Coefficient of variation for grouped data

Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution.

Formula

Coefficient of variation is given by

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

where,

• $\overline{x} =\dfrac{1}{N}\sum_{i=1}^{n}f_ix_i$ is the sample mean of $X$,
• $N$ total number of observations,
• $s_x=\sqrt{V(x)}$ is the standard deviation of $X$,
• $s_x^2=V(x)=\dfrac{1}{N}\sum_{i=1}^{n}f_ix_i^2 -(\overline{x})^2$ is the variance of $X$

Coefficient of variation of one data set is lower than the coefficient of variation of other data set, then the data set with lower coefficient of variation is more consistent than the other.

Example 1

Compute coefficient of variation for the following frequency distribution.

$x$ 2 3 4 5 6
$f$ 1 15 10 5 4

Solution

$x_i$ $f_i$ $f_i*x_i$ $f_i*x_i^2$
2 1 2 4
3 15 45 135
4 10 40 160
5 5 25 125
6 4 24 144
Total 35 136 568

Sample mean of $X$ is

\begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{136}{35}\\ &=3.8857 \end{aligned}

Sample variance of $X$ is

\begin{aligned} s_x^2 &=\frac{1}{N}\sum_{i=1}^n f_ix_i^2-(\overline{x})^2\\ &=\frac{1}{35}\big(568\big)-(3.8857)^2\\ &=16.2286-15.0987\\ &=1.1299 \end{aligned}

Sample standard deviation of $X$ is

\begin{aligned} s_x&=\sqrt{s_x^2}\\ &= \sqrt{1.1299}\\ &=1.063. \end{aligned}

Coefficient of variation is \begin{aligned} CV &=\frac{sx}{\overline{x}}\times 100\\ &=\frac{1.063}{3.8857}\times 100\\ &=27.3567 \end{aligned}

Example 2

Compute coefficient of variation for the following frequency distribution.

$x$ 5-8 9-12 13-16 17-20 21-24
$f$ 2 13 21 14 5

Solution

Class Interval Class Boundries mid-value ($x_i$) $f_i$ $f_i*x_i$ $f_ix_i^2$
5-8 4.5-8.5 6.5 2 13 84.5
9-12 8.5-12.5 10.5 13 136.5 1433.25
13-16 12.5-16.5 14.5 21 304.5 4415.25
17-20 16.5-20.5 18.5 14 259 4791.5
21-24 20.5-24.5 22.5 5 112.5 2531.25
Total 55 825.5 13255.75

Sample mean of $X$ is

\begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{825.5}{55}\\ &=15.0091 \end{aligned}

Sample variance of $X$ is

\begin{aligned} s_x^2 &=\frac{1}{N}\sum_{i=1}^n f_ix_i^2-(\overline{x})^2\\ &=\frac{1}{55}\big(13255.75\big)-(15.0091)^2\\ &=241.0136-225.2731\\ &=15.7406 \end{aligned}

Sample standard deviation of $X$ is

\begin{aligned} s_x&=\sqrt{s_x^2}\\ &= \sqrt{15.7406}\\ &=3.9674. \end{aligned}

Coefficient of variation is \begin{aligned} CV &=\frac{sx}{\overline{x}}\times 100\\ &=\frac{3.9674}{15.0091}\times 100\\ &=26.4333 \end{aligned}