Solution ch4

Solution ch4 - Comments If we recall exponential...

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Comments: If we recall exponential distribution, f(x) is an exponential pdf. User Y=X-4 Comments: we should also include F(x)=0 x<=4

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Comments: for (e) P(-x/4<Z<x/4)=P(Z<x/4)- P(Z<-x/4) here we use formula P(Z<-a)=1- P(Z<a) Then we have 2P(Z<x/4)-1=0.99 P(Z<x/4)=0.995 x/4=NORMSINV(0.995)= 2.575829 x=10.3 Comments: (a) also can be solved with Excel: =1-NORMDIST(240,260,50,TRUE) (b)25 quantile =NORMINV(0.75,260,50); 75 quantile=NORMINV(0.25,260,50) (c)It is to find x such that P(X>x)=0.95 so x=norminv(0.05,260,50) Comments: 1. It is necessary to state that X ~Binomial (1000, 0.02), because (1) Chips are independent; (2) Test for chip can only have TWO possible outcomes: defective or qualified. (3) Probability for A CHIP be defective is constant as 2%; 2. The procedure of normal approximation is
(1) First, it is necessary to check if the problem satisfy Binomial->Normal Condition: a. n * p= 20>5 ok b. n*p*(1-p)=19.6>5 ok c. p is a little close to 0, but condition a, b is more important. (2)
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Solution ch4 - Comments If we recall exponential...

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