🚀 go-pugleaf

It is true that the `usual estimator' for the standard deviation is biased
for the population SD.  However, it might remain popular for the following
reason:

Suppose we want to test the hypothesis  Null:  mean=0  vs.  Alt:  mean != 0, 
but we do not know the population variance.  Then our test takes the form:

 	reject the null if Xbar/sqrt(S^2) > some constant

where S^2 is the unbiased estimator for the variance and the constant is
chosen so that we reject with a certain probability.
So where am I going with this?  Well, under the assumption that the
data are normally distributed with mean 0, the test statistic above (sometimes
called T squared) has a t-distribution with n-1 degrees of freedom; this is 
only true, however, under the additional contstaint that we use the unbiased
estimator for the variance in forming the test-statistic.  Thus the "N-1".

That said, let me make two more notes:
(1)  S^2 is unbiased for population variance, regardless of the underlying
dsitribution (as long as we have finite variance).
(2)  The distribution of the T-statistic is somewhat robust against non-
normal underlying distributions.

I realize that this gets awat a bit from the original question, but I think
it rounds out the defense for using `N-1' in the SD estimate.

Chad