This problem involves determining the probability of accepting a shipment based on an acceptance sampling plan. Here’s the information given and the approach to solve it:

1. Given Data:

   • A pharmaceutical company receives shipments of aspirin tablets.
   • The acceptance sampling plan involves randomly selecting and testing 52 tablets from a shipment.
   • The shipment will be accepted if there are one or zero defective tablets in the sample of 52.
   • One shipment contains 4,000 aspirin tablets with a 4% defect rate.

2. Objective:

   • Find the probability that the shipment will be accepted.
   • Assess whether most shipments with a 4% defect rate will be accepted or rejected under this plan.

3. Solution Approach:

   • Define the probability of a defect in any given tablet as $p = 0.04$.
   • Let $X$ represent the number of defective tablets in a sample of 52. $X$ follows a binomial distribution: $X \sim \text{Binomial}(n = 52, p = 0.04)$
   • We want to calculate $P(X \leq 1)$, which is the probability that there are zero or one defective tablets in the sample.

4. Calculations:

   • The probability of exactly $k$ defective tablets in the sample can be found using the binomial probability formula: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$
   • Therefore:
     • $P(X = 0) = \binom{52}{0} (0.04)^0 (0.96)^{52}$
     • $P(X = 1) = \binom{52}{1} (0.04)^1 (0.96)^{51}$

5. Interpretation:

   • By calculating $P(X = 0) + P(X = 1)$, we will get the probability that the shipment will be accepted.
   • If this probability is high, then most shipments with a 4% defect rate will be accepted. If it is low, then many shipments will likely be rejected.

Let me proceed with the calculations.from scipy.stats import binom

Parameters

n = 52 # sample size p = 0.04 # defect rate

Calculating P(X = 0) and P(X = 1)

prob_acceptance = binom.pmf(0, n, p) + binom.pmf(1, n, p)

prob_acceptance0.37905869981471730.3790586998147173successThe probability that the shipment will be accepted (i.e., having one or zero defective tablets in the sample) is approximately 0.379 or 37.9%.

Interpretation:

With a 4% defect rate, only about 37.9% of such shipments will be accepted under this sampling plan. This suggests that most shipments with this defect rate would likely be rejected since there's a greater probability that more than one defective tablet would be found in the sample of 52 tablets.

Would you like further details on this solution, or do you have any specific questions?

Here are 5 related questions to expand on this topic:

1. What would be the acceptance probability if the defect rate were lower, say 2%?
2. How does changing the sample size from 52 to a smaller or larger number affect the acceptance probability?
3. What impact would allowing up to two defective tablets in the sample have on the acceptance probability?
4. Could we use a different probability distribution for similar calculations in quality control?
5. How does this sampling method compare to a continuous quality monitoring approach?

Tip: Understanding binomial distributions can be very helpful in quality control and reliability testing, as they model the probability of a given number of successes in a fixed number of trials.