Ans: D
Let $\mu \text{ marks}$ and $\sigma \text{ marks}$ be the mean and the standard deviation of the examination scores respectively.
Let $\mu \text{ marks}$ and $\sigma \text{ marks}$ be the mean and the standard deviation of the examination scores respectively.
$\left\{ \begin{array}{ll}
\dfrac{55 – \mu}{\sigma} = -3 & \ldots \unicode{x2460} \\
\dfrac{95 – \mu}{\sigma} = 2 & \ldots \unicode{x2461}
\end{array} \right.$
$\unicode{x2461} \div \unicode{x2460}$, we have
$\begin{array}{rcl}
\dfrac{95 – \mu}{\sigma} \div \dfrac{55 – \mu}{\sigma} & = & \dfrac{2}{-3} \\
\dfrac{95 – \mu}{55 – \mu} & = & \dfrac{2}{-3} \\
-285 + 3\mu & = & 110 – 2 \mu \\
5 \mu & = & 395 \\
\mu & = & 79
\end{array}$
Therefore, the mean marks is $79\text{ marks}$.
