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 $\bar{X}$ is

A
0
B
$\frac{\bar{X}}{n}$
C
$n \bar{X}$
D
none of these

Solution:

The algebraic sum of deviations about mean is

$\sum_{i = 1}^{n} (x_{i} - \bar{X}) = \sum_{i = 1}^{n} x_{i} - \sum_{i = 1}^{n} \bar{X} = n \bar{X} - n \bar{X} = 0$

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