# Positive Deviations = Negative Deviations?

• Mar 20th 2009, 11:23 PM
mibamars
Positive 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, 05:25 AM
HallsofIvy
Quote:

Originally Posted by mibamars
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?

I presume that by "deviation from the mean" you mean simply the value minus the mean.: $x_i- \mu$. Mean itself is defined by $\mu= \frac{\sum x_i}{n}$ so the sum of deviations is $\sum (x_i-\mu)= \sum x_i- \sum \mu$. But $\mu$ is a constant so $\sum \mu= n\mu=n\frac{\sum x_i}{n}= \sum x_i$. Putting those together, $\sum (x_i- \mu)= \sum x_i- \sum x_i= 0$. Since that sum is 0, the negative values and positive values must cancel.