1 \right )^{th} observation}{2}$

= 6thobservation+7thobservation2$\frac{6 ^{th}\; observation + 7^{th}\; observation}{2}$

= 13+142$\frac{13 + 14}{2}$

= 272$\frac{27}{2}$

= 13.5

The deviations of the respective observations from the median x¯$\bar{x}$,

xiM$x_{i} – M$ are –3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations,

|xiM|$\left |x_{i} – M \right |$ are 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is,

M.D.(M)=12i=1|xiM|12$M.D. \;\left (M\right ) = \frac{\sum_{i = 1}^{12} \left | x_{i} – M \right |}{12}$

= 3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.512$\frac{3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5}{12}$

= 2812$\frac{28}{12}$

= 2.33

Q4. Calculate the mean deviation about the median for the given data:

36, 46, 60, 53, 51, 72, 42, 45, 46, 49

Sol:

The given data is,

36, 46, 60, 53, 51, 72, 42, 45, 46, 49

There are 10 observations which is even number.

Arrange the data in ascending order:

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Median,

M=(102)thobservation+(102+1)thobservation2$M = \frac{\left (\frac{10}{2} \right )^{th} observation + \left ( \frac{10}{2} + 1 \right )^{th} observation}{2}$

= 5thobservation+6thobservation2$\frac{5 ^{th}\; observation + 6^{th}\; observation}{2}$

= 46+492$\frac{46 + 49}{2}$

= 952$\frac{95}{2}$

= 47.5

The deviations of the respective observations from the median x¯$\bar{x}$,

xiM$x_{i} – M$ are –11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations,

|xiM|$\left |x_{i} – M \right |$ are 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The required mean deviation about the median is,

M.D.(M)=10i=1|xiM|10$M.D. \;\left (M\right ) = \frac{\sum_{i = 1}^{10} \left | x_{i} – M \right |}{10}$

= 11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.510$\frac{11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5}{10}$

= 7010$\frac{70}{10}$

= 7

Q5. Calculate the mean deviation about the mean for the given data

 xi$x_{i}$ 5 10 15 20 25 fi$f_{i}$ 7 4 6 3 5

Sol:

 xi$x_{i}$ xi$x_{i}$ xi$x_{i}$ |xi–x¯|$|x_{i} – \bar{x}|$ fi|xi–x¯|$f_{i} |x_{i} – \bar{x}|$ 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

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

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

Therefore,

x¯=1N5i=1fixi$\bar{x} = \frac{1}{N}\sum_{i = 1}^{5}f_{i}x_{i}$

= 125×350$\frac{1}{25} \times 350$

= 14

MD(x¯)=1N5i=1fi|xix¯|$MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{5}f_{i}\left |x_{i} – \bar{x} \right |$

= 125×158$\frac{1}{25} \times 158$

= 6.32

Q6. Calculate the mean deviation about the mean for the given data

 xi$x_{i}$ 10 10 15 20 25 fi$f_{i}$ 7 4 6 3 5

Sol:

 xi$x_{i}$ xi$x_{i}$ xi$x_{i}$ |xi–x¯|$|x_{i} – \bar{x}|$ fi|xi–x¯|$f_{i} |x_{i} – \bar{x}|$ 10 4 35 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

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

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

Therefore,

x¯=1N5i=1fixi$\bar{x} = \frac{1}{N}\sum_{i = 1}^{5}f_{i}x_{i}$

= 180×4000$\frac{1}{80} \times 4000$

= 50

MD(x¯)=1N5i=1fi|xix¯|$MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{5}f_{i}\left |x_{i} – \bar{x} \right |$

= 180×1280$\frac{1}{80} \times 1280$

= 16

Q7. Calculate the mean deviation about the median for the given data

 xi$x_{i}$ 10 10 15 20 25 fi$f_{i}$ 7 4 6 3 5
 xi$x_{i}$ 5 7 9 10 12 15 fi$f_{i}$ 8 6 2 2 2 6

Sol:

The observations given are in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:

 xi$x_{i}$ fi$f_{i}$ c.f. 5 8 8 7 6 14 9 2 16 10 2 18 12 2 20 15 6 26

Here, N = 26, which is even.

Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Median=13thobservation+14thobservation2$Median = \frac{13^{th}\; observation + 14^{th} observation}{2}$

= 7+72$\frac{7 + 7}{2}$

= 7

The absolute values of the deviations from median,

 |xi–M|$|x_{i} – M|$ 2 0 2 3 5 8 fi$f_{i}$ 8 6 2 2 2 6 fi|xi–M|$f_{i}|x_{i} – M|$ 16 0 4 6 10 48

6i=1fi=26$\sum_{i = 1}^{6}f_{i} = 26$

6i=1fi|xiM|=84$\sum_{i = 1}^{6}f_{i}|x_{i} – M| = 84$

Therefore,

MD(M)=1N6i=1fi|xiM|$MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{6}f_{i}\left |x_{i} – M\right |$

= 126×84$\frac{1}{26} \times 84$

= 3.23

Q8. Calculate the mean deviation about the median for the given data

 xi$x_{i}$ 15 21 27 30 35 fi$f_{i}$ 3 5 6 7 8

Sol:

The observations given are in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:

 xi$x_{i}$ fi$f_{i}$ c.f. 15 3 3 21 5 8 27 6 14 30 7 21 35 8 29

Here, N = 29, which is odd.

Median=(29+12)thobservation$Median = \left (\frac{29 + 1}{2} \right )^{th} observation$

= 15th observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

Therefore,

Median = 30

The absolute values of the deviations from median,

 |xi–M|$|x_{i} – M|$ 15 9 3 0 5 fi$f_{i}$ 3 5 6 7 8 fi|xi–M|$f_{i}|x_{i} – M|$ 45 45 18 0 40

5i=1fi=29$\sum_{i = 1}^{5}f_{i} = 29$

5i=1fi|xiM|=148$\sum_{i = 1}^{5}f_{i}|x_{i} – M| = 148$

Therefore,

MD(M)=1N5i=1fi|xiM|$MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{5}f_{i}\left |x_{i} – M\right |$

= 129×148$\frac{1}{29} \times 148$

= 5.1

Q9. Calculate the mean deviation about the mean for the given data:

 Income per day Number 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

Sol:

The given table is formed.

 Income per day Number of persons fi$f_{i}$ Mid – point xi$x_{i}$ fi$f_{i}$ xi$x_{i}$ |xi–x¯|$|x_{i} – \bar{x}|$ fi|xi–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 700 – 800 3 750 2250 392 1176 50 17900 7896

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

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

Therefore,

x¯=1N8i=1fixi$\bar{x} = \frac{1}{N}\sum_{i = 1}^{8}f_{i}x_{i}$

= 150×17900$\frac{1}{50} \times 17900$

= 358

MD(x¯)=1N8i=1fi|xix¯|$MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{8}f_{i}\left |x_{i} – \bar{x} \right |$

= 150×7896$\frac{1}{50} \times 7896$

= 175.92

Q10. Calculate the mean deviation about the mean for the given data:

 Height in cm Number of boys 95 – 105 9 105 – 115 13 115 – 125 26 125 – 135 30 135 – 145 12 145 – 155 10

Sol:

The given table is obtained:

 Height in cm Number of boys fi$f_{i}$ Mid – point xi$x_{i}$ fi$f_{i}$ xi$x_{i}$ |xi–x¯|$|x_{i} – \bar{x}|$ fi|xi–x¯|$f_{i}|x_{i} – \bar{x}|$ 95 – 105 9 100 900 25.3 227.7 105 – 115 13 110 1430 15.3 198.9 115 – 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

N=6i=1fi=100$N = \sum_{i = 1}^{6}f_{i} = 100$

6i=1fixi=12530$\sum_{i = 1}^{6}f_{i}x_{i} = 12530$

Therefore,

x¯=1N6i=1fixi$\bar{x} = \frac{1}{N}\sum_{i = 1}^{6}f_{i}x_{i}$

= 1100×12530$\frac{1}{100} \times 12530$

= 125.3

MD(x¯)=1N6i=1fi|xix¯|$MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{6}f_{i}\left |x_{i} – \bar{x} \right |$

= 1100×1128.8$\frac{1}{100} \times 1128.8$

= 11.28

Q11. Calculate the mean deviation about the median for the given data:

 Marks Number of girls 0 – 10 6 10 – 20 8 20 – 30 14 30 – 40 16 40 – 50 4 50 – 60 2

Sol:

The given table is obtained:

 Marks Number of girls fi$f_{i}$ Cumulative frequency (c.f.) Mid – point xi$x_{i}$ |xi–M|$|x_{i} – M|$ fi|xi–x¯|$f_{i}|x_{i} – \bar{x}|$ 0 – 10 6 6 5 22.85 137.1 10 – 20 8 14 15 12.85 102.8 20 – 30 14 28 25 2.85 39.9 30 – 40 16 44 35 7.15 114.4 40 – 50 4 48 45 17.15 68.6 50 – 60 2 50 55 27.15 54.3 50 517.1

The class interval containing (N2)th$\left ( \frac{N}{2} \right )^{th}$ or 25th item is 20 – 30.

