Re: variance aggregate function incorrect? Reference Materials

Joseph Syjuco wrote:
> hi
> i needed the variance function ... i dont know if i introduced the wrong
> parameters or maybe this variance is not the variance that im looking
> for but it doesnt provide the right results
>
> variance=(nEx^2 - (Ex)^2)/(n(n-1))
> my sql statement
> select variance(answer) from tbl_answer (where answer is of type
> integer)
>
> because of this i need to create my own variance aggregate (unless
> someone enlightens me on whats happening with the variance aggregate) so
> i would like to know if theres anyone who knows a good create aggregate
> tutorial (with samples) ... all i got from google so far is the
> reference from ninthwonder.com

Uh, I tried this:

test=> select variance(relnatts) from pg_class;
variance
---------------
24.4259169884
(1 row)

Not sure if it is the right number or not. Have you tried taking the
square root and see if that number is correct.

The docs say:

The variance is the square of the standard deviation. The
supported data types and result types are the same as for
standard deviation.

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