Show that for ANY data set, the sum of the positive deviations from the eman must be exactly equal to the sum of negative deviations from the mean?

I can work this out when subbing in values, but can anyone prove it with formulae?

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- Mar 20th 2009, 10:23 PMmibamarsPositive Deviations = Negative Deviations?
Show that for ANY data set, the sum of the positive deviations from the eman must be exactly equal to the sum of negative deviations from the mean?

I can work this out when subbing in values, but can anyone prove it with formulae? - Mar 22nd 2009, 04:25 AMHallsofIvy
I presume that by "deviation from the mean" you mean simply the value minus the mean.: $\displaystyle x_i- \mu$. Mean itself is defined by $\displaystyle \mu= \frac{\sum x_i}{n}$ so the sum of deviations is $\displaystyle \sum (x_i-\mu)= \sum x_i- \sum \mu$. But $\displaystyle \mu$ is a constant so $\displaystyle \sum \mu= n\mu=n\frac{\sum x_i}{n}= \sum x_i$. Putting those together, $\displaystyle \sum (x_i- \mu)= \sum x_i- \sum x_i= 0$. Since that sum is 0, the negative values and positive values must cancel.