2013-II-45

Ans: C
Let $\bar{x}$ be the mean of the five numbers $x_1$, $x_2$, $x_3$, $x_4$ and $x_5$. Therefore, we have

$\begin{array}{rcl}{\displaystyle \frac{1}{5}}[(x_1-\bar{x}{)}^2+\dots +(x_5-\bar{x}{)}^2]& =& 13\end{array}$

The mean of the new five numbers

$\begin{array}{cl}=& {\displaystyle \frac{1}{5}}(3x_1+4+\dots +3x_5+4)\\ =& {\displaystyle \frac{1}{5}}[3(x_1+\dots x_5)+20]\\ =& 3\bar{x}+4\end{array}$

The variance of the new five numbers

$\begin{array}{cl}=& {\displaystyle \frac{1}{5}}\{[(3x_1+4-(3\bar{x}+4){]}^2+\dots +[(3x_5+4-(3\bar{x}+4){]}^2\}\\ =& {\displaystyle \frac{1}{5}}[(3x_1-3\bar{x}{)}^2+\dots +(3x_5-3\bar{x}{)}^2]\\ =& {\displaystyle \frac{1}{5}}\times 9[(x_1-\bar{x}{)}^2+\dots +(x_5-\bar{x}{)}^2]\\ =& 9\times 13\\ =& 117\end{array}$