Question 730638: A production process outputs items in lots of 50. Sampling plans exist in which lots are pulled aside periodically and exposed to a certain type of inspection. It is usually assumed that the proportion defective is very small. It is important to the company that lots containing defectives be a rare event. The current inspection plan is to periodically sample randomly 10 out of the 50 items in a lot and, if none are defective, to perform no intervention.
(a) Suppose in a lot chosen at random, 2 out of 50 are defective. Calculate the
probability that at least 1 in the sample of 10 from the lot is defective.
(b) Calculate the mean number of defects found out of 10 items sampled.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The current inspection plan is to periodically sample randomly 10 out of the 50 items in a lot and, if none are defective, to perform no intervention.
(a) Suppose in a lot chosen at random, 2 out of 50 are defective. Calculate the
probability that at least 1 in the sample of 10 from the lot is defective.
For each item p(defective) = 2/50 ; P(not defective) = 48/50
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P(at least one defective ) = 1 - P(none are defective)
= 1 - 48C10/50C10 = 0.6367
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(b) Calculate the mean number of defects found out of 10 items sampled.
mean = np = 10(2/50) = 2/5
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Cheers,
Stan H.
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