Comments: If we recall exponential distribution, f(x) is an exponential pdf.
User Y=X4
Comments: we should also include F(x)=0
x<=4
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View Full DocumentComments: 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: =1NORMDIST(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*(1p)=19.6>5
ok
c.
p is a little close to 0, but condition a, b is more important.
(2)
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 Spring '10
 Zhai
 Normal Distribution, Poisson Distribution, Probability theory, Binomial distribution, Exponential distribution, consecutive arrivals

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