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July 1st, 2011, 01:51 AM   #1
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normal approximation

Use a normal distribution to approximate Pr(X+Y>4)
Let X have a uniform distribition on [0,2] and let Y be independent of X with a uniform distribution over (0,3)
So I get:
E[X]=1 Var[X]=1/3
E[Y]=1.5 Var [Y]=3/4

Furthermore I think that E[X+Y]=2.5 and Var[X+Y]= 1/3+3/4=13/12, Since X and Y are independent.

Pr(X+Y>4)
Pr((X+Y-E[X+Y])/Var[X=Y]>(4-2.5)/13/12)
Pr(Z>1.3846)=0.083087

Is this correct. Can someone tell me if I make a mistake. since the solution according to the book is 0.0748
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July 1st, 2011, 01:14 PM   #2
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Re: normal approximation

The divisor (in the next to the last line) should be the standard deviation, not the variance.
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