3.In a statistical process control chart example, samples of 20 parts from a met

Question

3.In a statistical process control chart example, samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let x denote the number of parts in the sample of 20 that require rework. A process problem is suspected if x exceeds its mean by more than three standard deviations. (a) If the percentage of parts that require rework remains at 1%, what is the probability that x exceeds its mean by more than three standard deviations? (b) If the rework percentage increases to 4%, what is the probability that x exceeds 1? (c) If rework percentage increases to 4%, what is the probability that x exceeds 1 in at least one of the next 5 hours of samples

Explanation / Answer

Note that here we can useBinomial random variable

n = 20, p = 0.01, and thus E(X) =np= 0.2 and V(X)=np(1-p) = 0.198.

a. What is the probability that X exceeds its mean by more than three standard deviations?

Solution: Since =sqrt(np(1-p)) =sqrt(0.198)= 0.45,

P(X > 0.2 + 3) = P(X > 1.55) = P(X 2) = 1 P(X = 0) P(X = 1) = 1 (0.99) 20 20(0.99) 19(0.01) = 0.015.

b. If the rework percentage increases to 4%, what is the probability that X exceeds 1?

Solution: Note that = sqrt(np(1 p)) = sqrt( 20(0.04)(0.96)) = 0.88.

P(X > 1) = P(X 2) = 1 P(X = 0) P(X = 1) = 1 (0.96) 20 20(0.96) 19(0.04) = 0.19.

(c). If the rework percentage increases to 4%, what is the probability that X exceeds 1 in at least one of the next 5 hours of samples?

Solution: Since P(X > 1 in one hour) = 0.19,

P(X 1 in one hour) = 0.81.

Thus P(X > 1 in at least one of five hours) = 1 P(X 1 in any one of five hours)

=> 1 P(X 1 in one hour)^5 = 1 0.815 = 0.65.

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