Ans: (a) $2$ (b) $4$ (c) $\dfrac{9}{20}$
- Since the sector of $2$ in the pie chart is the largest, then the mode of the distribution is $2$.
- The angle of the sector of $5$
$\begin{array}{cl}
= & 360^\circ – 90^\circ – 144^\circ – 54^\circ \\
= & 72^\circ
\end{array}$Therefore, the mean of the distribution
$\begin{array}{cl}
= & \dfrac{2 \times 144^\circ + 3 \times 54^\circ + 5 \times 72^\circ + 7 \times 90^\circ}{360^\circ} \\
= & 4
\end{array}$ - The required probability
$\begin{array}{cl}
= & \dfrac{72^\circ + 90^\circ}{360^\circ} \\
= & \dfrac{9}{20}
\end{array}$
