# Coefficient of variation for ungrouped data

Coefficient of variation for ungrouped data

## 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.

Coefficient of variation formula 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}
Hope you enjoyed the step by step solution to find Coefficient of variation for ungrouped data with help of coefficient of variation formula.

Do read more about step by step solution to find Coefficient of variation for grouped data.

If you have any doubt or queries feel free to post them in the comment section.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.