A pharmaceutical company receives large shipments of ibuprofen tablets...

Question

A pharmaceutical company receives large shipments of ibuprofen tablets and uses this acceptance sampling plan: randomly select and test 29 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 1% rate of defects, what is the probability that this whole shipment will be accepted?

(Report answer as a decimal value accurate to four decimal places.) 

Answer 1

Hello Sela,

If the number of tablets in a shipment is sufficiently large, we can consider the testing procedure you describe (selecting a random sample of 29 ibuprofen tablets and counting the number of defective tablets in the sample) to be a binomial probability experiment.

Let X = (Number of defective ibuprofen tablets in a random sample of 29 tablets.) Then we may reasonably assume that X follows a binomial distribution. For each "trial" (that is, each tablet selected), there are two possible outcomes, "success" (the tablet is defective), or "failure" (the tablet is not defective). Note that the word success as used in the binomial distribution does not carry the connotation of being a "good" outcome. A "success" is simply the outcome you are interested in counting, which in this case is the number of defectives in the sample. The important parameters for a binomial distribution are n (the number of trials) and p (the probability of success on any one trial.) For this situation, we have p = .01 (since we are told 1% of the tablets in the shipment are defective) and n = 29. The general formula for the binomial distribution is

P(x) = _nC_x·p^x·(1 - p)^n-x

where x is the number of successes in n trials (In this problem, x is the number of defectives in a random sample of 29 tablets.) The symbol _nC_x is the "number of combinations of n things taken x at a time". This is the number ways of choosing x distinct objects from a set of n objects, without regard to order. It is given by

_nC_x = n!/[x!·(n-x)!]

where ! is the factorial symbol.

We want to find the probability that the shipment is accepted. We are told that a shipment will be accepted if at most one tablet doesn't meet the required specifications. That is, the number of defectives must be less than or equal to one.

P(Shipment accepted) = P(X ≤ 1). (In this case, X ≤ 1 if and only if X=0 or X=1)

= P(0) + P(1).

So we calculate P(0) (the probability of 0 defective tablets in the shipment), and P(1) (the probability of 1 defective.) Using the formula for the binomial distribution, we have

P(0) = ₂₉C₀·(.01)⁰·(1 - .01)^{29 - 0}.

But ₂₉C₀ = 1, so

P(0) = 1·(.01)⁰·(.99)²⁹ ≅ .74717

Also,

P(1) = ₂₉C₁·(.01)¹·(1 - .01)^{29 - 1}.

Calculating ₂₉C₁, we have

₂₉C₁ = 29!/[1!·(29-1)!] = 29, so

P(1) = 29·(.01)¹·(99)²⁸

P(1) ≅ .21887

Finally, then, we have

P(Shipment accepted) ≅ .74717 + .21887 ≅ .9660 (to 4 decimal places.)

Hope that helps! Let me know if need any further explanation.

William