## Coefficient of variation for ungrouped data

Coefficient of variation is an absolute measure of variation and is used for the comparison of variability between two or more frequency distribution.

Let $x_i, i=1,2, \cdots , n$ be $n$ observations.

## Formula

Coefficient of variation is given by

`$CV =\dfrac{s_x}{\overline{x}}\times 100$`

where,

`$\overline{x} =\dfrac{1}{n}\sum_{i=1}^{n}x_i$`

is the sample mean of $X$,`$n$`

total number of observations,`$s_x=\sqrt{V(x)}$`

is the standard deviation of $X$,`$s_x^2=V(x)=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2 -\dfrac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)$`

is the variance of $X$

## Example 1

The following data gives the hourly wage rates (in dollars) of 10 employees of a company.

20,21,24,25,18,22,24,22,20,22.

Find the coefficient of variation.

### Solution

$x_i$ | $x_i^2$ | |
---|---|---|

20 | 400 | |

21 | 441 | |

24 | 576 | |

25 | 625 | |

18 | 324 | |

22 | 484 | |

24 | 576 | |

22 | 484 | |

20 | 400 | |

22 | 484 | |

Total | 218 | 4794 |

**Sample mean**

The sample mean of $X$ is

`$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{218}{10}\\ &=21.8\text{ dollars} \end{aligned} $$`

The average of hourly wage rates is $21.8$ dollars.

**Sample variance**

Sample variance of $X$ is

`$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{9}\bigg(4794-\frac{(218)^2}{10}\bigg)\\ &=\dfrac{1}{9}\big(4794-\frac{47524}{10}\big)\\ &=\dfrac{1}{9}\big(4794-4752.4\big)\\ &= \frac{41.6}{9}\\ &=4.6222 \end{aligned} $$`

**Sample standard deviation**

The sample standard deviation is the positive square root of the sample variance.

The sample standard deviation is

`$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{4.6222}\\ &=2.1499 \text{ dollars} \end{aligned} $$`

Thus the sample standard deviation of hourly wage rates is $2.1499$ dollars.

**Coefficient of Variation**

The coefficient of variation is

`$$ \begin{aligned} CV &=\frac{s_x}{\overline{x}}\times 100\\ &=\frac{2.1499}{21.8}\times 100\\ &= 9.8621 \end{aligned} $$`

## Example 2

The age (in years) of 6 randomly selected students from a class are

22,25,24,23,24,20.

Compute Coefficient of variation.

### Solution

$x_i$ | $x_i^2$ | |
---|---|---|

22 | 484 | |

25 | 625 | |

24 | 576 | |

23 | 529 | |

24 | 576 | |

20 | 400 | |

Total | 138 | 3190 |

**Sample mean**

The sample mean of $X$ is

`$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{138}{6}\\ &=23\text{ years} \end{aligned} $$`

The average of age of students is $23$ years.

**Sample variance**

Sample variance of $X$ is

`$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{5}\bigg(3190-\frac{(138)^2}{6}\bigg)\\ &=\dfrac{1}{5}\big(3190-\frac{19044}{6}\big)\\ &=\dfrac{1}{5}\big(3190-3174\big)\\ &= \frac{16}{5}\\ &=3.2 \end{aligned} $$`

**Sample standard deviation**

The sample standard deviation is the positive square root of the sample variance.

The sample standard deviation is

`$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{3.2}\\ &=1.7889 \text{ years} \end{aligned} $$`

Thus the sample standard deviation of age of students is $1.7889$ years.

**Coefficient of Variation**

The coefficient of variation is

`$$ \begin{aligned} CV &=\frac{s_x}{\overline{x}}\times 100\\ &=\frac{1.7889}{23}\times 100\\ &= 7.7776 \end{aligned} $$`