Now suppose that the manufacturer decides to accept the shipment if there is at most one defective part in the sample. How large does K have to be to ensure that the probability that the manufacturer accepts an unacceptable shipment is less than 0.1?

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Answer:

hello your question is incomplete below is the missing parts of the question

answer : K = 80

Step-by-step explanation:

P [ x ≤ 1 ]  < 0.1

determine how large k have to be

note : n = k

sample proportion = 0.01

E = 0.05 - 0.01 = 0.04

applying this formula below

E = [tex]Z_{c}[/tex] [tex]\sqrt{\frac{pq}{n} }[/tex]  

0.04 = 1.645 [tex]\sqrt{\frac{0.05*0.95}{n} }[/tex]    make n subject of the equation

n = 80.3353

hence  the value of K ≈ 80

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