The Quality Manager for Graphics,Inc.is concerned about the quality of the lot of several thousand posters which her company printed this week and is now preparing to ship.If she uses an acceptance sampling plan of n = 10 and c = 2,what is the probability that,if this lot is 10% defective,it will be accepted for shipment without inspection?
A)0.9298
B)0.0702
C)0.8189
D)0.2198
E)0.5995

Question

Quizplus · Accepted Answer

The probability that a lot with 10% defective items will be accepted using an acceptance sampling plan with n = 10 and c = 2 can be calculated using the binomial distribution formula. The formula for the probability of x defects in n trials is P(x) = (nCx) * p^x * (1-p)^(n-x), where p is the probability of defect (0.1 in this case), n is the sample size (10), and x is the number of defects (ranging from 0 to c, which is 2). The probability of acceptance is the sum of the probabilities of finding 0, 1, or 2 defective items in the sample. Calculating this gives a probability of 0.9298, meaning there's a 92.98% chance the lot will be accepted if it's 10% defective.