Question

Sum of squares of the deviations is minimum when deviations are taken from
$A.$ Mean
$B.$ Median
$C.$ Mode
$D.$ Zero

Answer 1

Hint:
The arithmetic average of the series disperses the whole series into individual minimum values therefore the sum of their squares is minimum.

Complete step by step solution:
When we take deviations around any arbitrary constant and we square them and sum them it will be minimum that is, Let the observations,
$\begin{align}
  & X \\
 & {{X}_{1}} \\
 & {{X}_{2}} \\
 & . \\
 & . \\
 & . \\
 & {{X}_{n}} \\
\end{align}$
Then the arbitrary constants,
  $\begin{align}
  & X-A \\
 & {{X}_{1}}-A \\
 & {{X}_{2}}-A \\
 & . \\
 & . \\
 & . \\
 & {{X}_{n}}-A \\
\end{align}$
Then we have to square this we get,
$\begin{align}
  & {{(X-A)}^{2}} \\
 & {{({{X}_{1}}-A)}^{2}} \\
 & {{({{X}_{2}}-A)}^{2}} \\
 & . \\
 & . \\
 & . \\
 & {{({{X}_{n}}-A)}^{2}} \\
\end{align}$
Then we sum the squares value we get $\sum{{{(X-A)}^{2}}}$
Basically we need to minimize the term $\sum{{{(X-A)}^{2}}}$
This sum is minimum only when $A=\overline{X}$
Always the value of this summation is greater than the similar value around mean.
So we have,
$\Rightarrow \sum{{{(X-A)}^{2}}\ge \sum{(X-\overline{X})\forall A}}$
Also we can simply say,
$\Rightarrow \sum{{{(X-A)}^{2}}\succ \sum{{{(X-X)}^{2}},A\ne \overline{X}}}$
Both are the same value.
How we can prove this means let take,
$\Rightarrow Y=\sum{{{(X-A)}^{2}}}$
If we want to minimize the term we know that we need to differentiate the term with respect to $A$
Then differentiate the term we get,
$\Rightarrow \dfrac{\partial Y}{\partial A}=2\sum{(X-A)(-1)}$
We know the first order condition,
That is,
$\Rightarrow \dfrac{\partial Y}{\partial A}=0$
So we take,
$\Rightarrow 2\sum{(X-A)(-1)=0}$
We do further,
$\Rightarrow 2\sum{X=2\sum{A}}$
Then we have,
$\Rightarrow \sum{X=\sum{A}}$
Then we can say the summation of arbitrary constant as $nA$
$\Rightarrow \sum{X=nA}$
$\Rightarrow A=\dfrac{\sum{X}}{n}$
Now the first order condition is about two things one is maximum and another is minimum. So we go for second order conditions. So we get,
$\Rightarrow \dfrac{\partial Y}{\partial A}=2\sum{(X-A)(-1)}$
Here $X$ is constant, $A$ is variable.
$\Rightarrow \dfrac{{{\partial }^{2}}Y}{\partial {{A}^{2}}}=-2\sum{(-1)}$
$\Rightarrow \dfrac{{{\partial }^{2}}Y}{\partial {{A}^{2}}}=2n\succ 0$
Now we are minimized the term $\sum{{{(X-A)}^{2}}}$
So we conclude that,
Sum of squares of the deviations is minimum when deviations are taken from the mean.

Note:
The sum of the squares of deviations of a set of values is minimum when taken about arithmetic mean.