- Note that the mean is $2$, then
$\begin{array}{rcl}
\dfrac{1\times 8 +2\times 5 +3\times n +4\times 1}{8+5+n+1} & = & 2 \\
22+3n & = & 2(14+n) \\
22 +3n & = & 28 +2n \\
n & = & 6
\end{array}$Therefore, there are $20$ students in the class.
The median
$\begin{array}{cl}
= & \dfrac{\text{the $10$th datum} + \text{the $11$th datum}}{2} \\
= & \dfrac{2+2}{2} \\
= & 2
\end{array}$The inter-quartile range
$\begin{array}{cl}
= & Q_3 – Q_1 \\
= & \dfrac{\text{the $15$th datum} + \text{the $16$th datum}}{2} -\dfrac{\text{the $5$th datum} + \text{the $6$th datum}}{2} \\
= & \dfrac{3+3}{2} -\dfrac{1+1}{2} \\
= & 3 -1 \\
= & 2
\end{array}$The variance $=0.9$.
- Note that the mean is not affected after the withdrawal of the two students. Then the mean of the numbers of calculators owned by the two withdrawn students is $2$.
Therefore, the numbers of calculator owned by the withdrawn students are $1$ and $3$, or $2$ and $2$.
For these two cases, the largest datum and the smallest datum are still $4$ and $1$ respectively.
Therefore, the range of the distribution remains unchanged.
2023-I-11
Ans: (a) Median $=2$, inter-quartile range $=2$, variance $=0.9$ (b) No
