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Math Help - Simple proof

  1. #1
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    Simple proof

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

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

    Thanks in advance.
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  2. #2
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    Maybe use the fact that that \bar{x} = \frac{\sum x_j}{n}
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  3. #3
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    Hello, Iceflash234!

    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-\overline{x }) \;=\;0

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


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


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

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