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S=Sigma(I:1/7:N/7:1/7:0/7+1/7) The variables are S N e.g. if N=3 S=.4286 From N=1 to N=8 the answers are logical and consistent but from N=9 to N=1,000 etc. S=N-1 e.g. N=21 S=2.8571 Anybody out there with an explanation?

The answer is yes and no.

Remember your maths classes ?

S is the square root of the square of the sum of the differences with the mean divided by the sample (n)

But to get ONE difference, you need TWO values, to get 5 differences, you need 6 values. So statisticians developped (I don't know the exact word in English) a "corrected" standard deviation which is the same formula, but divided by n-1 instead of by n.

The largerpopulation, the less effect n or n-1 has on the standard deviation calculation.

I don't know the 17BII, but I do know the 19BII and no doubt that the used formula is this corrected standard deviation (divided by n-1). You want to get the normal standard deviation ? Nothing more easy : once your list entered, calculated the mean, and key in S+ (sigma plus). You'll add 1 to n, so n-1+1 = n, and the difference of this item with the mean is null... This is even explained in the 19BII manual, so I guess this problem and how to treat is is also explained in the 17BII manual.

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