Ans: D
For the group $\{a-7, a-1, a, a+2, a+4, a+8\}$.
For the group $\{a-7, a-1, a, a+2, a+4, a+8\}$.
The mean
$\begin{array}{cl}
= & \dfrac{(a-7)+(a-1)+a+(a+2)+(a+4)+(a+8)}{6} \\
= & a+1
\end{array}$
The median
$\begin{array}{cl}
= & \dfrac{a+(a+2)}{2} \\
= & a+1
\end{array}$
The range
$\begin{array}{cl}
= & (a+8)-(a-7) \\
= & 15
\end{array}$
For the group $\{a-9, a-2, a-1, a+3, a+4, a+6\}$.
The mean
$\begin{array}{cl}
= & \dfrac{(a-9)+(a-2)+(a-1)+(a+3)+(a+4)+(a+6)}{6} \\
= & \dfrac{6a+1}{6}
\end{array}$
The median
$\begin{array}{cl}
= & \dfrac{(a-1)+(a+3)}{2} \\
= & a+1
\end{array}$
The range
$\begin{array}{cl}
= & (a+6) – (a-9) \\
= & 15
\end{array}$
Therefore, the two means are not equal, the two medians and ranges are the same.
