Question

If the variance of $1,2,3,4,...,10$ is $\frac{99}{12}$, then the standard deviation of $3,6,9,12,...,30$ is

• A. $\frac{297}{4}$
• B. $\frac{3}{2}\sqrt{33}$
• C. $\frac{3}{2}\sqrt{99}$
• D. $\frac{\sqrt{99}}{12}$
• E. $\frac{3\sqrt{3}}{2}$

Answer 1

The correct option is B

$\frac{3}{2}\sqrt{33}$

Explanation for the correct option :

Find the standard deviation,

Each term in the series $3,6,9,12,...,30$ is three times the corresponding term in the series $1,2,3,4,...,10$.

So the variance of $3,6,9,12,...,30$ is $3^2$ times the variance of $1,2,3,4,...,10$.

The variance of $1,2,3,4,...,10$ is $\frac{99}{12}$.

So the variance of the series $3,6,9,12,...,30$ is: $3^2\times \frac{99}{12}$.

It is known that standard deviation is the square root of the variance.

So the standard deviation of the series $3,6,9,12,...,30$ is:

$\begin{array}{rcl}\sqrt{3^2\times \frac{99}{12}}& =& \sqrt{3^2}\times \sqrt{\frac{33}{4}}\\ & =& 3\times \frac{\sqrt{33}}{\sqrt{2^2}}\\ & =& \frac{3}{2}\sqrt{33}\end{array}$

Hence, the correct option is B.