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Newbie here. Curious why standard deviation subtracts x from xbar and then ^2's them instead of skipping hte squaring/square-rooting and instead takes the ABS value of each x-xbar Thanks mods

user308710
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  • I'll expand on this is an answer, but the gist is that $\sqrt{a^2 + b^2} \ne \vert a \vert+ \vert b \vert$. Try it out with $a=3$ and $b=4$: $\sqrt{3^2 + 4^2} = 5 \ne \vert 3 \vert + \vert 4 \vert = 7$. – Dave Jan 19 '21 at 18:51
  • @Dave thank you. but, in this example, we don't subtract a nor b from any value (xbar or whatever), so of course | | absolute values become a moot function.Would you be willing to clarify why a2+b2−−−−−−√≠|a|+|b| holds true in this scenario?

    Thank you

     – user308710 Jan 19 '21 at 18:57
  • Let $x_1=2$ and $x_2 = 4$. Then $\bar{x} = 3$. Run those numbers through the equations. You will get different results. (It would be helpful if you posted what you propose as an alternative equation for standard deviation.) – Dave Jan 19 '21 at 18:59
  • I believe what you're proposing is called "mean absolute deviation", there's a previous question about the differences here: https://stats.stackexchange.com/questions/81986/mean-absolute-deviation-vs-standard-deviation – Max S. Jan 19 '21 at 19:15

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