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 xi yi=xi4510 y2i fiyi fiy2i
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¯

= A+7i=1xiN×h

= 45+(6)150×10

= 45 – 0.4

= 44.6

Variance,

(σ21)

= h2N2[N7i=1fiy2i(7i=1fiyi)2]

= 1021102[150×342(6)2]

= 10022500[51264]

= 1225×51264

= 227.84

Standard deviation,

(σ1)

= 227.84

= 15.09

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

Marks fi xi yi=xi4510 y2i fiyi fiy2i
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¯

= A+7i=1xiN×h

= 45+(6)150×10

= 45 – 0.4

= 44.6

Variance,

(σ22)

= h2N2[N7i=1fiy2i(7i=1fiyi)2]

= 1021102[150×366(6)2]

= 10022500[54864]

= 1225×54864

= 243.84

Standard deviation,

(σ1)

= 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¯

= 1N10i=1xi

= 110×510

= 51

The following table is obtained corresponding to shares X.

xi (xix¯) (xix¯)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)

= 1N10i=1(xix¯)2

= 110×350

= 35

Standard deviation,

(σ1)

= 35

= 5.91

C.V.

σ1x×100

= 5.9151×100

= 11.58

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

Mean,

y¯

= 1N10i=1yi

= 110×1050

= 105

The following table is obtained corresponding to shares Y.

yi (yiy¯) (yiy¯)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)

= 1N10i=1(yiy¯)2

= 110×40

= 4

Standard deviation,

(σ2)

= 4

= 2

C.V.

σ2y×100

= 2105×100

= 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) = 100

Therefore,

Standard deviation of the distribution of wages in firm A,

A(σ1)=100 = 10

Variance of the distribution of wages in firm B,

B(σ22) = 121

Therefore,

Standard deviation of the distribution of wages in firm B,

B(σ2)=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 x2i fix2i
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

= 5025

= 2

Therefore, the mean of both the teams is same.

Standard deviation,

σ

= 1NNfix2i(fixi)2

= 12525×130(50)2

= 125750

= 125×27.38

= 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

50i=1x2i=902.8

50i=1yi=261

50i=1y2i=1457.6

 

Which is more varying, the length or weight?

 

Sol:

Length

50i=1xi=212 50i=1x2i=902.8

Here, N = 50

Mean,

x¯

= 50i=1yiN

= 21250

= 4.24

Variance,

σ21

= 1N50i=1(xix¯)2

= 15050i=1(xi4.24)2

= 15050i=1[x2i8.48xi+17.97]

= 150[50i=1x2i8.4850i=1xi+17.97×50]

= 150[902.88.48×(212)+898.5]

= 150[1801.31797.76]

= 150×3.54

= 0.07

Standard deviation,

σi(Length)

= 0.07

= 0.26

C.V. (Length)

= StandarddeviationMean×100

= 0.264.24×100

= 6.13

Weight

50i=1yi=261 50i=1y2i=1457.6

Mean,

y¯

= 50i=1yiN

= 26150

= 5.22

Variance,

σ22

= 1N50i=1(yiy¯)2

= 15050i=1(yi5.22)2

= 15050i=1[y2i10.44yi+27.24]

= 150[50i=1y2i10.4450i=1yi+27.24×50]

= 150[1457.610.44×(261)+1362]

= 150[2819.62724.84]

= 150×94.76

= 1.89

Standard deviation,

σ2(Weight)

= 1.89

= 1.37

C.V. (Weight)

= StandarddeviationMean×100

= 1.375.22×100

= 26.24

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