Statistics – Exercise 6.2 – Class X

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Statistics – Exercise 6.2

1. Calculate the coefficient of variation of the following data: 40, 36, 64, 48, 52
2. If the coefficient of variation of a collection of data is 45 and its standard deviation is 2.5, then find the mean.
3. 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?
4. In n = 10, x = 12 and ⅀x2 = 1530 then calculate the coefficient of variation.
5. 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?
6. 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.
7. 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:

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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

1. 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