I must be true. Since $x_1 = a_{50}$, then for the first group of numbers,
$\begin{array}{rcl}
\dfrac{a_1 + a_2 + \cdots + a_{49} + x_1}{50} & = & x_1 \\
a_1 + a_2 + \cdots + z_{49} & = & 49x_1
\end{array}$
For the second group of numbers,
$\begin{array}{rcl}
\dfrac{a_1 + a_2 + \cdots + a_{49}}{49} & = & x_2 \\
\dfrac{49x_1}{49} & = & x_2 \\
x_1 & = & x_2
\end{array}$
II may not be true. Without loss of generality, assume that $a_1\le a_2\le \ldots \le a_{49}$. Let $a_{23} = x_1$. For the first group of numbers,
$\begin{array}{rcl}
y_1 & = & \dfrac{a_{24} + a_{25}}{2}
\end{array}$
For the second group of numbers,
$\begin{array}{rcl}
y_2 & = & a_{25}
\end{array}$
Note that $a_{24} \le a_{25}$. Therefore
$\begin{array}{rcl}
\dfrac{a_{24} + a_{25}}{2} & \le & a_{25} \\
y_1 & \le & y_2
\end{array}$
III must be true. Since $x_1=a_{50}$, then for the first group of numbers,
$\begin{array}{rcl}
z_1 & = & \dfrac{(a_1 – x_1)^2 + \cdots + (a_{49}-x_1)^2 + (x_1 – x_1)^2}{50} \\
z_1 & = & \dfrac{(a_1 – x_1)^2 + \cdots + (a_{49}-x_1)^2}{50} \ \ldots \unicode{x2460}
\end{array}$
Since $x_1=x_2$, then for the second group of numbers,
$\begin{array}{rcl}
z_2 & = & \dfrac{(a_1 – x_2)^2 + \cdots + (a_{49}-x_2)^2}{49} \\
z_2 & = & \dfrac{(a_1 – x_1)^2 + \cdots + (a_{49}-x_1)^2}{49} \ \ldots \unicode{x2461}
\end{array}$
By comparing $\unicode{x2460}$ and $\unicode{x2461}$, we have $z_1\le z_2$.
