If $M$ is the mean of $x_1, x_2, x_3, x_4, x_5$ and $x_6$, prove that

$(x_1 - M) + (x_2 - M) + (x_3 - M) + (x_4 - M) + (x_5 - M) + (x_6 - M) = 0$.

Given:

$M$ is the mean of $x_1, x_2, x_3, x_4, x_5$ and $x_6$.

To do:

We have to prove that $(x_1 - M) + (x_2 - M) + (x_3 - M) + (x_4 - M) + (x_5 - M) + (x_6 - M) = 0$.

Solution:

We know that,

Mean $\overline{X}=\frac{Sum\ of\ the\ observations}{Number\ of\ observations}$

$M=\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}$

$\Rightarrow x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=6 \mathrm{M}$

LHS $=(x_{1}-\mathrm{M})+(x_{2}-\mathrm{M})+(x_{3}-\mathrm{M})+(x_{4}-\mathrm{M})+(x_{5}-\mathrm{M})+(x_{6}-\mathrm{M})$

$=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}-\mathrm{M}-\mathrm{M}-\mathrm{M}-\mathrm{M}-\mathrm{M}-\mathrm{M}$

$=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}-6 \mathrm{M}$

$=6 \mathrm{M}-6 \mathrm{M}$

$=0$

$=$ RHS

Tutorialspoint

Updated on: 10-Oct-2022

33 Views

Kickstart Your Career

Get certified by completing the course

Get Started