Therefore, 20 -30 is the median class.

Now,

Median=l+N2Cf×h$Median = l + \frac{\frac{N}{2} – C}{f} \times h$

Where, l = 20

C = 14

f = 14

h = 10

N = 50

Median=20+251414×10$Median = 20 + \frac{25 – 14}{14} \times 10$

= 20+11014$20 + \frac{110}{14}$

= 20 + 7.85

= 27.85

Mean deviation about the median is,

MD(M)=1N6i=1fi|xiM|$MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{6}f_{i}\left |x_{i} – M\right |$

= 150×517.1$\frac{1}{50} \times 517.1$

= 10.34

Q12.  Calculate the mean deviation about the median for the given data:

 Age Number 16 – 20 5 21 – 25 6 26 – 30 12 31 – 35 14 36 – 40 26 41 – 45 12 46 – 50 16 51 – 55 9

Sol:

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is as given below:

 Age Number fi$f_{i}$ Cumulative frequency (c.f.) Mid – point xi$x_{i}$ |xi–M|$|x_{i} – M|$ fi|xi–x¯|$f_{i}|x_{i} – \bar{x}|$ 16 – 20 5 5 18 20 100 21 – 25 6 11 23 15 90 26 – 30 12 23 28 10 120 31 – 35 14 37 33 5 70 36 – 40 26 63 38 0 0 41 – 45 12 75 43 5 60 46 – 50 16 91 48 10 160 51 – 55 9 100 53 15 135 100

The class interval containing (N2)th$\left ( \frac{N}{2} \right )^{th}$ or 50th item is 35.5 – 40.5. Therefore, 35.5 – 40.5 is the median class.

Now,

Median=l+N2Cf×h$Median = l + \frac{\frac{N}{2} – C}{f} \times h$

Where, l = 35.5

C = 37

f = 26

h = 5

N = 100

Median=35.5+503726×5$Median = 35.5 + \frac{50 – 37}{26} \times 5$

= 35.5+13×526$35.5 + \frac{13 × 5}{26}$

= 35.5 + 2.5

= 38

Mean deviation about the median is,

MD(M)=1N8i=1fi|xiM|$MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{8}f_{i}\left |x_{i} – M\right |$

= 1100×735$\frac{1}{100} \times 735$

= 7.35

Exercise 15.2

Q1. Calculate mean and variance for the given data 6, 7, 10, 12, 13, 4, 8, 12

Sol:

6, 7, 10, 12, 13, 4, 8, 12

Mean,

x¯=8i=1xin$\bar{x} = \frac{\sum_{i = 1}^{8}x_{i}}{n}$

= 6+7+10+12+13+4+8+128$\frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8}$

= 728$\frac{72}{8}$

= 9

The given table is obtained,

 xi$x_{i}$ (xi–x¯)$\left ( x_{i} – \bar{x} \right )$ (xi–x¯)2$\left ( x_{i} – \bar{x} \right )^{2}$ 6 -3 9 7 -2 4 10 -1 1 12 3 9 13 4 16 4 -5 25 8 -1 1 12 3 9 74

Variance,

(σ2)=1n8i=1(xix¯)2$\left ( \sigma ^{2} \right ) = \frac{1}{n}\sum_{i = 1}^{8}\left ( x_{i} – \bar{x} \right )^{2}$

= 18×74$\frac{1}{8} \times 74$

= 9.25

Q2. Calculate the mean and variance for the first n natural numbers.

