Question

Calculate the coefficient of variation for the given data (36, 15, 25, 10, 14).

Answer 1

Hint: In this question, we are given data and we have to find the coefficient of variation. In order to calculate the coefficient of variation, the standard deviation of the series is divided by the mean of the series and then multiplied by 100. For calculating mean, we will divide the sum of given terms by the number of terms. After that, we take a deviation from the mean in the frequency distribution table and then square those deviations. Using all these columns we will find standard deviation of the data using formula \[\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}}\].
Where $\sum{{{x}^{2}}}$ is the sum of squares of deviation and N is the number of terms. At last, we will apply the formula for calculating the coefficient of variation given as:
\[CV=\dfrac{\sigma }{\overline{X}}\times 100\]
Where, $\sigma $ is the standard deviation, $\overline{X}$ is the mean and CV is the coefficient of variation.

Complete step-by-step solution:
Let us solve this sum step by step:
Step 1: We are given data as 36, 15, 25, 10, 14.
Here, total number of terms = 5. Hence N = 5.
Now to calculate mean, let us add these terms. Therefore,
\[\sum{x}=36+15+25+10+14=100\]
As we know, \[\text{Mean}=\dfrac{\sum{x}}{N}\]
Therefore, \[\text{Mean}=\overline{X}=\dfrac{100}{5}=20\]
Step 2: Let us draw frequency distribution table having columns as X, x, ${{x}^{2}}$ where X represents given terms, x represents deviation from mean $\left( x=X-\overline{X} \right)$ and ${{x}^{2}}$ is square of deviation.

X	$\left( x=X-\overline{X} \right)$$\overline{X}=20$	${{x}^{2}}={{\left( X-\overline{X} \right)}^{2}}$
36	16	256
15	-5	25
25	5	25
10	-10	100
14	-6	36
Sum=	0	442

Now, adding all ${{x}^{2}}$ we get:
\[\sum{{{x}^{2}}}=442\]
Step 3: Let us calculate standard deviation. Standard deviation $\sigma $ is given as:
\[\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{N}}\]
We have calculated earlier that $\sum{{{x}^{2}}}=442$ and N = 5, therefore
\[\begin{align}
  & \sigma =\sqrt{\dfrac{442}{5}} \\
 & \sigma =\sqrt{88.4} \\
 & \sigma =9.4 \\
\end{align}\]
Hence, standard deviation is 9.4
Step 4: Now, let us calculate the coefficient of variation. Coefficient of variation is given as:
\[CV=\dfrac{\sigma }{\overline{X}}\times 100\]
We have calculated earlier that $\sigma =9.4$ and $\text{Mean}=\overline{X}=20$, Therefore,
\[\begin{align}
  & CV=\dfrac{9.4}{20}\times 100 \\
 & \Rightarrow 47 \\
\end{align}\]
Hence, coefficient of variation for given data is 47.

Note: Students should note that the sum of deviation from mean is always zero. They should not ignore the negative signs in the deviation column. Students should try to use calculators for calculating squares and square roots for proper calculations. If calculators are not allowed they can leave the answer in roots, if they don't know how to solve the root. Students should not get confused in $\sum{{{x}^{2}}}\text{ and }{{\left( \sum{x} \right)}^{2}}$. In $\sum{{{x}^{2}}}$ we have added squared terms and in ${{\left( \sum{x} \right)}^{2}}$ we added the terms and then squared the answer.