Question

The sum of squares of deviations for 10 observations taken from mean 50 is 250. The coefficient of variation is\[\]
A. 10%\[\]
B. 40%\[\]
C. 50%\[\]
D. None of these

Answer 1

Hint: Use the given data to calculate the value standard deviation using the basic definition and then use the relation among mean(given in the question), standard deviation and coefficient of variance.

Complete step by step answer:
If there are $n$ number of data points in the sample represented as ${{x}_{1}},{{x}_{2}},{{x}_{3}},...,{{x}_{n}}$ their arithmetic mean is denoted by $\mu $ and is is given by
\[\mu = \dfrac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n} = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \]
The arithmetic mean or simply mean , abbreviated as AM is a measure of central tendency in other words $\mu $ tells us how the data is tending towards a particular centre. \[\]
The quantity that tells us about how the data is dispersed from the centre or mean is variance. The variance is denoted as ${{\sigma }^{2}}$and defined as the sum of squares of difference each data point from the mean. It is given by
\[{{\sigma }^{2}}=\dfrac{{{\left( {{x}_{1}}-\mu \right)}^{2}}+{{\left( {{x}_{2}}-\mu \right)}^{2}}+{{\left( {{x}_{3}}-\mu \right)}^{2}}+...+{{\left( {{x}_{n}}-\mu \right)}^{2}}}{n}=\sum\limits_{i=1}^{n}{\dfrac{1}{n}{{({{x}_{i}}-\mu )}^{2}}}\]
The quantity that tells us about how much the data from the centre is deviated is called standard deviation. It is denoted by $\sigma $and defined as the positive square root of variance
\[\sigma =\sqrt{{{\sigma }^{2}}}=\sqrt{\sum\limits_{i=1}^{n}{\dfrac{1}{n}{{({{x}_{i}}-\mu )}^{2}}}}\]
When $\sigma $approaches to 0 the data points are closer to each other and when $\sigma $becomes greater than 0 , the data points are farther from each other. \[\]
The coefficient of variation is a standardized and practical measure of data dispersion. It is given in percentage of the ratio of standard deviation to arithmetic mean. It is denoted by $CV$and given by
 \[\text{CV}=\dfrac{\sigma }{\mu }\times 100\]
If the value of CV is lower then there is less variability and more stability in the data and if CV is higher then there is more variability and more volatility in the data. \[\]
As given in the question, the size of the sample is $n=10$ and the mean is $\mu =50$. It is also given that the sum of squares of deviation is 250. So,
\[{{\left( {{x}_{1}}-\mu \right)}^{2}}+{{\left( {{x}_{2}}-\mu \right)}^{2}}+{{\left( {{x}_{3}}-\mu \right)}^{2}}+...+{{\left( {{x}_{n}}-\mu \right)}^{2}}=\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\mu )}^{2}}}=250\]
We find the standard deviation using the formula
\[\sigma =\sqrt{\sum\limits_{i=1}^{n}{\dfrac{1}{n}{{({{x}_{i}}-\mu )}^{2}}}}=\sqrt{\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\mu )}^{2}}}}=\sqrt{\dfrac{1}{10}\times 250}=5\]
The negative value is rejected as $\sigma $ is a positive quantity.
So the coefficient of variation is given by
$\text{CV}=\dfrac{5}{50}\times 100=10\%$\[\]

So, the correct answer is “Option A”.

Note: The question tests your knowledge of basic statistical terms. Careful substitution and use of formula will lead us to the correct result. The question can also be framed to find out the variance. The coefficient of variation used in risk analysis.