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Variance of a weighted population

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    KEnHA.png


    2. Relevant equations



    3. The attempt at a solution

    Was able to get the population mean of 5.3 with 4*.2 + 5*.4 + 6*.3 + 7*.1
    But for some reason i'm drawing blank for the variance (which i guess is .81)
     
  2. jcsd
  3. Mar 29, 2012 #2
    Variance is defined as:
    [tex]var\{x\}=E\{(x-E\{x\})^2\}[/tex]
    So if you wanted to directly apply the equation above, you would compute it the same way as mean, except instead of doing a weighted sum of x, you would do a weighted sum of (x-E{x})^2. So for each x, before doing the weighted sum, you would subtract off the mean you found already and square it.

    But there is a popular reworking of the equation:
    [tex]var\{x\}=E\{x^2-2xE\{x\}+E^2\{x\}\}=E\{x^2\}-E^2\{x\}[/tex]

    So in this case, you would compute the expected value of x^2 (the same way you did the mean except use x^2 instead of x). Then, subtract the square of the expectation you already found.
     
  4. Mar 29, 2012 #3
    Got it! thanks, that was easier
     
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