# Why doesn't Stdev take absolute value of x- xbar? [duplicate]

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

• 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
Commented Jan 19, 2021 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 Commented Jan 19, 2021 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
Commented Jan 19, 2021 at 18:59
• I believe what you're proposing is called "mean absolute deviation", there's a previous question about the differences here: stats.stackexchange.com/questions/81986/… Commented Jan 19, 2021 at 19:15