# Class 11 Maths Ncert Solutions Ex 15.2

## Class 11 Maths Ncert Solutions Chapter 15 Ex 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=1fiyi)2]$\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{5}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{5}f_{i}y_{i} \right )^{2}\right ]$

= 421002[100×199(25)2]$\frac{4^{2}}{100^{2}}\left [ 100 \times 199 – \left ( 25 \right )^{2}\right ]$

= 1610000[19900625]$\frac{16}{10000}\left [ 19900 – 625 \right]$

= 1610000×19275$\frac{16}{10000} \times 19275$

= 30.84

Standard deviation,

(σ)$\left (\sigma \right )$

= 30.84$\sqrt{30.84}$

= 5.55