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,
$\begin{array}{rl}& X\\ & X_1\\ & X_2\\ & .\\ & .\\ & .\\ & X_n\end{array}$
Then the arbitrary constants,
  $\begin{array}{rl}& X-A\\ & X_1-A\\ & X_2-A\\ & .\\ & .\\ & .\\ & X_n-A\end{array}$
Then we have to square this we get,
$\begin{array}{rl}& {(X-A)}^2\\ & {(X_1-A)}^2\\ & {(X_2-A)}^2\\ & .\\ & .\\ & .\\ & {(X_n-A)}^2\end{array}$
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 $A=\bar{X}$
Always the value of this summation is greater than the similar value around mean.
So we have,
$\Rightarrow \sum {(X-A)}^2\ge \sum (X-\bar{X})\mathrm{\forall }A$
Also we can simply say,
$\Rightarrow \sum {(X-A)}^2\succ \sum {(X-X)}^2,A\ne \bar{X}$
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,
$\Rightarrow {\displaystyle \frac{\partial Y}{\partial A}}=2\sum (X-A)(-1)$
We know the first order condition,
That is,
$\Rightarrow {\displaystyle \frac{\partial Y}{\partial A}}=0$
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 $nA$
$\Rightarrow \sum X=nA$
$\Rightarrow A={\displaystyle \frac{\sum X}{n}}$
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,
$\Rightarrow {\displaystyle \frac{\partial Y}{\partial A}}=2\sum (X-A)(-1)$
Here $X$ is constant, $A$ is variable.
$\Rightarrow {\displaystyle \frac{{\partial }^{2}Y}{\partial A^2}}=-2\sum (-1)$
$\Rightarrow {\displaystyle \frac{{\partial }^{2}Y}{\partial A^2}}=2n\succ 0$
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.