Question

Sum of deviation of variate fron their A.M is always zero that is $\sum (x_i-\overline{x})=0$ sum of square of deviation of variate from their A.M is minimum that is $\sum (x_i-\overline{x})=0^2$ is minimum

Answer 1

consider 3 number
3,4,5
$\overline{x}=\frac{3+4+5}{3}=4$
$\sum (x_i-\overline{x})=(3-4)+(4-4)+(5-4)$
$=-1+0+1$
$=0$ (always correct)
$\sum (x_i-\overline{x}{)}^2=(-1{)}^2+o^2+1^2$
$=2$(minimum)