Class 11 Maths Ncert Solutions Chapter 15 Ex 15.3 Statistics PDF

# Class 11 Maths Ncert Solutions Ex 15.3

## Class 11 Maths Ncert Solutions Chapter 15 Ex 15.3

Q1. From the data given below state which group is more variable, A or B

 Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Group A 9 17 32 33 40 10 9 Group B 10 20 30 25 43 15 7

Sol:

The standard deviation of group A is calculated as given in the table below.

 Marks fi$f_{i}$ xi$x_{i}$ yi=xi−4510$y_{i}=\frac{x_{i}-45}{10}$ y2i$y_{i}^{2}$ fiyi$f_{i}y_{i}$ fiy2i$f_{i}y_{i}^{2}$ 10 – 20 9 15 -3 9 -27 81 20 – 30 17 25 -2 4 -34 68 30 – 40 32 35 -1 1 -32 32 40 – 50 33 45 0 0 0 0 50 – 60 40 55 1 1 40 40 60 – 70 10 65 2 4 20 40 70 – 80 9 75 3 9 27 81 150 -6 342

N = 150

h = 10

A = 45

Mean,

x¯$\bar{x}$

= A+7i=1xiN×h$A + \frac{\sum_{i = 1}^{7}x_{i}}{N} \times h$

= 45+(6)150×10$45 + \frac{(-6)}{150} \times 10$

= 45 – 0.4

= 44.6

Variance,

(σ21)$\left ( \sigma _{1}^{2} \right )$

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

= 1021102[150×342(6)2]$\frac{10^{2}}{110^{2}}\left [ 150 \times 342 – \left ( -6 \right )^{2}\right ]$

= 10022500[51264]$\frac{100}{22500}\left [ 51264 \right]$

= 1225×51264$\frac{1}{225} \times 51264$

= 227.84

Standard deviation,

(σ1)$\left (\sigma _{1}\right )$

= 227.84$\sqrt{227.84}$

= 15.09

The standard deviation of group A is calculated as given in the table below.

 Marks fi$f_{i}$ xi$x_{i}$ yi=xi−4510$y_{i}=\frac{x_{i}-45}{10}$ y2i$y_{i}^{2}$ fiyi$f_{i}y_{i}$ fiy2i$f_{i}y_{i}^{2}$ 10 – 20 10 15 -3 9 -30 90 20 – 30 20 25 -2 4 -40 80 30 – 40 30 35 -1 1 -30 30 40 – 50 25 45 0 0 0 0 50 – 60 43 55 1 1 43 43 60 – 70 15 65 2 4 30 60 70 – 80 7 75 3 9 21 63 150 -6 366

N = 150

h = 10

A = 45

Mean,

x¯$\bar{x}$

= A+7i=1xiN×h$A + \frac{\sum_{i = 1}^{7}x_{i}}{N} \times h$

= 45+(6)150×10$45 + \frac{(-6)}{150} \times 10$

= 45 – 0.4

= 44.6

Variance,

(σ22)$\left ( \sigma _{2}^{2} \right )$

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

= 1021102[150×366(6)2]$\frac{10^{2}}{110^{2}}\left [ 150 \times 366 – \left ( -6 \right )^{2}\right ]$

= 10022500[54864]$\frac{100}{22500}\left [ 54864 \right]$

= 1225×54864$\frac{1}{225} \times 54864$

= 243.84

Standard deviation,

(σ1)$\left (\sigma _{1}\right )$

= 243.84$\sqrt{243.84}$

= 15.61

As the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group B has more variability in the marks.

Q2. From the prices of shares X and Y below, find out which is more stable in value.

 X 35 54 52 53 56 58 52 50 51 49 Y 108 107 105 105 106 107 104 103 104 101

Sol:

The prices of the shares X are 35, 54, 52, 53, 56, 58, 52, 50, 51, and 49.

Here, the number of observations, N = 10

Mean,

x¯$\bar{x}$

= 1N10i=1xi$\frac{1}{N}\sum_{i = 1}^{10}x_{i}$

= 110×510$\frac{1}{10} \times 510$

= 51

The following table is obtained corresponding to shares X.

 xi$x_{i}$ (xi−x¯)$(x_{i}-\bar{x})$ (xi−x¯)2$(x_{i}-\bar{x})^{2}$ 35 -16 256 54 3 9 52 1 1 53 2 4 56 5 25 58 7 49 52 1 1 50 -1 1 51 0 0 49 -2 4 350

Variance,

(σ21)$\left ( \sigma _{1}^{2} \right )$

= 1N10i=1(xix¯)2$\frac{1}{N}\sum_{i = 1}^{10}\left ( x_{i} – \bar{x} \right )^{2}$

= 110×350$\frac{1}{10} \times 350$

= 35

Standard deviation,

(σ1)$\left (\sigma _{1}\right )$

= 35$\sqrt{35}$

= 5.91

C.V.

