Coefficient of Dispersion: Definition, Formula and Examples

Collecting the data and expressing it in the form of measures of data is an essential concept for us. The measure of the spread shows how much variation is there in data. It shows how the data is spread, and scattered, and what is the deviation, and variance of the data. These values describe the data in a better way and help the analyst to analyze the data in a better way and take out the insights from it.

Coefficients of Dispersion

The measure of variability which is independent of units is called coefficient of dispersion.

An important characteristic of any set of data is the variation in the data. The degree to which the numerical data tends to vary about an average value is called the dispersion or scatteredness of the data.

The following are the measures of dispersion:

1. Range

2. Mean Deviation

3. Standard deviation and Variance

1. Coefficients of Range

Range is the difference between the highest and the lowest value in a set of observations.
The range of data gives us a rough idea of variability or scatter but does not tell about the dispersion of the data from a measure of central tendency.
It equals $\frac{x_{\max }+x_{\min }}{x_{\max }+x_{\min }}$
Where $x_{\max }$ is the highest observation, and $x_{\min }$ is the lowest observation

2. Coefficients of Standard Deviation

The standard deviation is a number that measures how far data values are from their mean.

$
\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2}
$

The positive square root of the variance is called the standard deviation. The standard deviation is usually denoted by $\sigma$ and it is given by It equals $\frac{\sigma}{\bar{x}}$
where $\sigma$ and $\bar{x}$ are the standard deviation and mean of the data.

3. Coefficients of Variance (denoted as C.V.)

The mean of the squares of the deviations from the mean is called the variance and is denoted by $\sigma^2$ (read as sigma square).
Variance is a quantity that leads to a proper measure of dispersion.
The variance of $n$ observations $x_1, x_2, \ldots, x_n$ is given by

$
\sigma^2=\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2
$

The coefficient of variation is defined as

$
\text { C.V. }=\frac{\sigma}{\bar{x}} \times 100, \bar{x} \neq 0
$

where $\sigma$ and $\bar{x}$ are the standard deviation and mean of the data.It is consistent than the other and thus is considered better.

Solved Examples Based on Coefficient of Dispersion:

Example 1: Coefficients of variation of two distributions are $50$ and $60$ , and their arithmetic means are $30$ and $25$ respectively. Difference of their $S.D.$ is?
1) $0$
2) $1$
3) $1.5$
4) $2.5$

Solution
$
\begin{aligned}
& \frac{\sigma_1}{\bar{x}_1} \times 100=50 \Rightarrow \sigma_1=\frac{\bar{x}}{2}=\frac{30}{2}=15 \\
& \frac{\sigma_2}{\bar{x}_2} \times 100=60 \Rightarrow \sigma_1=\frac{60}{4}=15 \\
& \left|\sigma_1-\sigma_2\right|=0
\end{aligned}
$
Hence, the answer is the option 1 .

Example 2: The following are the weights (in kg ) of $8$ students in a class: $62,65,58,60,70,55,68$, and $72$ . What is the coefficient of variation $(CV)$ of the weights?
1) $10.3 \%$
2) $9.9 \%$
3) $23.1 \%$
4) $27.5 \%$

Solution
To calculate the coefficient of variation $(CV)$, we first need to calculate the mean weight and standard deviation of the weights:

$
\begin{aligned}
& \text { Mean weight }=\frac{62+65+58+60+70+55+68+72}{8}=64.5 \\
& \begin{aligned}
\text { Standard deviation } & =\sqrt{\frac{(62-64.5)^2+(65-64.5)^2+(58-64.5)^2+(60-64.5)^2+(70-64.5)^2+(55-64.5)^2+(68-64.5)^2+(72-64.5)^2}{7}} \\
& =6.39
\end{aligned}
\end{aligned}
$
Now we can calculate the coefficient of variation $(CV)$:

$
\frac{6.39}{64.5} \times 100 \%=9.9 \%
$
Hence the answer is option (2)

Example 3: If the coefficient of variation $(CV)$ of a set of data is $0.5$ , what can be said about the variability of the data compared to the mean?
1) The data is highly variable compared to the mean.
2) The data is moderately variable compared to the mean.
3) The data is slightly variable compared to the mean.
4) The variability of the data cannot be determined from the coefficient of variation.

Solution mean, which indicates that the data is highly variable compared to the mean. In general, a $CV$ greater than $1$ indicates high variability, while a $CV$ less than $1$ indicates low variability.
Hence, the answer is option (1).

Example 4: Which of the following is a drawback of using the coefficient of quartile deviation as a measure of dispersion?
1) It does not take into account extreme values.
2) It is sensitive to changes in the units of measurement.
3) It is difficult to calculate.
4) It is affected by the size of the dataset.

Solution

The coefficient of quartile deviation $(CQD)$ is a measure of relative dispersion that is based on the quartiles of a dataset. However, a drawback of $CQD$ is that it does not take into account extreme values, such as outliers, which can have a significant impact on the dispersion of the data. Therefore, $CQD$ may not accurately reflect the dispersion of the dataset if it contains extreme values.
Hence, the answer is option (1).

Summary
The measure of variability which is independent of units is called coefficient of dispersion.The standard deviation is a number that measures how far data values are from their mean.
$
\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2}
$
The mean of the squares of the deviations from the mean is called the variance.
The coefficient of variation is defined as

$
\text { C.V. }=\frac{\sigma}{\bar{x}} \times 100, \bar{x} \neq 0
$

where $\sigma$ and $\bar{x}$ are the standard deviation and mean of the data.