A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 14 tablets. The entire shipment is accepted if at most 2 tablets do not meet the required specifications. If a particular shipment of thousands of aspirin tablets actually has a 4.0 % rate of defects, what is the probability that this whole shipment will be accepted?

Answer: the probability that this whole shipment will be accepted is 0.673

Step-by-step explanation:

This is a binomial distribution because the probabilities are either that of success or failure

The probability that a particular shipment of thousands of aspirin tablets actually has defects is 4% = 0.04. Then the probability that there would be no defect is 1 - 0.04 = 0.96

Since the shipment is accepted with at most even when there is defect, then the probability of success is 0.04 and that of failure is 0.96

The formula is expressed as

P (x = r) = nCr * p^r * q^ (n - r)

Where

x represent the number of successes.

p represents the probability of success.

q = (1 - p) represents the probability of failure.

n represents the number of trials or sample.

From the information given,

p = 0.04

q = 0.96

n = 14

Since entire shipment is accepted if at most 2 tablets do not meet the required specifications, then the probability is

P (x ≤ 2) = P (x = 0) + P (x = 1) + P (x = 2)

P (x = 0) = 14C0 * 0.04^0 * 0.96^ (14 - 0) = 0.56

P (x = 1) = 14C1 * 0.04^1 * 0.96^ (14 - 1) = 0.024

P (x = 2) = 14C2 * 0.04^2 * 0.96^ (14 - 2) = 0.089

P (x ≤ 2) = 0.56 + 0.024 + 0.089 = 0.673