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 $\bar{X}_{A}=\bar{X}_{B}$
$\therefore 25 \sigma_{A}=50 \sigma_{B}$
$\Rightarrow \sigma_{A}=2 \sigma_{B}$