**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 ⅀x^{2}= 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 | x^{2} |

40 | 1600 |

36 | 1296 |

64 | 4096 |

48 | 2304 |

52 | 2704 |

⅀x = 240 | ⅀x^{2} = 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 ⅀x^{2}= 1530 then calculate the coefficient of variation.

Solution:

Given, n = 10, *x * = 12 and ⅀x^{2} = 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**