Sol:

The mean of first n natural number is,

Mean=SumofallobservationsNumberofobservations$Mean = \frac{Sum \; of \; all \; observations}{Number \; of \; observations}$ Mean=n(n+1)2n$Mean = \frac{\frac{n\left ( n + 1 \right )}{2}}{n}$

= n+12$\frac{ n + 1 }{2}$

Variance,

(σ2)$\left ( \sigma ^{2} \right )$

= 1nni=1(xix¯)2$\frac{1}{n}\sum_{i = 1}^{n}\left ( x_{i} – \bar{x} \right )^{2}$

= 1nni=1[xi(n+12)]2$\frac{1}{n}\sum_{i = 1}^{n}\left [ x_{i} – \left ( \frac{n + 1}{2} \right ) \right ]^{2}$

= 1nni=1x2i1nni=12(n+12)xi+1nni=1(n+12)2$\frac{1}{n}\sum_{i = 1}^{n} x_{i}^{2} – \frac{1}{n}\sum_{i = 1}^{n}2\left ( \frac{n + 1}{2} \right )x_{i} + \frac{1}{n}\sum_{i = 1}^{n}\left ( \frac{n + 1}{2} \right )^{2}$

= 1nn(n+1)(2n+1)6(n+1n)[n(n+1)2]+(n+1)24n×n$\frac{1}{n}\frac{n\left ( n + 1 \right )\left ( 2n + 1 \right )}{6} – \left ( \frac{n + 1}{n} \right )\left [ \frac{n\left ( n + 1 \right )}{2} \right ] + \frac{\left ( n + 1 \right )^{2}}{4n} \times n$

= (n+1)(2n+1)6(n+1)22+(n+1)24$\frac{\left ( n + 1 \right )\left ( 2n + 1 \right )}{6} – \frac{\left (n + 1 \right )^{2}}{2} + \frac{\left ( n + 1 \right )^{2}}{4}$

= (n+1)(2n+1)6(n+1)24$\frac{\left ( n + 1 \right )\left ( 2n + 1 \right )}{6} – \frac{\left ( n + 1 \right )^{2}}{4}$

= (n+1)[4n+23n312]$\left ( n + 1 \right )\left [\frac{ 4n + 2 – 3n – 3 }{12}\right ]$

= (n+1)(n1)12$\frac{\left ( n + 1 \right )\left ( n – 1 \right )}{12}$

= n2112$\frac{n^{2} – 1}{12}$

Q3. Calculate the mean and variance for the first 10 multiples of 3.

Sol:

The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Number of observations,

n = 10

Mean,

x¯=10i=110$\bar{x} = \frac{\sum_{i = 1}^{10}}{10}$

= 16510$\frac{165}{10}$

= 16.5

The given table is obtained,

 xi$x_{i}$ (xi–x¯)$(x_{i} – \bar{x})$ (xi–x¯)2$(x_{i} – \bar{x})^{2}$ 3 -13.5 182.25 6 -10.5 110.25 9 -7.5 56.25 12 -4.5 20.25 15 -1.5 2.25 18 1.5 2.25 21 4.5 20.25 24 7.5 56.25 27 10.5 110.25 30 13.5 182.25 742.5

Variance,

(σ2)$\left ( \sigma ^{2} \right )$

= 1n10i=1(xix¯)2$\frac{1}{n}\sum_{i = 1}^{10}\left ( x_{i} – \bar{x} \right )^{2}$

= 110×742.5$\frac{1}{10} \times 742.5$

= 74.25

Q4. Calculate the mean and variance for the data

 xi$x_{i}$ 6 10 14 18 24 28 30 fi$f_{i}$ 2 4 7 12 8 4 3

Sol:

The data obtained is given in the tabular form:

 xi$x_{i}$ xi$x_{i}$ fixi$f_{i}x_{i}$ xi–x¯$x_{i} – \bar{x}$ (xi−x¯)2$(x_{i}-\bar{x})^{2}$ fi(xi−x¯)2$f_{i}(x_{i}-\bar{x})^{2}$ 6 2 12 -13 169 338 10 4 40 -9 81 324 14 7 98 -5 25 175 18 12 216 -1 1 12 24 8 192 5 25 200 28 4 112 9 81 324 30 3 90 11 121 363 40 760 1736

Here,

N = 40

7i=1fixi$\sum_{i = 1}^{7}f_{i}x_{i}$ = 760

x¯=7i=1fixiN$\bar{x} = \frac{\sum_{i = 1}^{7}f_{i}x_{i}}{N}$

= 76040$\frac{760}{40}$

= 19

Variance,

(σ2)$\left ( \sigma ^{2} \right )$

= 1n7i=1(xix¯)2$\frac{1}{n}\sum_{i = 1}^{7}\left ( x_{i} – \bar{x} \right )^{2}$

= 140×1736$\frac{1}{40} \times 1736$

= 43.4

Q5. Calculate the mean and variance for the data

 xi$x_{i}$ 92 93 97 98 102 104 109 fi$f_{i}$ 3 2 3 2 6 3 3

Sol:

