A machine produces 1000 steel Orings per day Each ring has p

-A machine produces 1000 steel O-rings per day. Each ring has probability 0.9 of meeting a thickness specification.

a) What is the probability that on a given day, fewer than 900 O-rings meet the specification?

b) Find the 60th percentile of the number of O-rings that meet the specification.

c) If the machine operates for 5 days, what is the probability that fewer than 900 O-rings meet the specification on 4 or more of those days?

SHOW SOLUTION STEP BY STEP PLEASE.

mean= n*p=1000*0.9 =900

standard deviation =sqrt(n*p*(1-p)) =sqrt(1000*0.9*0.1) =9.486833

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a) What is the probability that on a given day, fewer than 900 O-rings meet the specification?

P(X<900) = P((X-mean)/s <(900-900)/9.486833)

=P(Z<0) = 0.5 (from standard normal table)

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b) Find the 60th percentile of the number of O-rings that meet the specification.

P(X<x)=0.6

--> P(Z<(x-900)/9.486833) =0.6

--> (x-900)/9.486833 = 0.25(from standard normal table)

So x= 900+0.25*9.486833 =902.3717

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c) If the machine operates for 5 days, what is the probability that fewer than 900 O-rings meet the specification on 4 or more of those days?

Given X follows Binomial distribution with n=5 and p=0.5

P(X=x)=5Cx*(0.5^x) for x=0,1,2,...,5

So P(X>=4) = P(X=4)+P(X=5)

=5Cx*(0.5^4)+5Cx*(0.5^5)

=0.1875