York Steel Corporation produces a special bearing that must meet rigid specifica

Question

York Steel Corporation produces a special bearing that must meet rigid specifications. When the production process is running properly, 10% of the bearings fail to meet the required specifications. Sometimes problems develop with the production process that causes the rejection rate to exceed 10%. To guard against this higher rejection rate, samples of 15 bearings are taken periodically and carefully inspected. If more than 2 bearings in a sample of 15 fail to meet the required specifications, production is suspended for necessary adjustments. If the true rate of rejection is 10% (that is, the production process is working properly), what is the probability that the production will be suspended based on a sample of 15 bearings? What assumptions did you make in part a?

Explanation / Answer

Part a

We are given p = 10% = 0.10

Sample size = n = 15

We have to find P(X>2)

P(X>2) = 1 – P(X2)

P(X2) = P(X=0) + P(X=1) + P(X=2)

P(X=x) = nCx*p^x*q^(n – x)

Where, q = 1 – p = 1 – 0.10 = 0.90

P(X=0) = 15C0*0.10^0*0.90^15 = 0.205891

P(X=1) = 15C1*0.10^1*0.90^14 = 0.343152

P(X=2) = 15C2*0.10^2*0.90^13 = 0.266896

P(X2) = P(X=0) + P(X=1) + P(X=2) = 0.815939

P(X>2) = 1 – P(X2) = 1 – 0.815939

P(X>2) = 0.184061

Required probability = 0.184061

Part b

For the above part, we assume that the production process have only two outcomes as fail or do not fail. This means the random variable has binomial outcomes and therefore it is follow binomial distribution.

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