The data obtained is given in the tabular form:

 xi$x_{i}$ fi$f_{i}$ fixi$f_{i}x_{i}$ xi–x¯$x_{i} – \bar{x}$ (xi−x¯)2$(x_{i}-\bar{x})^{2}$ fi(xi−x¯)2$f_{i}(x_{i}-\bar{x})^{2}$ 92 3 276 -8 64 192 93 2 186 -7 49 98 97 3 291 -3 9 27 98 2 196 -2 4 8 102 6 612 2 4 24 104 3 312 4 16 48 109 3 327 9 81 243 22 2200 640

Here,

N = 22

7i=1fixi$\sum_{i = 1}^{7}f_{i}x_{i}$ = 2200

x¯=7i=1fixiN$\bar{x} = \frac{\sum_{i = 1}^{7}f_{i}x_{i}}{N}$

= 220022$\frac{2200}{22}$

= 100

Variance,

(σ2)$\left ( \sigma ^{2} \right )$

= 1n7i=1(xix¯)2$\frac{1}{n}\sum_{i = 1}^{7}\left ( x_{i} – \bar{x} \right )^{2}$

= 122×640$\frac{1}{22} \times 640$

= 29.09

Q6. Calculate the mean and standard deviation using short-cut method.

 xi$x_{i}$ 60 61 62 63 64 65 66 67 68 fi$f_{i}$ 2 1 12 29 25 12 10 4 5

Sol:
The data is obtained in tabular form as follows.

 xi$x_{i}$ fi$f_{i}$ fi=xi−641$f_{i}=\frac{x_{i}-64}{1}$ y2i$y_{i}^{2}$ fiyi$f_{i}y_{i}$ fiy2i$f_{i}y_{i}^{2}$ 60 2 -4 16 -8 32 61 1 -3 9 -3 9 62 12 -2 4 -24 48 63 29 -1 1 -29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80 100 220 0 286

Mean,

x¯=A9i=1fiyiN×h$\bar{x} = A\frac{\sum_{i = 1}^{9}f_{i}y_{i}}{N} \times h$

= 64+0100×1$64 + \frac{0}{100} \times 1$

= 64 + 0

= 64

Variance,

(σ2)$\left ( \sigma ^{2} \right )$

= h2N2[N9i=1fiy2i(9i=1fiyi)2]$\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{9}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{9}f_{i}y_{i} \right )^{2}\right ]$

= 11002[100×2860]$\frac{1}{100^{2}}[ 100 \times 286 – 0]$

= 2.86

Standard deviation,

σ=2.86$\sigma = \sqrt{2.86}$

= 1.69

Q7. Calculate the mean and variance for the given frequency distribution.

 Classes 0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 150 – 180 180 – 210 Frequencies 2 3 5 10 3 5 2

Sol:

 Class fi$f_{i}$ xi${x_{i}}$ yi=xi−10530$y_{i}=\frac{x_{i}-105}{30}$ y2i$y_{i}^{2}$ fiyi$f_{i}y_{i}$ fiy2i$f_{i}y_{i}^{2}$ 0 – 30 2 15 -3 9 -6 18 30 – 60 3 45 -2 4 -6 12 60 – 90 5 75 -1 1 -5 5 90 – 120 10 105 0 0 0 0 120 – 150 3 135 1 1 3 3 150 – 180 5 165 2 4 10 20 180 – 210 2 195 3 9 6 18 30 2 76

Mean,

x¯$\bar{x}$

= A+7i=1fiyiN×h$A + \frac{\sum_{i = 1}^{7}f_{i}y_{i}}{N} \times h$

= 105+230×30$105 + \frac{2}{30} \times 30$

= 105 + 2

= 107

Variance,

(σ2)$\left ( \sigma ^{2} \right )$

= h2N2[N7i=1fiy2i(7i=1fiyi)2]$\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{7}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{7}f_{i}y_{i} \right )^{2}\right ]$

= 302302[30×76(2)2]$\frac{30^{2}}{30^{2}}\left [ 30 \times 76 – \left ( 2 \right )^{2}\right ]$

= 2280 – 4

= 2276

Q8. Calculate the mean and variance for the following frequency distribution.

