The algebraic sum of deviation of a set of $n$ observation from their means is

0

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\frac{n (n + 1)}{2}

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\frac{n (n - 1)}{2}

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\frac{n + 1}{2}

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Solution

The correct option is A $0$
Let $a_{1}, a_{2}, . . . . . a_{n}$ be the $n$ observations.
Therefore,
$¯ x = \frac{a_{1} + a_{2} + . . . . . + a_{n}}{n}$
$\Rightarrow a_{1} + a_{2} + . . . . . + a_{n} = n ¯ x$
Now,
$M = (a_{1} - ¯ x) + (a_{2} - ¯ x) + . . . . . + (a_{n} - ¯ x)$
$M = (a_{1} + a_{2} + . . . . . + a_{n}) - (¯ x + ¯ x + . . . . . n times)$
$\Rightarrow M = n ¯ x - n ¯ x = 0$

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