2013-I-10

$\begin{array}{cl}
= & 31
\end{array}$

The mode

$\begin{array}{cl}
= & 23
\end{array}$

1. Since the range of the distribution is $47$, then we have

   $\begin{array}{rcl}
   (60 + b) – (20 + a) & = & 47 \\
   b-a & = & 7
   \end{array}$

   Therefore, the possible answers are $\left\{ \begin{array}{l}
   a=0 \\
   b=7
   \end{array}\right.$
   or $\left\{ \begin{array}{l}
   a=1 \\
   b=8
   \end{array}\right.$
   or $\left\{ \begin{array}{l}
   a=2 \\
   b=9
   \end{array}\right.$.

2. The required probability

   $\begin{array}{cl}
   = & \dfrac{2}{20}\times\dfrac{3}{13} + \dfrac{7}{20}\times\dfrac{2}{13} +\dfrac{3}{20}\times\dfrac{4}{13} \\
   = & \dfrac{8}{65}
   \end{array}$