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 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.

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