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# If the variance 1,2,3,4,5....10 is $\large\frac{99}{12}$ then the standard deviation of 3,6,9,12....30 is

$\begin {array} {1 1} (A)\;\large\frac{297}{4} & \quad (B)\;\large\frac{3}{2}\sqrt{33} \\ (C)\;\large\frac{3}{2}\sqrt{99} & \quad (D)\;\sqrt{\large\frac{99}{12}} \end {array}$

Given $\sigma^2_{10}=\large\frac{99}{12}$
$= \large\frac{33}{4}$
$\sigma_{10} = \large\frac{\sqrt{33}}{2}$
SD of required series = $3\sigma_{10}$
$= \large\frac{3}{2}$$\sqrt{33}$