Previous exercise – Statistics – Exercise 6.1 – Class X
Statistics – Exercise 6.2
- Calculate the coefficient of variation of the following data: 40, 36, 64, 48, 52
- If the coefficient of variation of a collection of data is 45 and its standard deviation is 2.5, then find the mean.
- A group of 100 candidates attending a physical test or recruitment have their average height as 163.8 cm with coefficient of variation 3.2 What is the standard deviation of their heights?
- In n = 10, x = 12 and ⅀x2 = 1530 then calculate the coefficient of variation.
- The coefficient of variation of two series are 58 and 69. Their standard deviations are 21.2 and 51.6 what are their arithmetic means?
- Batsman A gets an average of 64 runs per innings with standard deviation of 18 runs, while batsman B gets an average score 43 runs with standard deviation of 9 runs in an equal number of innings. Discuss the efficiency and consistency of both the batsman.
- In two construction companies A and B, the average weekly wages in rupees and standard deviations are as follows:
Determine which factory has greater variability in individual wages?
| Company | Average of wages | Standard deviation in Rs. |
| A | 3450 | 6.21 |
| B | 2850 | 4.56 |
Statistics – Exercise 6.2 – Solution:
- Calculate the coefficient of variation of the following data: 40, 36, 64, 48, 52
Solution:
Given: the coefficient of variation of the following data: 40, 36, 64, 48, 52.
We can tabulate the data given in the following way:
| x | x2 |
| 40 | 1600 |
| 36 | 1296 |
| 64 | 4096 |
| 48 | 2304 |
| 52 | 2704 |
| ⅀x = 240 | ⅀x2 = 12000 |
Mean = 40+36+64+48+52/5 = 48
We know, coefficient of variation = standard deviation/mean x 100
= 9.8/48 x 100
= 20.416
- If the coefficient of variation of a collection of data is 45 and its standard deviation is 2.5, then find the mean.
Solution:
Given coefficient of variation, C. V. = 45 and standard deviation is 2.5/ Then we need to find mean.
We know, coefficient of variation = standard deviation/mean x 100
Mean = 100 x standard deviation/ coefficient of variation
= 100 x 2.5/45
= 5.55
- A group of 100 candidates attending a physical test or recruitment have their average height as 163.8 cm with coefficient of variation 3.2. What is the standard deviation of their heights?
Solution:
Given, a group of 100 candidates attending a physical test or recruitment have their average height as 163.8 cm with coefficient of variation 3.2
n = 100
coefficient of variation = 3.2
average height as = mean = 163.8 cm
We know, coefficient of variation = standard deviation/mean x 100
Standard deviation = coefficient of variation x mean/100
= 3.2 x 163.8/100
= 5.2
- In n = 10, x = 12 and ⅀x2 = 1530 then calculate the coefficient of variation.
Solution:
Given, n = 10, x = 12 and ⅀x2 = 1530 then we need to calculate the coefficient of variation
We know, coefficient of variation = standard deviation/mean x 100
To find coefficient of variation we need to find standard deviation,
Mean = ⅀x/n ⇒12 = ⅀x/10
⅀x = 12×10 = 120
We know, coefficient of variation = standard deviation/mean x 100
Coefficient of variation = 3/12 x 100
Coefficient of variation = 25
- The coefficient of variation of two series are 58 and 69. Their standard deviations are 21.2 and 51.6 what are their arithmetic means?
Solution:
Given, the coefficient of variation of two series are 58, 69 and their standard deviations are 21.2 and 51.6
| coefficient of variation | Standard deviation | |
| A | 58 | 21.2 |
| B | 69 | 51.6 |
We need to find the arithmetic means of two series
We know, coefficient of variation = standard deviation/mean x 100
Then, Mean = 100 x standard deviation/ coefficient of variation
Arithmetic mean of A = 100 x 21.2/58 = 36.55
Arithmetic mean of B = 100 x 51.6/69 = 74.78
Therefore, arithmetic means are 36.55 and 74.78
- Batsman A gets an average of 64 runs per innings with standard deviation of 18 runs, while batsman B gets an average score 43 runs with standard deviation of 9 runs in an equal number of innings. Discuss the efficiency and consistency of both the batsman.
Solution:
Given, batsman A gets an average of 64 runs per innings with standard devation of 18 runs, while batsman B gets an average score 43 runs with standard deviation of 9 runs in an equal number of innings.
We can tabulate the given data as follows:
| Average | Standard deviation | |
| A | 64 | 18 |
| B | 43 | 9 |
We need to find the efficiency and consistency of both the batsman.
We know, coefficient of variation = standard deviation/mean x 100
Coefficient of variation of A = 18/64 x 100 = 28.125
Coefficient of variation of B = 9/43 x 100 = 20.93
- In two construction companies A and B, the average weekly wages in rupees and standard deviations are as follows:
Determine which factory has greater variability in individual wages?
| Company | Average of wages | Standard deviation in Rs. |
| A | 3450 | 6.21 |
| B | 2850 | 4.56 |
Solution:
Data given:
| Company | Average of wages | Standard deviation in Rs. |
| A | 3450 | 6.21 |
| B | 2850 | 4.56 |
We have to find out which factory has greater variability in individual wages.
So for this we have to find coefficient of variation.
We know, coefficient of variation = standard deviation/mean x 100
Coefficient of variation of A = 6.21/3450 x 100 = 0.18
Coefficient of variation of B = 4.56/2850 x 100 = 0.16
Therefore, factory A has greater variability in individual wages.
Next exercise – Statistics – Exercise 6.3 – Class X
