Ans: B
Let $\sigma$ be the standard deviation of the test.
Let $\sigma$ be the standard deviation of the test.
Since Peter gets $46$ marks and his standard score is $-2.2$, we have
$\begin{array}{rcl}
\dfrac{46 – 68}{\sigma} & = & -2.2 \\
\sigma & = & 10
\end{array}$
Hence, the standard score of Susan
$\begin{array}{cl}
= & \dfrac{52 – 68}{10} \\
= & -1.6
\end{array}$
