..the first part (using it to find averages) and it worked.

Presumably that was with $n1 = 1 otherwise I can't see how $esum0=$esum0/$n2 would work.

Shouldn't it be $eavg0 = $esum0/($n2-$n1+1); ?

poj

In reply to Re: standard deviation help by poj
in thread standard deviation help by vka725

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