2013-I-10

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

眾數

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

1. 因為該分佈的分佈域為 $47$，所以

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

   因此，可能的答案為 $\left\{ \begin{array}{l}
   a=0 \\
   b=7
   \end{array}\right.$
   或 $\left\{ \begin{array}{l}
   a=1 \\
   b=8
   \end{array}\right.$
   或 $\left\{ \begin{array}{l}
   a=2 \\
   b=9
   \end{array}\right.$。

2. 所求的概率為

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