Question

Two teams $A$ and $B$ have the same mean and their coefficients of variation are $4,2$ , respectively. If ${\sigma }_A,{\sigma }_B$ are the standard deviations of teams $A,B$ respectively, then the relation between them is

• A. ${\sigma }_A={\sigma }_B$
• B. ${\sigma }_B=2{\sigma }_A$
• C. ${\sigma }_A=2{\sigma }_B$
• D. ${\sigma }_B=4{\sigma }_A$

Answer 1

Answer: (C)

We have,
${\bar{X}}_A={\bar{X}}_B\text{ and }C{V}_A=4,C{V}_B=2$
Let ${\sigma }_A$ and ${\sigma }_B$ be the standard deviations of teams $A$ and $B$ respectively.
Then, $CV=\frac{\sigma }{\bar{X}}\times 100$ or $\sigma =\frac{CV\times \bar{X}}{100}$
$\therefore {\sigma }_A=\frac{4{\bar{X}}_A}{100}$ and ${\sigma }_B=\frac{2{\bar{X}}_B}{100}$
$\Rightarrow {\bar{X}}_A=25{\sigma }_A$ and ${\bar{X}}_B=50{\sigma }_B$
But ${\bar{X}}_A={\bar{X}}_B$
$\therefore 25{\sigma }_A=50{\sigma }_B$
$\Rightarrow {\sigma }_A=2{\sigma }_B$