*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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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
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.) = (σ/ x̅) × 100
Where, σ = standard deviation, x̅ = mean
For Group A,
| Marks | Group A
fi |
Mid-point
Xi |
Yi = (xi – A)/h | (Yi)2 | fiyi | fi(yi)2 |
| 10 – 20 | 9 | 15 | ((15 – 45)/10) = -3 | (-3)2
= 9 |
– 27 | 81 |
| 20 – 30 | 17 | 25 | ((25 – 45)/10) = -2 | (-2)2
= 4 |
– 34 | 68 |
| 30 – 40 | 32 | 35 | ((35 – 45)/10) = – 1 | (-1)2
= 1 |
– 32 | 32 |
| 40 – 50 | 33 | 45 | ((45 – 45)/10) = 0 | 02 | 0 | 0 |
| 50 – 60 | 40 | 55 | ((55 – 45)/10) = 1 | 12
= 1 |
40 | 40 |
| 60 – 70 | 10 | 65 | ((65 – 45)/10) = 2 | 22
= 4 |
20 | 40 |
| 70 – 80 | 9 | 75 | ((75 – 45)/10) = 3 | 32
= 9 |
27 | 81 |
| Total | 150 | -6 | 342 |
Where A = 45,
and yi = (xi – A)/h
Here, h = class size = 20 – 10
h = 10
So, x̅ = 45 + ((-6/150) × 10)
= 45 – 0.4
= 44.6
σ2 = (102/1502) [150(342) – (-6)2]
= (100/22500) [51,300 – 36]
= (100/22500) × 51264
= 227.84
Hence, standard deviation = σ = √227.84
= 15.09
∴ C.V for group A = (σ/ x̅) × 100
= (15.09/44.6) × 100
= 33.83
Now, for group B.
| Marks | Group B
fi |
Mid-point
Xi |
Yi = (xi – A)/h | (Yi)2 | fiyi | fi(yi)2 |
| 10 – 20 | 10 | 15 | ((15 – 45)/10) = -3 | (-3)2
= 9 |
– 30 | 90 |
| 20 – 30 | 20 | 25 | ((25 – 45)/10) = -2 | (-2)2
= 4 |
– 40 | 80 |
| 30 – 40 | 30 | 35 | ((35 – 45)/10) = – 1 | (-1)2
= 1 |
– 30 | 30 |
| 40 – 50 | 25 | 45 | ((45 – 45)/10) = 0 | 02 | 0 | 0 |
| 50 – 60 | 43 | 55 | ((55 – 45)/10) = 1 | 12
= 1 |
43 | 43 |
| 60 – 70 | 15 | 65 | ((65 – 45)/10) = 2 | 22
= 4 |
30 | 60 |
| 70 – 80 | 7 | 75 | ((75 – 45)/10) = 3 | 32
= 9 |
21 | 63 |
| Total | 150 | -6 | 366 |
Where A = 45
h = 10
So, x̅ = 45 + ((-6/150) × 10)
= 45 – 0.4
= 44.6
σ2 = (102/1502) [150(366) – (-6)2]
= (100/22500) [54,900 – 36]
= (100/22500) × 54,864
= 243.84
Hence, standard deviation = σ = √243.84
= 15.61
∴ C.V for group B = (σ/ x̅) × 100
= (15.61/44.6) × 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 x̅ = ∑xi/n
Where n = Number of terms
= 510/10
= 51
Then, Variance for x =
= (1/102)[(10 × 26360) – 5102]
= (1/100) (263600 – 260100)
= 3500/100
= 35
WKT Standard deviation = √variance
= √35
= 5.91
So, the co-efficient of variation = (σ/ x̅) × 100
= (5.91/51) × 100
= 11.58
Now, we have to calculate the Mean for y.
Mean ȳ = ∑yi/n
Where, n = Number of terms
= 1050/10
= 105
Then, Variance for y =
= (1/102)[(10 × 110290) – 10502]
= (1/100) (1102900 – 1102500)
= 400/100
= 4
WKT Standard deviation = √variance
= √4
= 2
So, the co-efficient of variation = (σ/ x̅) × 100
= (2/105) × 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 × 5253
= Rs 3078258
Mean monthly wages of firm B = Rs 5253
Number of wage earners = 648
Then,
Total amount paid = 648 × 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 (σ)= √100
=10
Variance of firm B = 121
Then,
Standard deviation (σ)=√(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.
∴ 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.
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