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

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

Answer 1

[James Tursa]

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

[Paul Camerino]

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]

Sometimes it can greatly help when you put spaces within your line to make the parentheses separations more obvious.

Answer 2

[Adam]

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

[Paul Camerino]

Nvm, fixed it. Thanks!