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Using the normal distribution. Let $X\sim \mathcal{N}(1, 2)$ and $Y\sim \mathcal{N}(2, 3)$ where $\mathcal{N}(μ, \sigma^2$) denotes the normal distribution with mean $\mu$ and variance $\sigma^2$. X and Y are independent. Let $U = 2X + 3Y$.

What is the mean of U? What is the variance of U? What is $P(6<=U<=7.5)$?

For the mean of U, I calculated $2(1)+3(2)=8$. For the variance of U I calculated $ 2(2)+3(3)=13$.

Then I found pnorm(7.5,mean=8,sd=sqrt13)-pnorm(6,mean=8,sd=sqrt13) [1] 0.1553036 for P(6<=U<=7.5).

Did I do this correctly? Thanks in advance!

Peter Flom
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dataznkid1
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  • I vote to close this as a duplicate. Did you read @Corone's [edited answer](http://stats.stackexchange.com/a/50581/6633) to your [other question](http://stats.stackexchange.com/a/31328/6633)? – Dilip Sarwate Feb 21 '13 at 23:11
  • Yes, I did. But this question has the U=2x+3Y involved. Also, I did not get a confirmation that my 67% was correct? If it is possible, could you verify that I took the correct course of action? – dataznkid1 Feb 22 '13 at 00:12

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