# Variance of a weighted population

by rogo0034
Tags: population, variance, weighted
 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)
P: 280
 Quote by rogo0034 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:
$$var\{x\}=E\{(x-E\{x\})^2\}$$
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:
$$var\{x\}=E\{x^2-2xE\{x\}+E^2\{x\}\}=E\{x^2\}-E^2\{x\}$$

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.
 P: 37 Got it! thanks, that was easier

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