Effect of adding and removing data on variance Consider a set of distinct numbers. After removing both the max and the min from the set and adding the median to the set, the set of numbers obviously becomes less dispersed and the variance should decrease. How do we prove this result formally?
I have attempted to work with the definition and the expansion seems a mess. It does not appear a feasible approach at all.

Thank you very much for @whuber♦'s detailed explanation. But I actually meant to include the median once only, so that we are moving from a set of $n$ numbers to a set of $n-1$ numbers. I attempted to follow your argument and consider the $X=\{{{x}_{2}},\cdots ,{{x}_{n-1}}\}$, $Y=\{{{x}_{1}},{{x}_{50}}\}$ and $Y'=\{{{x}_{M}}\}$, where ${{\mu }_{1}}(X,Y)$ is assumed to 0. Then I obtained
$\begin{align}
  & {{\Delta }_{X}}(Y,Y')=\operatorname{Var}(X,Y)-\text{Var}(X,Y') \\ 
 & =[{{\mu }_{2}}(X,Y)-{{\mu }_{1}}{{(X,Y)}^{2}}]-[{{\mu }_{2}}(X,Y')-{{\mu }_{1}}{{(X,Y')}^{2}}] \\ 
 & =\left[ \frac{\sum\limits_{i=1}^{n}{x_{i}^{2}}}{n}-0 \right]-\left[ \frac{x_{M}^{2}+\sum\limits_{i=2}^{n-1}{x_{i}^{2}}}{n-1}-{{\left( \frac{{{x}_{M}}+\sum\limits_{i=2}^{n-1}{{{x}_{i}}}}{n-1} \right)}^{2}} \right] \\ 
 & =\frac{\sum\limits_{i=1}^{n}{x_{i}^{2}}}{n}-\frac{x_{M}^{2}+\sum\limits_{i=2}^{n-1}{x_{i}^{2}}}{n-1}+\frac{{{({{x}_{M}}-{{x}_{1}}-{{x}_{n}})}^{2}}}{{{(n-1)}^{2}}} \\ 
 & =\frac{(n-1)\sum\limits_{i=1}^{n}{x_{i}^{2}}-nx_{M}^{2}-n\sum\limits_{i=2}^{n-1}{x_{i}^{2}}}{n(n-1)}+\frac{{{({{x}_{M}}-{{x}_{1}}-{{x}_{n}})}^{2}}}{{{(n-1)}^{2}}} \\ 
 & =\frac{(n-1)(x_{1}^{2}+x_{n}^{2})-nx_{M}^{2}-\sum\limits_{i=2}^{n-1}{x_{i}^{2}}}{n(n-1)}+\frac{{{({{x}_{M}}-{{x}_{1}}-{{x}_{n}})}^{2}}}{{{(n-1)}^{2}}}  
\end{align}$
It is not obvious to me how this messy expression can be further simplified in order to establish its non-negativity. Would you mind pointing it out? Thank you.
 A: Breaking the problem down into conceptually distinct parts makes it soluble.  This is one general approach to dealing with variances.

Framing the problem
We may view the data as a collection of numbers (which needn't be distinct) $x_1, x_2, \ldots, x_n$ where $n\ge 2,$ $x_1$ is the smallest, and $x_n$ is the largest.  Partition this into two groups: $X=(x_2, x_3, \ldots, x_{n-1})$ of $n-2$ numbers and $Y=(x_1,x_n)$ of $2$ numbers.  The problem asks what happens when $Y$ is replaced by $Y^\prime=(x_M)$ where $x_M$ is the median of all $n$ values.

Simplifying the problem with preliminary manipulations and inequalities
Because the variance does not change when all values are shifted, we may assume the midrange of all the values is $0$: that is, $x_1 = -x_n.$
Choose a unit of measure in which $x_n=1.$  This is always possible unless $x_1=x_n=0,$ in which case it is obvious that the initial and final variances are both zero.
Note that since now all values lie between $-1$ and $1$, the median $x_M$ and the mean $\bar x$ of $X$ also lie between $-1$ and $1.$  That is,
$$-1=x_1 \le \bar x \le x_n = 1.\tag{*}$$
Also, the variance $\sigma^2$ of $X$ cannot exceed $1.$  Let $\delta=x_M-\bar x$ be the difference between the median and mean. It is a simple exercise to show that 
$$\delta^2 \le \sigma^2 \le 1.\tag{**}$$
Familiar Definitions
Moments
The (raw) moment of degree $k$ of a collection of numbers is the arithmetic mean of their $k^\text{th}$ powers.  For convenience, when $Z$ denotes a collection of numbers $(z_1, z_2, \ldots, z_m),$ let's write
$$\mu_k(Z) = \frac{1}{m}\sum_{i=1}^m z_i^k$$
for their $k^\text{th}$ moment.
Behavior of moments under partitioning
When a collection of numbers $Z$ is partitioned into $Z = (X,Y)$ with $X=(x_1,\ldots,x_n)$ and $Y=(y_1,\ldots,y_m),$ the foregoing formula breaks into two separate sums, yielding
$$\mu_k(Z) = \frac{1}{n+m}\left(n\,\mu_k(X) + m\,\mu_k(Y)\right).$$
Variance
One formula for the variance of a set of numbers is their second moment minus the square of their first moment:
$$\operatorname{Var}(Z) = \mu_2(Z) - \mu_1(Z)^2.$$

Solution

Step 1: The formula
We need to find a simple formula for $\Delta_X(Y,Y^\prime)=\operatorname{Var}(X,Y) - \operatorname{Var}(X,Y^\prime)$ and then analyze it: the problem is to show this is never negative.
The preceding formulas algebraically simplify, without any trouble, to
$$\eqalign{
n^2(n-1)^2\Delta_X(Y,Y^\prime) &= n(2 + n(n-2)(2-\delta^2))\\
&+ (n-1)(n-2)\left(2(n-1)(\bar x)^2 - n\sigma^2\right).
}$$

Step 2: Analysis of the formula
Apply the inequalities in $(*)$ and $(**)$ by substituting $\delta^2=1,$ $\sigma^2=1,$ and $\bar x = 0$ to obtain
$$\eqalign{&n^2(n-1)^2\Delta_X(Y,Y^\prime)\\& \ge n(2 + n(n-2)(2-1^2)) + (n-1)(n-2)(2(n-1)(0)^2 - n(1)) \\
&=n^2 \gt 0,
}$$
QED.
