Coefficient of variation is a statistical measure of the dispersion associated with data points in a data series across the mean. 

A useful figure for comparing the degree of variation collected from one of data series completely to another, even if the means are drastically not the same as each other. 

The lower the ratio associated with standard deviation to mean return, better your risk-return tradeoff. Realize that if the expected return within the denominator of the calculations is negative or zero, the ratio will not seem sensible. the coefficient of variation (CV) is a normalized measure of dispersion of your probability distribution or volume (frequency) distribution.

The coefficient of variation (CV), generally known as “relative variability”, equals the conventional deviation divided by your mean of a dataset. It can be expressed either like a fraction or a %. It shows the extent of variability regarding mean in the inhabitants. 

The coefficient of variation should be computed only for data measured on a ratio scale, as these include measurements that can just take non-negative valuations. The coefficient connected with variation may have no meaning for data by using an interval size. Coefficient connected with variation is unitless.
In the mathematical form it is expressed as:

CV = $\frac{\sigma}{\mu}$ $\times$ 100%

where $\sigma$ : Standard deviation

$\mu$ : Mean, the expected return
It shows the extent of variability in relation to mean of the population.
Coefficient associated with variation (CV) is usually calculated and interpreted in two distinct settings:

1) Analyzing just one variable and interpreting a model.
The conventional formulation of the CV, the ratio from the standard deviation towards mean, applies in the single variable environment.

2) In the modeling environment, the CV is calculated since the ratio of the root mean squared error (RMSE) towards mean of the dependent variable.
Inside both settings, the CV can often be presented as the given ratio multiplied by 100. The CV for the single variable aims to spell it out the dispersion from the variable in a way that does not depend on the variable's measurement unit. Greater the CV, the harder the dispersion inside variable. The CV for any model aims to spell out the model fit regarding the relative sizes of the squared residuals along with outcome values.

The low the CV, small the residuals relative to the predicted worth.
The only advantage is which it lets you assess the scatter regarding variables expressed in several units might add up to compare the two CV values.
Sometimes we should compare variations from different sets of data. Comparing variations in information isn’t a problem if you are comparing two sets connected with IQ scores from related classes, or two models of SAT scores coming from incoming freshmen. But in order to compare two sets of data that contain different units (like two tests on different scales), then you certainly need the coefficient connected with variation (CV). It’s accustomed to compare the standard deviations connected with two sets of data that contain significantly different means.
Some simple steps to be followed for calculating coefficient of variation are explained below:
1) For the given set of data, calculate the mean using the formula:

Mean($\mu$) = $\frac{\sum{x}}{n}$

2) Note down the deviations (d = X - $\bar{X}$)

3) Square the deviations (d$^{2}$).

4) Use the coefficient of variable formula.

Coefficient of Variation(c.v) = $\frac{Standard\ Deviation(\sigma)}{Mean(\mu)}$ $\times$ 100%
With the help of above simple steps one can easily solve the coefficient of variation problems. Some of the problems based on coefficient of variation are solved below, to help you out.
Example 1:
 
Find coefficient of variation for the following data set. $2, 5, 8, 12, 15$

Solution:

Calculate Mean

The formula to calculate mean is

$\bar{x}$ = $\frac{\sum x}{n}$

$\bar{x}$ = $\frac{42}{5}$

= 8.4

Find the sample standard deviation

Formula for sample standard deviation is

S = $\sqrt{\frac{\sum (x-\bar{x})^{2}}{n-1}}$

 $x-\bar{x}$ $(x-\bar{x})^{2}$
 2  - 6.4
 40.96
 5  - 3.4  11.56
 8  - 0.4
 0.16
 12  3.6
 12.96
 15  6.6
 43.56
 $\sum x$ = 42      $\sum (x-\bar{x})^{2}$ =109.2

S = $\sqrt{\frac{109.2}{4}}$ = 5.225

(C. V) = $\frac{Standard\ Deviation(\sigma)}{Mean(\mu)}$ $\times$ 100%

 = $\frac{5.225}{8.4}$

= 0.62 $\times$ 100%

= 62.20%
Example 2: 

The number of employees, wage per employee and the variance of the wages per employee for two factories are given below:


Factory A
 Factory B
 Number of Employees
 50  100
 Average wages per employee per month  $\$$120  $\$$85
 Variance of wages per employee per month  $\$$9  $\$$16

In which factory is there greater variation in the distribution of wages per employee. Which factory pays more wages?

Solution: 

Factory A

$\sigma$ = $\sqrt{Variance}$

= $\sqrt{9}$

= 3

c.v = $\frac{\sigma}{\bar{x}}$ $\times$ 100

= $\frac{3}{120}$ $\times$ 100

= 2.5%

$\sum$ x = 120 $\times$ 50

= $\$$ 6000

= Total wages

Factory B

$\sigma$ = $\sqrt{Variance}$

= $\sqrt{16}$

= 4

c.v = $\frac{\sigma}{\bar{x}}$ $\times$ 100

= $\frac{4}{85}$ $\times$ 100

= 4.71%

$\sum$ x = 100 $\times$ 85

= $\$$ 8500

= Total wages

Therefore, in factory B there is greater variation. and

Factory B pays more wages.