Variance of a weighted population


by rogo0034
Tags: population, variance, weighted
rogo0034
rogo0034 is offline
#1
Mar29-12, 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)
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RoshanBBQ
RoshanBBQ is offline
#2
Mar29-12, 09:23 PM
P: 280
Quote Quote by rogo0034 View Post
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)
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.
rogo0034
rogo0034 is offline
#3
Mar29-12, 10:31 PM
P: 37
Got it! thanks, that was easier


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