Sum of Deviations from Mean

Theorem
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a set of real numbers.

Let $\overline x$ denote the arithmetic mean of $S$.

Then:


 * $\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x} = 0$

Proof
For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.

Then:

Also see

 * Sum of Squared Deviations from Mean