A whole lot of DEVIATION
Author
Let standard deviation = sqrt((sum(x^2) - sum(x)^2)/n)
Anyone have a good algorithm for calculating Standard Devatiion "on the fly" so that you don''t overflow on summing the x^2?
Thanks,
Steve
Hi,
Isn't your formula a bit wrong?
Shouldn't it be:
std = sqrt(1/sqrt(N-1)) * sqrt( sum(x^2)- sum(x)^2/N )
N = [total number of observations]
Anyway, why don't you use this formula:
std = 1/sqrt(N-1) * sum((x(i) - mean(x))^2)
where x(i) is the i:th element.
Then you ought not to get an overflow.
/Mankind gave birth to God.
Edited by - silvren on December 31, 2001 3:31:36 PM
Isn't your formula a bit wrong?
Shouldn't it be:
std = sqrt(1/sqrt(N-1)) * sqrt( sum(x^2)- sum(x)^2/N )
N = [total number of observations]
Anyway, why don't you use this formula:
std = 1/sqrt(N-1) * sum((x(i) - mean(x))^2)
where x(i) is the i:th element.
Then you ought not to get an overflow.
/Mankind gave birth to God.
Edited by - silvren on December 31, 2001 3:31:36 PM
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