
#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



Register to reply 
Related Discussions  
Expected variance of subset of population  Set Theory, Logic, Probability, Statistics  1  
Variance of Sampling Distribution VS Sample Variance  Calculus & Beyond Homework  0  
to calculate variance and CV from multiple (weighted) variables  Set Theory, Logic, Probability, Statistics  0  
Infinite Variance VS Zero Variance  General Math  4  
Weighted average  Introductory Physics Homework  5 