Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!ncar!tank!paideia!vevea
From: vevea@paideia.uchicago.edu (Jack L. Vevea)
Newsgroups: comp.lang.c
Subject: Re:  Standard Deviation
Message-ID: <2437@tank.uchicago.edu>
Date: 24 Mar 89 05:36:32 GMT
References: <2221@maccs.McMaster.CA>
Sender: news@tank.uchicago.edu
Reply-To: vevea@paideia.uchicago.edu (Jack L. Vevea)
Organization: University of Chicago, Dept of Education
Lines: 52

In article <2221@maccs.McMaster.CA> cs3b3aj@maccs.UUCP (Stephen M. Dunn) writes:
>
>   Well, of course, the standard textbook methods of calculating the
>standard deviation are as follows:
>
>1)  assuming values are in array x, there are n values, and the mean is 
>    known and is in the variable mean: 
>
>temp = 0;
>for (i = 1; i <= n; i++)
>   temp += (x [i] - mean) ^ 2;
>std_dev = sqrt (temp / n);

And this is precisely what Doug Gwynn was warning against in his
earlier posting; if your numbers have variations of small magnitude,
you will end up with more rounding error than meaning.  Don't use this.



>2)  assuming that the SQUARES of the values are in array x, there are n 
>    values, and the mean is known and is in the variable mean: 
>
>temp = 0;
>for (i = 1; i <= n; i++)
>   temp += x [i] ^ 2;
>std_dev = sqrt ((temp - (mean ^ 2) / n) / n);
>

And if the array already contains the squares, why square them again, pray?

Try this:

for(i=0;i<n;i++) {
	temp1 += x * x;
	temp2 += x; }

sd = sqrt(temp1 - (temp2 * temp2 / n)) / (n-1);


(Note division by n-1, not n; although the original poster didn't
give much information on what he planned to do with this, he almost
certainly wants _sample_ sd's, not population as the above quoted
formulae assume.)








Saepe fidelis.