Statistics - Co-efficient of Variation
Coefficient of Variation
Standard variation is an absolute measure of dispersion. When comparison has to be made between two series then the relative measure of dispersion, known as coeff.of variation is used.
Coefficient of Variation, CV is defined and given by the following function:
Formula
${CV = \frac{\sigma}{X} \times 100 }$
Where −
Example
Problem Statement:
From the following data. Identify the risky project, is more risky:
| Year | 1 | 2 | 3 | 4 | 5 |
|---|
| Project X (Cash profit in Rs. lakh) | 10 | 15 | 25 | 30 | 55 |
|---|
| Project Y (Cash profit in Rs. lakh) | 5 | 20 | 40 | 40 | 30 |
|---|
Solution:
In order to identify the risky project, we have to identify which of these projects is less consistent in yielding profits. Hence we work out the coefficient of variation.
| Project X | Project y |
|---|
| ${X}$ | ${X_i - \bar X}$
${x}$ | ${x^2}$ | ${Y}$ | ${Y_i - \bar Y}$
${y}$ | ${y^2}$ |
|---|
| 10 | -17 | 289 | 5 | -22 | 484 |
| 15 | -12 | 144 | 20 | -7 | 49 |
| 25 | -2 | 4 | 40 | 13 | 169 |
| 30 | 3 | 9 | 40 | 13 | 169 |
| 55 | 28 | 784 | 30 | 3 | 9 |
| ${\sum X = 135}$ | | ${\sum x^2 = 1230}$ | ${\sum Y = 135}$ | | ${\sum y^2 = 880}$ |
Project X
${Here\ \bar X= \frac{\sum X}{N} \\[7pt]
= \frac{\sum 135}{5} = 27 \\[7pt]
and\ \sigma_x = \sqrt {\frac{\sum X^2}{N}} \\[7pt]
\Rightarrow \sigma_x = \sqrt {\frac{1230}{5}} \\[7pt]
= \sqrt{246} = 15.68 \\[7pt]
\Rightarrow CV_x = \frac{\sigma_x}{X} \times 100 \\[7pt]
= \frac{15.68}{27} \times 100 = 58.07}$
Project Y
${Here\ \bar Y= \frac{\sum Y}{N} \\[7pt]
= \frac{\sum 135}{5} = 27 \\[7pt]
and\ \sigma_y = \sqrt {\frac{\sum Y^2}{N}} \\[7pt]
\Rightarrow \sigma_y = \sqrt {\frac{880}{5}} \\[7pt]
= \sqrt{176} = 13.26 \\[7pt]
\Rightarrow CV_y = \frac{\sigma_y}{Y} \times 100 \\[7pt]
= \frac{13.25}{27} \times 100 = 49.11}$
Since coeff.of variation is higher for project X than for project Y, hence despite the average profits being same, project X is more risky.
