Class 11 Maths Ncert Solutions 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	$f_{i}$	$x_{i}$	$y_{i}=\frac{x_{i}-45}{10}$	$y_{i}^{2}$	$f_{i}y_{i}$	$f_{i}y_{i}^{2}$
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,

$\bar{x}$

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

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

= 45 – 0.4

= 44.6

Variance,

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

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

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

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

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

= 227.84

Standard deviation,

$\left (\sigma _{1}\right )$

= $\sqrt{227.84}$

= 15.09

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

Marks	$f_{i}$	$x_{i}$	$y_{i}=\frac{x_{i}-45}{10}$	$y_{i}^{2}$	$f_{i}y_{i}$	$f_{i}y_{i}^{2}$
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,

$\bar{x}$

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

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

= 45 – 0.4

= 44.6

Variance,

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

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

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

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

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

= 243.84

Standard deviation,

$\left (\sigma _{1}\right )$

= $\sqrt{243.84}$

= 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,

$\bar{x}$

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

= $\frac{1}{10} \times 510$

= 51

The following table is obtained corresponding to shares X.

$x_{i}$	$(x_{i}-\bar{x})$	$(x_{i}-\bar{x})^{2}$
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,

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

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

= $\frac{1}{10} \times 350$

= 35

Standard deviation,

$\left (\sigma _{1}\right )$

= $\sqrt{35}$

= 5.91

C.V.

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

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

= 11.58

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

Mean,

$\bar{y}$

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

= $\frac{1}{10} \times 1050$

= 105

The following table is obtained corresponding to shares Y.

$y_{i}$	$(y_{i}-\bar{y})$	$(y_{i}-\bar{y})^{2}$
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,

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

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

= $\frac{1}{10} \times 40$

= 4

Standard deviation,

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

= $\sqrt{4}$

= 2

C.V.

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

= $\frac{2}{105} \times 100$

= 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,

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

Therefore,

Standard deviation of the distribution of wages in firm A,

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

Variance of the distribution of wages in firm B,

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

Therefore,

Standard deviation of the distribution of wages in firm B,

$B\left ( \sigma _{2} \right ) = \sqrt{121}$ = 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	$f_{i}x_{i}$	$x_{i}^{2}$	$f_{i}x_{i}^{2}$
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,

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

= $\frac{50}{25}$

= 2

Therefore, the mean of both the teams is same.

Standard deviation,

$\sigma$

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

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

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

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

= 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:

 

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

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

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

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

 

Which is more varying, the length or weight?

 

Sol:

Length

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

Here, N = 50

Mean,

$\bar{x}$

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

= $\frac{212}{50}$

= 4.24

Variance,

$\sigma _{1}^{2}$

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

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

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

= $\frac{1}{50} \left [\sum_{i = 1}^{50}x_{i}^{2} – 8.48\sum_{i = 1}^{50}x_{i} + 17.97 \times 50 \right ]$

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

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

= $\frac{1}{50} \times 3.54$

= 0.07

Standard deviation,

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

= $\sqrt{0.07}$

= 0.26

C.V. (Length)

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

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

= 6.13

Weight

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

Mean,

$\bar{y}$

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

= $\frac{261}{50}$

= 5.22

Variance,

$\sigma _{2}^{2}$

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

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

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

= $\frac{1}{50} \left [\sum_{i = 1}^{50}y_{i}^{2} – 10.44 \sum_{i = 1}^{50}y_{i} + 27.24 \times 50 \right ]$

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

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

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

= 1.89

Standard deviation,

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

= $\sqrt{1.89}$

= 1.37

C.V. (Weight)

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

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

= 26.24

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

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