Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.7
Question 1. Two plants A and B of a factory show the following results about the number of workers and the wages paid to them
| Plant A | Plant B |
No. of workers | 5000 | 6000 |
Average monthly wages | ₹2500 | ₹2500 |
The variance of distribution of wages | 81 | 100 |
In which plant A or B is there greater variability in individual wages?
Solution:
Variation of the distribution of wages in plant A (σ^{2} =18)
So, Standard deviation of the distribution A (σ – 9)
Similarly, the Variation of the distribution of wages in plant B (σ^{2} =100)
So, Standard deviation of the distribution B (σ – 10)
And, Average monthly wages in both the plants is 2500,
Since, the plant with a greater value of SD will have more variability in salary.
∴ Plant B has more variability in individual wages than plant A
Question 2. The means and standard deviations of heights and weights of 50 students in a class are as follows:
| Weights | Heights |
Mean | 63.2 Kg | 63.2 inch |
Standard deviation | 5.6 Kg | 11.5 inch |
Which shows more variability, heights or weights?
Solution:
We observe that the average weights and height for the 50 students is same i.e. 63.2.
Therefore, the parameter with greater variance will have more variability.
Thus, height has greater variability
Question 3. The coefficient of variation of two distribution are 60% and 70%, and their standard deviations are 21 and 16 respectively. What is their arithmetic means?
Solution:
Coefficient of variation =
So, we have:
∴ Means are 35 and 22.85
Question 4. Calculate coefficient of variation from the following data:
Income(in ₹): | 1000 – 1700 | 1700 – 2400 | 2400 – 3100 | 3100 – 3800 | 3800 – 4500 | 4500 – 5200 |
No. of families: | 12 | 18 | 20 | 25 | 35 | 10 |
Solution:
Class
F_{i}
x_{i }
f_{i}u_{i }
f_{i}u_{i}^{2 }
1000 – 1700
12
1350
-2
-24
48
1700 – 2400
18
2050
-1
-18
18
2400 – 3100
20
2750
0
0
0
3100 – 3800
25
3450
1
25
25
3800 – 4500
35
4150
2
70
140
4500 – 5200
10
4850
3
30
90
Now,
N = 120,
Mean,
Variance = 1076332.64
Standard Deviation,
Coefficient of variation =
= 32.08
∴ The coefficient variation is 32.08
Question 5. An analysis of the weekly 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 |
Average weekly wages | ₹52.5 | ₹47.5 |
The variance of the distribution of wages | 100 | 121 |
(i) Which firm A or B pays out the larger amount as weekly wages?
(ii) Which firm A or B has greater variability in individual wages?
Solution:
(i) Average weekly wages =
Total weekly wages = (Average weekly wages) × (No. of workers)
Total weekly wages of Firm A = 52.5 × 586 = Rs 30765
Total weekly wages of Firm B = 47.5 × 648 = Rs 30780
Firm B pays a larger amount as Firm A
(ii) Here,
S.D (Firm A) = 10 and S.D (Firm B) = 11
Coefficient variance (Firm A) =
= 19.04
Coefficient variance (Firm B) =
= 23.15
∴ Coefficient variance of Firm B is greater than that of Firm A, Firm B has greater variability in individual wages.
Question 6. The following are some particulars of the distribution of weights of boys and girls in a class:
Boys | Girls | |
Number | 100 | 50 |
Mean weight | 60 Kg | 45 Kg |
Variance | 9 | 4 |
Which of the distributions is more variable?
Solution:
Given:
S.D (Boys) is 3 and S.D (Girls) is 2
Coefficient variance (Boys) =
= 5
Coefficient variance (Girls) =
= 4.4
∴ Coefficient variance of Boys is greater than Coefficient variance of girls, and then the distribution of weights of boys is more variable than that of girls.
Question 7. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:
Subject Mean | Mathematics 42 | Physics 32 | Chemistry 40.9 |
Standard deviation | 12 | 15 | 20 |
Solution:
In order to compare the variability of marks in Math, Physics and Chemistry.
We have to calculate their coefficient of variation.
Let σ_{1}, σ_{2} and σ_{3} denote the standard deviation of marks in Math, Physics and Chemistry respectively. Further, Let be the mean scores in Math, Physics and Chemistry respectively.
We have
⇒ σ_{1} = 12 σ_{2} = 15 σ_{3} = 20
Now,
Coefficient of variation in Maths =
Coefficient of variation in Physics =
Coefficient of variation in Chemistry =
Clearly, coefficient of variation in marks is greatest in Chemistry and lowest in Math.
So, marks in chemistry show highest variability and marks in maths show lowest variability.
Question 8. From the data given below state which group is more variable, G_{1} or G_{2}?
Marks Group G_{1} | 10 – 20 9 | 20 – 30 30 – 40 17 32 | 40 – 50 33 | 50 – 60 60 – 70 40 10 | 70 – 80 9 |
Group G_{2} | 10 | 20 30 | 25 | 43 15 | 7 |
Solution:
Let’s first find the coefficient of variable for group G_{1}
CI f
10 – 20 9
20 – 30 17
x u=(x – A)/h
15 -3
25 -2
fu u^{2 }
-27 9
-34 4
fu^{2 }
81
68
30 – 40 32
40 – 50 33
35 -1
45 0
-32 1
0 0
32
0
50 – 60 40 55 1 40 1 40 60 – 70 10
70 – 80 9
65 2
75 3
20 4
27 9
40
81
150 -6 342 Here, N = 150, A = 45, and h = 10
∴ Mean =
Coefficient of variation =
Now, lets find the coefficient of variable for group G_{2}
CI f
10 – 20 10
20 – 30 20
x u=(x – A)/h
15 -3
25 -2
fu u^{2}
-30 9
-40 4
fu^{2}
90
80
30 – 40 30
40 – 50 25
35 -1
45 0
-30 1
0 0
30
0
50 – 60 43 55 1 43 1 43 60 – 70 15
70 – 80 7
65 2
75 3
30 4
21 9
60
63
150 -6 366 Here, N = 150, A = 45, and h = 10
∴ Mean =
Coefficient of variation =
Group G_{2} is more variable
Question 9. Find the coefficient of variation for the following data:
Size (in cms): 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 |
No. of items: 2 | 8 | 20 | 35 | 20 | 15 |
Solution:
CI f x
10 – 15 2 12.5
15 – 20 8 17.5
u=(x – A)/h fu u^{2}
-2 -4 4
-1 -8 1
fu^{2}
8
8
20 – 25 20 22.5
25 – 30 35 27.5
0 0 0
1 35 1
0
35
30 – 35 20 32.5 2 40 4 80 35 – 40 15 37.5 3 45 9 135 100 108 266 Here, N = 100, A = 22.5, and h = 5
∴ Mean =
Coefficient of variation =
Question 10. From the prices of shares X and Y given 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:
x d = (x – Mean)
35 -13
24 -24
d^{2}
169
576
52 4
53 5
16
25
56 8 64 58 10
52 4
100
16
50 2 4 51 3 9 49 1 1 480 980 ∴ Mean =
Coefficient of variation =
x d = (x – Mean)
35 -13
24 -24
d^{2}
169
576
52 4
53 5
16
25
56 8 64 58 10
52 4
100
16
50 2 4 51 3 9 49 1 1 480 980 ∴ Mean =
Coefficient of variation =
Since the coefficient of variation for share Y is smaller than the coefficient of variation for shares X, they are more stable.