1. Simple proof

I need help proving that the sum of deviations of a set of measurements about their mean is 0.

$\displaystyle \sum_{j=1}^n (x_j-$$\displaystyle ) = 0$

2. Maybe use the fact that that $\displaystyle \bar{x} = \frac{\sum x_j}{n}$

3. Hello, Iceflash234!

I need help proving that the sum of deviations
of a set of measurements about their mean is 0.

. . $\displaystyle \displaystyle \sum_{j=1}^n (x_j-\overline{x }) \;=\;0$

$\displaystyle \displaystyle \text{We have: }\;S \;\;=\;\;\sum ^n_{j=1}(x_j - \overline{x}) \;\;=\;\;\sum^n_{j=1}x_j - \sum^n_{j=1} \overline{x}$

$\displaystyle \displaystyle \text{Since }\overline{x} \text{ is a constant, we have: }\; S \;\;=\;\;\sum^n_{j=1}x_j - n\overline{x}$

$\displaystyle \displaystyle \text{Multiply the sum by }\dfrac{n}{n}\!:\;\;\;S \;\;=\;\; n\,\cdot$ $\displaystyle \underbrace{\begin{array}{c}_n \;\;\;\; \\ \sum x_j\\ ^{j=1}\;\;\; \\ \hline n \end{array}}_{\text{This is }\overline{x}} -\; n\overline{x}$

Therefore: . $\displaystyle S \;\;=\;\;n\overline{x} - n\overline{x}\;\;=\;\;0$