Ans: C
Let $\overline{x}$ be the mean of the five numbers $x_1$, $x_2$, $x_3$, $x_4$ and $x_5$. Therefore, we have
Let $\overline{x}$ be the mean of the five numbers $x_1$, $x_2$, $x_3$, $x_4$ and $x_5$. Therefore, we have
$\begin{array}{rcl}
\dfrac{1}{5}[(x_1-\overline{x})^2 + \ldots + (x_5-\overline{x})^2] & = & 13
\end{array}$
The mean of the new five numbers
$\begin{array}{cl}
= & \dfrac{1}{5} (3x_1+4 +\ldots +3x_5+4) \\
= & \dfrac{1}{5} [3(x_1+\ldots x_5) +20] \\
= & 3\overline{x} +4
\end{array}$
The variance of the new five numbers
$\begin{array}{cl}
= & \dfrac{1}{5}\{ [ (3x_1+4-(3\overline{x}+4)]^2 + \ldots + [ (3x_5+4-(3\overline{x}+4)]^2\} \\
= & \dfrac{1}{5}[(3x_1-3\overline{x})^2 + \ldots +(3x_5-3\overline{x})^2] \\
= & \dfrac{1}{5} \times 9 [(x_1-\overline{x})^2 + \ldots +(x_5-\overline{x})^2] \\
= & 9\times 13 \\
= & 117
\end{array}$
