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Variance of a weighted population 
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#1
Mar2912, 08:45 PM

P: 37

1. The problem statement, all variables and given/known data
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
Mar2912, 09:23 PM

P: 280

[tex]var\{x\}=E\{(xE\{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 (xE{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^22xE\{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. 


#3
Mar2912, 10:31 PM

P: 37

Got it! thanks, that was easier



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