Ans: (a) $60\text{ marks}$ (b) Yes
- Let $\overline{x}$ and $\sigma$ be the mean mark and the standard deviation respectively. According to Susan’s score and standard score, we have
$\begin{array}{rcl}
\dfrac{90-\overline{x}}{\sigma} & = & 3 \\
\sigma & = & \dfrac{90-\overline{x}}{3}~\ldots\unicode{x2460}
\end{array}$According to Tom’s score and standard score, we have
$\begin{array}{rcl}
\dfrac{65-\overline{x}}{\sigma} & = & 0.5 \\
\sigma & = & \dfrac{65-\overline{x}}{0.5}~\ldots\unicode{x2461}
\end{array}$Sub. $\unicode{x2461}$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
\dfrac{90-\overline{x}}{3} & = & \dfrac{65-\overline{x}}{0.5} \\
90-\overline{x} & = & 390-6\overline{x} \\
5\overline{x} & = & 300 \\
\overline{x} & = & 60
\end{array}$Therefore, the mean is $60$ marks.
- Note that the median is $55$ marks. Since the mean marks is higher than the median (i.e. $60>55$), then there are at least half of the students having marks below mean. Note also that the standard score of the mean below mean is negative. Therefore, the standard scores of at least half of the students in the test are negative. I agree with Susan.
