Ans: (a) $12$ (b) $19$ minutes (c) $\dfrac{3}{5}$
- Note that $a=3$.
$\begin{array}{rcl}
x & = & 9 +a \\
x & = & 9 +3 \\
x & = & 12
\end{array}$ -
$\begin{array}{rcl}
y +3 & = & 20 \\
y & = & 17
\end{array}$Also,
$\begin{array}{rcl}
a+9+b & = & y \\
3 +9 +b & = & 17 \\
b & = & 5
\end{array}$Hence, the mean of the distribution
$\begin{array}{cl}
= & \dfrac{3\times 12+9 \times 17 +5 \times 22 +3 \times 27}{3+9+5+3} \\
= & 19 \text{ minutes}
\end{array}$ - The required probability
$\begin{array}{cl}
= & \dfrac{x}{3+9+5+3} \\
= & \dfrac{12}{20} \\
= & \dfrac{3}{5}
\end{array}$
