Standard deviation formula:


If I just want to measure the amount of dispersion from the mean, why can't it be just an average of the absolute differences of each term from the mean?

Something like this:

( |x1 - μ| + |x2 - μ| + ... + |xN - μ| ) / N

Why is this not a good measure of amount of dispersion from the mean? Why is standard deviation better?

  • $\begingroup$ people do use en.wikipedia.org/wiki/Geometric_median $\endgroup$ – Will Jagy Oct 19 '14 at 18:52
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    $\begingroup$ The main reason why statisticians use $\cdot^{2}$ instead of $\left|\cdot\right|$ is because the calculus is much more clean. $\endgroup$ – Clarinetist Oct 19 '14 at 18:53
  • $\begingroup$ As described, if one only wants some measure of the dispersion, then the average of absolute differences is such a measure. However the mean $\mu$ minimizes the sum of squares measure, not the sum of absolute differences measure, and for related reasons the square norm of deviations is often convenient. $\endgroup$ – hardmath Oct 19 '14 at 18:54

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