If 8 of 100 homes violate the building code. & What is the probability that a ho

Question

If 8 of 100 homes violate the building code. & What is the probability that a housing inspector, randomly selecting 50 of them, finds that: a) none of the houses violates the building code b) one violates the building code c) two violate the building code d ) At least three violate the building code If 8 of 100 homes violate the building code. & What is the probability that a housing inspector, randomly selecting 50 of them, finds that: a) none of the houses violates the building code b) one violates the building code c) two violate the building code d ) At least three violate the building code a) none of the houses violates the building code b) one violates the building code c) two violate the building code d ) At least three violate the building code a) none of the houses violates the building code b) one violates the building code c) two violate the building code d ) At least three violate the building code

Explanation / Answer

Answers to the questionin detail with calculations:

p ( voilate) = 8/100 = .08

n = 50

So, we know params of the Binomial distribution:

a. P(X=0) = 50C0(.08^0)(.92^50) = .01547

b. P(X=1) = 50C1(.08^1)(.92^49) = .06725

c. P(X=2) = 50C2(.08^2)(.92^48) = .14326

d. P(X>=3) = 1- P(X=1,2,3) = 1-(.01547+.06725+.14326) = 0.7740

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