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.
