2006-II-34

答案：B
I 為錯誤。第 1 組數的平均值

$\begin{array}{cl}
= & \dfrac{(x-6)+(x-3)+(x+4)+(x+5)}{4} \\
= & x
\end{array}$

第 2 組數的平均值

$\begin{array}{cl}
= & \dfrac{(x-8)+(x-1)+(x+2)+(x+9)}{4} \\
= & x + \dfrac{1}{2}
\end{array}$

II 為正確。第 1 組數的中位數

$\begin{array}{cl}
= & \dfrac{1}{2}[(x-3) + (x+4)] \\
= & x + \dfrac{1}{2}
\end{array}$

第 2 組數的中位數

$\begin{array}{cl}
= & \dfrac{1}{2}[(x-1) + (x+2)] \\
= & x + \dfrac{1}{2}
\end{array}$

III 為錯誤。第 1 組數的分佈域

$\begin{array}{cl}
= & (x+5) – (x-6) \\
= & 11
\end{array}$

第 2 組數的分佈域

$\begin{array}{cl}
= & (x+9) – (x-8) \\
= & 17
\end{array}$