How can I calculate the standard deviation of a LARGE set of values without using the' std' or 'mean' commands?

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Paul Camerino
Paul Camerino 2016-11-1
评论： James Tursa 2016-11-1
I tried entering
endStandDevx = sqrt(sum((x-(AvgX))).^2/(length(x))
(since we can't use mean I found the average by finding sum over total # of vals)
However, this line gives me a far different value than
std(x)
and i'm not really sure why...
Any help would be lovely :D

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James Tursa
James Tursa 2016-11-1
编辑：James Tursa 2016-11-1
You've got the ^2 in the wrong place in your calculation. Also, the std function uses n-1 in the denominator by default. Try this:
>> x = rand(1,10000);
>> AvgX = sum(x)/length(x);
>> endStandDevx = sqrt(sum((x-AvgX).^2)/(length(x)-1))
endStandDevx =
0.2895
>> std(x)
ans =
0.2895
  2 个评论
Paul Camerino
Paul Camerino 2016-11-1
AWESOME! Thanks so much, mate. That really helped. I guess my parentheses and everything was just way out of whack, so your direction really helped and gave me the correct value. Thanks so much!
James Tursa
James Tursa 2016-11-1
Sometimes it can greatly help when you put spaces within your line to make the parentheses separations more obvious.

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更多回答（1 个）

Adam
Adam 2016-11-1
编辑：Adam 2016-11-1
std divides by n-1 rather than n. You didn't post what kind of difference you are getting so I don't know if that is the only difference but it will be a difference, obviously less noticeable with a bigger sample size though. Since you say your result is far away and your sample size is large I assume this is not the only issue.
Is your average calculation correct and matching the output of mean?
The formula for std is on its help page
doc std

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