Ans: (a) $7$ (b) $40$ (c) $\dfrac{1}{10}$ (d) Pie chart: no; Bar chart: yes, double the height of each bar
- Since the numbers of students and the angles at centre of the pie chart are in proportion, then we have
$\begin{array}{rcl}
\dfrac{17}{153^\circ} & = & \dfrac{k}{63^\circ} \\
k & = & \dfrac{17 \times 63^\circ}{153^\circ} \\
k & = & 7
\end{array}$ - The number of students in class $A$
$\begin{array}{cl}
= & 17 \div \dfrac{153^\circ}{360^\circ} \\
= & 40
\end{array}$ - The number of students having $1$ key
$\begin{array}{cl}
= & 40 – 12 – 17 – 7 \\
= & 4
\end{array}$Therefore, the required probability
$\begin{array}{cl}
= & \dfrac{4}{40} \\
= & \dfrac{1}{10}
\end{array}$ - There is no modification need for the pie chart. And for the bar chart, the lengths of bars should be doubled.