σ1x×100$\frac{\sigma _{1}}{x} \times 100$

= 5.9151×100$\frac{5.91}{51} \times 100$

= 11.58

The prices of share Y are 108, 107, 105, 105, 106, 107, 104, 103, 104, and 101.

Mean,

y¯$\bar{y}$

= 1N10i=1yi$\frac{1}{N}\sum_{i = 1}^{10}y_{i}$

= 110×1050$\frac{1}{10} \times 1050$

= 105

The following table is obtained corresponding to shares Y.

 yi$y_{i}$ (yi−y¯)$(y_{i}-\bar{y})$ (yi−y¯)2$(y_{i}-\bar{y})^{2}$ 108 3 9 107 2 4 105 0 0 105 0 0 106 1 1 107 2 4 104 -1 1 103 -2 4 104 -1 1 101 -4 16 40

Variance,

(σ22)$\left ( \sigma _{2}^{2} \right )$

= 1N10i=1(yiy¯)2$\frac{1}{N}\sum_{i = 1}^{10}\left ( y_{i} – \bar{y} \right )^{2}$

= 110×40$\frac{1}{10} \times 40$

= 4

Standard deviation,

(σ2)$\left (\sigma _{2}\right )$

= 4$\sqrt{4}$

= 2

C.V.

σ2y×100$\frac{\sigma _{2}}{y} \times 100$

= 2105×100$\frac{2}{105} \times 100$

= 1.9

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.

Therefore, the prices of shares Y are more stable than the prices of shares X.

Q3. 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?

Sol:

(i)

Monthly wages of firm A = Rs. 5253

Number of wage earners in firm A = 586

Therefore,

Total amount paid = Rs. 5253 × 586

Monthly wages of firm B = Rs. 5253

Number of wage earners in firm B = 648

Therefore,

Total amount paid = Rs. 5253 × 648

Hence, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii)

Variance of the distribution of wages in firm A,

A(σ21)$A\left ( \sigma _{1}^{2} \right )$ = 100

Therefore,

Standard deviation of the distribution of wages in firm A,

A(σ1)=100$A\left ( \sigma _{1} \right ) = \sqrt{100}$ = 10

Variance of the distribution of wages in firm B,

B(σ22)$B\left ( \sigma _{2}^{2} \right )$ = 121

Therefore,

Standard deviation of the distribution of wages in firm B,

B(σ2)=121$B\left ( \sigma _{2} \right ) = \sqrt{121}$ = 11

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.

Therefore, firm B has greater variability in the individual wages.

Q4. The following is the record of goals scored by team A in a football session:

 Number of goals scored 0 1 2 3 4 Number of matches 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with standard deviation 1.25 goals. Find which team may be considered more consistent?

Sol:
The mean and the standard deviation of goals scored by team A are calculated as follows.

 No. of goals scored No. of matches fixi$f_{i}x_{i}$ x2i$x_{i}^{2}$ fix2i$f_{i}x_{i}^{2}$ 0 1 0 0 0 1 9 9 1 9 2 7 14 4 28 3 5 15 9 45 4 3 12 16 48 23 50 130

Mean,

5i=1fixi5i=1fi$\frac{\sum_{i = 1}^{5}f_{i}x_{i}}{\sum_{i = 1}^{5}f_{i}}$

= 5025$\frac{50}{25}$

= 2

Therefore, the mean of both the teams is same.

Standard deviation,

σ$\sigma$

= 1NNfix2i(fixi)2$\frac{1}{N}\sqrt{N\sum f_{i}x_{i}^{2} – \left ( \sum f_{i}x_{i} \right )^{2}}$

= 12525×130(50)2$\frac{1}{25}\sqrt{25 \times 130 – \left ( 50 \right )^{2}}$

= 125750$\frac{1}{25}\sqrt{750}$

= 125×27.38$\frac{1}{25} \times 27.38$

= 1.09

The standard deviation of team B is 1.25 goals.

The average number of goals scored by both the teams is same that is 2. Therefore, the team with lower standard deviation will be more consistent.

Therefore, team A is more consistent than team B.