 Classes 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 Frequencies 5 8 15 16 6

Sol:

 Class Frequency fi$f_{i}$ Mid – point xi$x_{i}$ yi=xi−2510$y_{i}=\frac{x_{i}-25}{10}$ y2i$y_{i}^{2}$ fiyi$f_{i}y_{i}$ fiy2i$f_{i}y_{i}^{2}$ 0 – 10 5 5 -2 4 -10 20 10 – 20 8 15 -1 1 -8 8 20 – 30 15 25 0 0 0 0 30 – 40 16 35 1 1 16 16 40 – 50 6 45 2 4 12 24 50 10 68

Mean,

x¯$\bar{x}$

= A+5i=1fiyiN×h$A + \frac{\sum_{i = 1}^{5}f_{i}y_{i}}{N} \times h$

= 25+1050×10$25 + \frac{10}{50} \times 10$

= 25 + 2

= 27

Variance,

(σ2)$\left ( \sigma ^{2} \right )$

= h2N2[N5i=1fiy2i(5i=1fiyi)2]$\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{5}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{5}f_{i}y_{i} \right )^{2}\right ]$

= 102502[50×68(10)2]$\frac{10^{2}}{50^{2}}\left [ 50 \times 68 – \left ( 10 \right )^{2}\right ]$

= 125[3400100]$\frac{1}{25}\left [ 3400 – 100 \right ]$

= 330025$\frac{3300}{25}$

= 132

Q9. Calculate the mean, variance and standard deviation using short – cut method:

 Height in cm Number of children 70 – 75 3 75 – 80 4 80 – 85 7 85 – 90 7 90 – 95 15 95 – 100 9 100 – 105 6 105 – 110 6 110 – 115 3

Sol:

 Class fi$f_{i}$ xi$x_{i}$ yi=xi−92.55$y_{i}=\frac{x_{i}-92.5}{5}$ y2i$y_{i}^{2}$ fiyi$f_{i}y_{i}$ fiy2i$f_{i}y_{i}^{2}$ 70 – 75 3 72.5 -4 16 -12 48 75 – 80 4 77.5 -3 9 -12 36 80 – 85 7 82.5 -2 4 -14 28 85 – 90 7 87.5 -1 1 -7 7 90 – 95 15 92.5 0 0 0 0 95 – 100 9 97.5 1 1 9 9 100 – 105 6 102.5 2 4 12 24 105 – 110 6 107.5 3 9 18 54 110 – 115 3 112.5 4 16 12 48 60 6 254

Mean,

x¯$\bar{x}$

= A+9i=1fiyiN×h$A + \frac{\sum_{i = 1}^{9}f_{i}y_{i}}{N} \times h$

= 92.5+660×5$92.5 + \frac{6}{60} \times 5$

= 92.5 + 0.5

= 93

Variance,

(σ2)$\left ( \sigma ^{2} \right )$

= h2N2[N9i=1fiy2i(9i=1fiyi)2]$\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{9}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{9}f_{i}y_{i} \right )^{2}\right ]$

= 52602[60×254(6)2]$\frac{5^{2}}{60^{2}}\left [ 60 \times 254 – \left ( 6 \right )^{2}\right ]$

= 253600[155204]$\frac{25}{3600}\left [ 155204 \right ]$

= 105.58

Standard deviation,

(σ)$\left (\sigma \right )$

= 105.58$\sqrt{105.58}$

= 10.27

Q10. The diameters of circles (in mm) drawn in a design are given below:

 Diameters Number of Children 33 – 36 15 37 – 40 17 41 – 44 21 45 – 48 22 49 – 52 25

Sol:

 Class fi$f_{i}$ xi$x_{i}$ yi=xi−42.54$y_{i}=\frac{x_{i}-42.5}{4}$ y2i$y_{i}^{2}$ fiyi$f_{i}y_{i}$ fiy2i$f_{i}y_{i}^{2}$ 32.5 – 36.5 15 34.5 -2 4 -30 60 36.5 – 40.5 17 38.5 -1 1 -17 17 40.5 – 44.5 21 42.5 0 0 0 0 44.5 – 48.5 22 46.5 1 1 22 22 48.5 – 52.5 25 50.5 2 4 50 100 100 25 199

N = 100

h = 4

Let us assume that mean A be 42.5

Mean,

x¯$\bar{x}$

= A+5i=1fiyiN×h$A + \frac{\sum_{i = 1}^{5}f_{i}y_{i}}{N} \times h$

= 42.5+25100×4$42.5 + \frac{25}{100} \times 4$

= 43.5

Variance,

(σ2)$\left ( \sigma ^{2} \right )$

= h2N2[N5i=1fiy2i(5i=1fiy