My Math Forum normal approximation

 July 1st, 2011, 01:51 AM #1 Member   Joined: Dec 2010 From: Drachten, the Netherlands Posts: 49 Thanks: 0 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
 July 1st, 2011, 01:14 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Re: normal approximation The divisor (in the next to the last line) should be the standard deviation, not the variance.

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