Of the people passing through an airport metal detector 05 a

Of the people passing through an airport metal detector, 0.5% activate it. Let X= the number among a randomly selected group of 500 who activate the detector.

a). What is the approximate pmf of X?

b). Compute Pr(X=5)

c). Compute Pr(5<=X)

Solution

(1)(i) By using Normal distriubution,

X follows normal distribution with mean=n*p=500*0.005=2.5 and standard deivation =sqrt(n*p*(1-p))=sqrt(500*0.005*(1-0.005)) =1.57718

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(ii) X follows Binomial distrubtion with n=500 and p=0.005

P(X=x)=500Cx*(0.005^x)*(0.995^(500-x))

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(2) Uising (i):

P(X=5) = P(4.5<X<5.5) (by using continuous correction)

=P((4.5-2.5)/1.57718 <(X-mean)/s <(5.5-2.5)/1.57718)

=P(1.27<Z<1.9) =0.0733 (from standard normal table)

Using (ii):

P(X=5) =500C5*(0.005^5)*(0.995^(500-5)) =0.06671626

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(3) Uising (i):
P(X<=5) = P(Z<(5-2.5)/1.57718) = P(Z<1.59) =0.9441 (from standard normal table)

Uising (ii):
P(X<=5) = P(X=0)+P(X=1)+...+P(X=5)

=500C0*(0.005^0)*(0.995^(500-0))+...500C5*(0.005^5)*(0.995^(500-5))

=0.9583972