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.

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

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

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

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

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