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# Weighted Standard deviation

Posted on 2013-09-12
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I found the attached 'weightsd.pdf' formula for weighted standard deviation from NIST.

Wolfram  shows the variance of the weighted mean in equation 24 in the 'maximum likelihood' attachment.

The link to Wikipedia below states the same formula in hte Wolfram article for determining the weighted standard deviation on the link below:

https://controls.engin.umich.edu/wiki/index.php/Basic_statistics:_mean,_median,_average,_standard_deviation,_z-scores,_and_p-value

My question is, I have yet to be able to reconcile the formula in the NIST attachment with the Wolfram and Wikipedia results, and was wondering which was the more apropriate approach to use (NIST or Wolfram) and why.

Thank you
weightsd.pdf
Maximum-Likelihood----from-Wolfr.pdf
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Question by:aasikolo
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LVL 27

Accepted Solution

aburr earned 1000 total points
ID: 39488597
I do not blame you for being confused. The notation and sentence construction in wiki is difficult.
Not that the capital values (X and Xbar) are already weighted while the x and xi in the NIST equation are not.
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Author Comment

ID: 39488964
What is xbar sub w if it is not weighted in the NIST equation?  I see the xi but no x in the NIST equation.  Did you mean x bar sub w and xi are not weighted in the NIST equation?

To confirm, If I manage to decipher this mess, both formulas should yield the same result?

Thank you
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Author Closing Comment

ID: 39488984
Examples with calculations help.  I'll send calculations when they aren't proprietary.
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LVL 27

Expert Comment

ID: 39489068
"Did you mean x bar sub w and xi are not weighted in the NIST equation?"
the xi are not weighted but the x (or as I should have said) x bar sub w is.
Use the nist. It is clearer.
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