Question

Prove that : ${\sum }_{i=1}^n(x_i-\overline{x})=0$

Answer 1

Proof :

Let $x_1,x_2,.........x_n$ are a set of $n$ measurements. Now we have to show that the sum of the deviations of the set about their mean is zero.

Now the mean of the above set of ${}^{\prime }{n}^{\prime }$ observations/measurements is

$\Rightarrow \overline{x}=\frac{x_1+x_2+x_3......+x_n}{n}$

$\Rightarrow \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n}$ $.......\left(1\right)$

Now the deviations of the 'n' measurements $x_1,x_2,x_3.....x_n$ are $(x_1-\overline{x}),(x_2-\overline{x}),(x_3-\overline{x}),......(x_n-\overline{x})$ respectively.

Sum of these deviations is

$\Rightarrow {\sum }_i(x_i-\overline{x})=x_1-\overline{x}+x_2-\overline{x}+x_3-\overline{x}+x_4-\overline{x}.....+x_n-\overline{x}$

$=(x_1+x_2+x_3+x_4........x_n)-n\overline{x}$

from $\left(1\right)$ we can write $x_1+x_2+x_3.......x_n=n\overline{x},$

$\Rightarrow {\sum }_i(x_i-\overline{x})=n\overline{x}-n\overline{x}=0$

Hence, proved.