答案:(a) $2$ (b) $4$ (c) $\dfrac{9}{20}$
- 由於 $2$ 的扇形在整個圓形圖中為最大,所以該分佈的眾數為 $2$。
- $5$ 的扇形的圓心角
$\begin{array}{cl}
= & 360^\circ – 90^\circ – 144^\circ – 54^\circ \\
= & 72^\circ
\end{array}$所以,該分佈的平均值
$\begin{array}{cl}
= & \dfrac{2 \times 144^\circ + 3 \times 54^\circ + 5 \times 72^\circ + 7 \times 90^\circ}{360^\circ} \\
= & 4
\end{array}$ - 所求的概率
$\begin{array}{cl}
= & \dfrac{72^\circ + 90^\circ}{360^\circ} \\
= & \dfrac{9}{20}
\end{array}$