# NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 13.

NCERT Solutions for Class 11 Maths Chapter 15 Statistics contain solutions for all Exercise 15.3 questions. Chapter 15 Statistics of Class 11 Maths is categorised under the CBSE Syllabus for the sessions 2023-24. All the exercise questions are solved by subject experts using a step-by-step approach. By practising NCERT Solutions, students will be able to clear their doubts about statistics and solve problems at their own pace. Download these NCERT Solutions of Class 11 Maths now and start practising to score well in the board exams.

## NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

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### Access Other Exercise Solutions of Class 11 Maths Chapter 15 Statistics

Exercise 15.1 Solutions: 12 Questions

Exercise 15.2 Solutions: 10 Questions

Miscellaneous Exercise Solutions: 7 Questions

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NCERT Solutions for Class 11 Maths Chapter 15

NCERT Solutions for Class 11

#### Access Solutions for Class 11 Maths Chapter 15 Exercise 15.3

1. From the data given below, the 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

Solution:-

To compare the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

Co-efficient of variation (C.V.) = (Ïƒ/ xÌ…) Ã— 100

Where, Ïƒ = standard deviation, xÌ… = mean

For Group A,

 Marks Group A fi Mid-point Xi Yi = (xi â€“ A)/h (Yi)2 fiyi fi(yi)2 10 â€“ 20 9 15 ((15 â€“ 45)/10) = -3 (-3)2 = 9 â€“ 27 81 20 â€“ 30 17 25 ((25 â€“ 45)/10) = -2 (-2)2 = 4 â€“ 34 68 30 â€“ 40 32 35 ((35 â€“ 45)/10) = â€“ 1 (-1)2 = 1 â€“ 32 32 40 â€“ 50 33 45 ((45 â€“ 45)/10) = 0 02 0 0 50 â€“ 60 40 55 ((55 â€“ 45)/10) = 1 12 = 1 40 40 60 â€“ 70 10 65 ((65 â€“ 45)/10) = 2 22 = 4 20 40 70 â€“ 80 9 75 ((75 â€“ 45)/10) = 3 32 = 9 27 81 Total 150 -6 342

Where A = 45,

and yi = (xi â€“ A)/h

Here, h = class size = 20 â€“ 10

h = 10

So, xÌ… = 45 + ((-6/150) Ã— 10)

= 45 â€“ 0.4

= 44.6

Ïƒ2 = (102/1502) [150(342) â€“ (-6)2]

= (100/22500) [51,300 â€“ 36]

= (100/22500) Ã— 51264

= 227.84

Hence, standard deviation = Ïƒ = âˆš227.84

= 15.09

âˆ´Â C.V for group A = (Ïƒ/ xÌ…) Ã— 100

= (15.09/44.6) Ã— 100

= 33.83

Now, for group B.

 Marks Group B fi Mid-point Xi Yi = (xi â€“ A)/h (Yi)2 fiyi fi(yi)2 10 â€“ 20 10 15 ((15 â€“ 45)/10) = -3 (-3)2 = 9 â€“ 30 90 20 â€“ 30 20 25 ((25 â€“ 45)/10) = -2 (-2)2 = 4 â€“ 40 80 30 â€“ 40 30 35 ((35 â€“ 45)/10) = â€“ 1 (-1)2 = 1 â€“ 30 30 40 â€“ 50 25 45 ((45 â€“ 45)/10) = 0 02 0 0 50 â€“ 60 43 55 ((55 â€“ 45)/10) = 1 12 = 1 43 43 60 â€“ 70 15 65 ((65 â€“ 45)/10) = 2 22 = 4 30 60 70 â€“ 80 7 75 ((75 â€“ 45)/10) = 3 32 = 9 21 63 Total 150 -6 366

Where A = 45

h = 10

So, xÌ… = 45 + ((-6/150) Ã— 10)

= 45 â€“ 0.4

= 44.6

Ïƒ2 = (102/1502) [150(366) â€“ (-6)2]

= (100/22500) [54,900 â€“ 36]

= (100/22500) Ã— 54,864

= 243.84

Hence, standard deviation = Ïƒ = âˆš243.84

= 15.61

âˆ´Â C.V for group B = (Ïƒ/ xÌ…) Ã— 100

= (15.61/44.6) Ã— 100

= 35

By comparing the C.V. of group A and group B,

C.V. of Group B > C.V. of Group A.

So, Group B is more variable.

2. 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

Solution:-

From the given data,

Let us make the table of the given data and append other columns after calculations.

 X (xi) Y (yi) Xi2 Yi2 35 108 1225 11664 54 107 2916 11449 52 105 2704 11025 53 105 2809 11025 56 106 8136 11236 58 107 3364 11449 52 104 2704 10816 50 103 2500 10609 51 104 2601 10816 49 101 2401 10201 Total = 510 1050 26360 110290

We have to calculate the Mean for x.

Mean xÌ… = âˆ‘xi/n

Where n = Number of terms

= 510/10

= 51

Then, Variance for x =

= (1/102)[(10 Ã— 26360) â€“ 5102]

= (1/100) (263600 â€“ 260100)

= 3500/100

= 35

WKT Standard deviation = âˆšvariance

= âˆš35

= 5.91

So, the co-efficient of variation = (Ïƒ/ xÌ…) Ã— 100

= (5.91/51) Ã— 100

= 11.58

Now, we have to calculate the Mean for y.

Mean È³ = âˆ‘yi/n

Where, n = Number of terms

= 1050/10

= 105

Then, Variance for y =

= (1/102)[(10 Ã— 110290) â€“ 10502]

= (1/100) (1102900 â€“ 1102500)

= 400/100

= 4

WKT Standard deviation = âˆšvariance

= âˆš4

= 2

So, the co-efficient of variation = (Ïƒ/ xÌ…) Ã— 100

= (2/105) Ã— 100

= 1.904

By comparing the C.V. of X and Y,

C.V. of X > C.V. of Y.

So, Y is more stable than X.

3. 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 wages earners 586 648 Mean of monthly wages Rs 5253 Rs 5253 The variance of the distribution of wages 100 121

(i) Which firm, A or B, pays a larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

Solution:-

(i) From the given table,

Mean monthly wages of firm A = Rs 5253

and Number of wage earners = 586

Then,

Total amount paid = 586 Ã— 5253

= Rs 3078258

Mean monthly wages of firm B = Rs 5253

Number of wage earners = 648

Then,

Total amount paid = 648 Ã— 5253

= Rs 34,03,944

So, firm B pays a larger amount as monthly wages.

(ii) Variance of firm A = 100

We know that standard deviation (Ïƒ)= âˆš100

=10

Variance of firm B = 121

Then,

Standard deviation (Ïƒ)=âˆš(121 )

=11

Hence, the standard deviation is more in the case of Firm B, which that means in firm B, there is greater variability in individual wages.

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

 No. of goals scored 0 1 2 3 4 No. of matches 1 9 7 5 3

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

Solution:-

From the given data,

Let us make the table of the given data and append other columns after calculations.

 Number of goals scored xi Number of matches fi fixi Xi2 fixi2 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 Total 25 50 130

The C.V. of firm B is greater.

âˆ´Â Team A is more consistent.

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

Which is more varying, the length or weight?

Solution:-

First, we have to calculate the Mean for Length x.