A company purchases shipments of machine components and uses this acceptance sampling plan: Randomly select and test 20 components and accept the whole batch if there are fewer than 3 defectives.If a particular shipment of thousands of components actually has a 4% rate of defects,what is the probability that this whole shipment will be accepted?
A)0.9561
B)0.0439
C)0.0364
D)0.5141
E)0.1458

Question

Quizplus · Accepted Answer

This problem can be solved using the binomial distribution formula, which is P(X = k) = (nCk) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes (in this case, defective components), p is the probability of success on a single trial, and nCk is the binomial coefficient. The question asks for the probability of finding fewer than 3 defective components in a sample of 20 when the defect rate is 4% (or 0.04). This means we need to calculate the probability of finding 0, 1, or 2 defective components and sum those probabilities.P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)Calculating each of these probabilities using the binomial formula and the given p = 0.04, n = 20:P(X = 0) = (20C0) * (0.04)^0 * (0.96)^20P(X = 1) = (20C1) * (0.04)^1 * (0.96)^19P(X = 2) = (20C2) * (0.04)^2 * (0.96)^18Summing these probabilities gives the total probability of the shipment being accepted, which matches choice A, 0.9561. This calculation involves using the binomial distribution to account for the probability of 0, 1, or 2 defects in a sample of 20, given a 4% defect rate.

A Company Purchases Shipments of Machine Components and Uses This