Q5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are as given below:

50i=1xi=212$\sum_{i = 1}^{50}x_{i} = 212$

50i=1x2i=902.8$\sum_{i = 1}^{50}x_{i}^{2} = 902.8$

50i=1yi=261$\sum_{i = 1}^{50}y_{i} = 261$

50i=1y2i=1457.6$\sum_{i = 1}^{50}y_{i}^{2} = 1457.6$

Which is more varying, the length or weight?

Sol:

Length

50i=1xi=212$\sum_{i = 1}^{50}x_{i} = 212$ 50i=1x2i=902.8$\sum_{i = 1}^{50}x_{i}^{2} = 902.8$

Here, N = 50

Mean,

x¯$\bar{x}$

= 50i=1yiN$\frac{\sum_{i = 1}^{50}y_{i}}{N}$

= 21250$\frac{212}{50}$

= 4.24

Variance,

σ21$\sigma _{1}^{2}$

= 1N50i=1(xix¯)2$\frac{1}{N}\sum_{i = 1}^{50}\left ( x_{i} – \bar{x} \right )^{2}$

= 15050i=1(xi4.24)2$\frac{1}{50}\sum_{i = 1}^{50}\left ( x_{i} – 4.24 \right )^{2}$

= 15050i=1[x2i8.48xi+17.97]$\frac{1}{50}\sum_{i = 1}^{50} \left [x_{i}^{2} – 8.48x_{i} + 17.97 \right ]$

= 150[50i=1x2i8.4850i=1xi+17.97×50]$\frac{1}{50} \left [\sum_{i = 1}^{50}x_{i}^{2} – 8.48\sum_{i = 1}^{50}x_{i} + 17.97 \times 50 \right ]$

= 150[902.88.48×(212)+898.5]$\frac{1}{50} \left [902.8 – 8.48 \times \left (212 \right ) + 898.5 \right ]$

= 150[1801.31797.76]$\frac{1}{50} \left [1801.3 – 1797.76 \right ]$

= 150×3.54$\frac{1}{50} \times 3.54$

= 0.07

Standard deviation,

σi(Length)$\sigma _{i}\left ( Length \right )$

= 0.07$\sqrt{0.07}$

= 0.26

C.V. (Length)

= StandarddeviationMean×100$\frac{Standard \; deviation}{Mean} \times 100$

= 0.264.24×100$\frac{0.26}{4.24} \times 100$

= 6.13

Weight

50i=1yi=261$\sum_{i = 1}^{50}y_{i} = 261$ 50i=1y2i=1457.6$\sum_{i = 1}^{50}y_{i}^{2} = 1457.6$

Mean,

y¯$\bar{y}$

= 50i=1yiN$\frac{\sum_{i = 1}^{50}y_{i}}{N}$

= 26150$\frac{261}{50}$

= 5.22

Variance,

σ22$\sigma _{2}^{2}$

= 1N50i=1(yiy¯)2$\frac{1}{N}\sum_{i = 1}^{50}\left ( y_{i} – \bar{y} \right )^{2}$

= 15050i=1(yi5.22)2$\frac{1}{50}\sum_{i = 1}^{50}\left ( y_{i} – 5.22 \right )^{2}$

= 15050i=1[y2i10.44yi+27.24]$\frac{1}{50}\sum_{i = 1}^{50} \left [y_{i}^{2} – 10.44 y_{i} + 27.24 \right ]$

= 150[50i=1y2i10.4450i=1yi+27.24×50]$\frac{1}{50} \left [\sum_{i = 1}^{50}y_{i}^{2} – 10.44 \sum_{i = 1}^{50}y_{i} + 27.24 \times 50 \right ]$

= 150[1457.610.44×(261)+1362]$\frac{1}{50} \left [1457.6 – 10.44 \times \left (261 \right ) + 1362 \right ]$

= 150[2819.62724.84]$\frac{1}{50} \left [2819.6 – 2724.84 \right ]$

= 150×94.76$\frac{1}{50} \times 94.76$

= 1.89

Standard deviation,

σ2(Weight)$\sigma _{2}\left ( Weight \right )$

= 1.89$\sqrt{1.89}$

= 1.37

C.V. (Weight)

= StandarddeviationMean×100$\frac{Standard \; deviation}{Mean} \times 100$

= 1.375.22×100$\frac{1.37}{5.22} \times 100$

= 26.24

Therefore, C.V. of weights is greater than the C.V. of lengths. Hence, weights vary more than the lengths.