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I'm trying to really understand what SD is telling me, but the information out there is sparse and confusing.

This thread and some others as well provide the answer that SD is L2 distance in euclidian space, and that each of the variables are actually dimensions.

But the formula for distance is sqr((q1-p1)^2+(q2-p2)^2+...+(qn-pn)^2).

if we put our variables and mean into it, we'd get: sqr((x1-m)^2+(x2-m)^2+...+(xn-m)^2)

which is close, but not quite, since SD=sqr(((x1-m)^2+(x2-m)^2+...+(xn-m)^2)/(n-1))

What does that (n-1) do in terms of euclidian distance? I mean, aren't we measuring a distance from one point(x1, x2, ...xn) to (m,m,..,m)?

What does /(n-1) before sqrrooting do in terms of distance?

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  • The formula you quote, although often referred to in brief as a "standard deviation," actually is not a standard deviation: it is an *estimator* of a standard deviation. I also notice in the thread you reference that one of the highest-voted answers states the SD is [a "version of" the L2 distance](http://stats.stackexchange.com/a/121/919) rather than the L2 distance itself. Regardless, since dividing by $n$ or by $n-1$ before taking the root is algebraically identical to dividing by $\sqrt{n}$ or $\sqrt{n-1}$ after taking the root, doesn't that fact alone fully answer your question? – whuber Apr 20 '15 at 16:13
  • Well, frankly, no :-( I get L2 distance over sqrt(n-1). Why sqrt here? – zapp0 Apr 20 '15 at 19:49

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