elementary statistics

Question

A pharmaceutical company receives large shipments of ibuprofen tablets and uses this acceptance sampling plan: randomly select and test 21 tablets, then accept the whole batch if there is at most one that doesn't meet the required specifications. If a particular shipment of thousands of ibuprofen tablets actually has a 11% rate of defects, what is the probability that this whole shipment will be accepted?

Answer 1

This problem can be solved using the binomial distribution. The binomial distribution is appropriate when we have a situation involving trials that can result in success/ failure. In this case, the trials would be the 21 randomly selected tablets and the 'success' in this case is actually that a pill does NOT meet the specs (you could set it up the other way as well, but we'll define it this way since not meeting specs is more interesting in this case).

The probability mass function for the binomial distribution is :

P(X = k) = _nC_k p^k (1-p)^n-k ,

where k is the number of 'successes', n is the total number of trials, and p is the probability of 'success'. In our case, n = 21 and p = 0.11.

Since we want to know what the probability of accepting the batch is if the true defect rate is 11%, we want to know the probability that we find 1 or fewer pills that don't meet specs from our random sample. Therefore we want to know P(X <=1).

P(X <=1) = P(X = 0) + P(X = 1)

P(X = 0) = ₂₁C₀ (0.11)⁰ (1 - 0.11)^21-0 = 0.087

P(X = 1) = ₂₁C₁ (0.11)¹ (1 - 0.11)^21-1 = 0.225

P(X <= 1) = 0.087 + 0.225 = 0.312

This means that the probability of accepting the whole shipment in this case is 0.312, or 31.2%. Looks like they might want to take a few more samples for quality control!