If X is the mean of  x1,x2,x3…xn Then, the algebraic sum of the deviations about mean X¯ is

# If X is the mean of  ${x}_{1},{x}_{2},{x}_{3}\dots {x}_{n}$ Then, the algebraic sum of the deviations about mean $\overline{X}$ is

1. A

0

2. B

$\frac{\overline{X}}{n}$

3. C

$n\overline{X}$

4. D

none of these

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### Solution:

The algebraic sum of deviations about mean is

$\sum _{i=1}^{n} \left({x}_{i}-\overline{X}\right)=\sum _{i=1}^{n} {x}_{i}-\sum _{i=1}^{n} \overline{X}=n\overline{X}-n\overline{X}=0$

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