Acceptance sampling on a 500-unit component lot
A buyer receives lots of 500 identical machined components from a supplier whose process has historically run at a 4% defective rate. The agreed sampling plan draws 20 components at random, without replacement, from each lot and rejects the lot if more than one defective is found in the sample; accepted lots go straight to the assembly line. Inspection costs Rs. 60 per component sampled. A defective component that reaches the line causes Rs. 1,800 of rework and downtime. The supplier has complained that the plan rejects acceptable lots far too often and has asked for the acceptance number to be relaxed from one to two. The buyer's quality manager is willing to accept at most a 5% chance of rejecting a lot that is genuinely 4% defective, but is worried about what happens to lots that are 12% defective.
1. Compute the probability that a lot which is genuinely 4% defective will be rejected under the current plan (accept on 0 or 1 defective).
Read the worked answer
With n = 20 and p = 0.04, the sample count is hypergeometric; because 20 out of 500 is only 4% of the lot, the binomial with p = 0.04 is an accurate working approximation (and errs slightly on the conservative side).
- P(0) = 0.96^20 = 0.4420
- P(1) = 20 × 0.04 × 0.96^19 = 0.8 × 0.4604 = 0.3683
- P(accept) = 0.8103
- P(reject) = 1 − 0.8103 = 0.1897
So the producer's risk is about 19%, meaning roughly one in five good lots is rejected - a real cost to the supplier and a real source of disputes, which is exactly what the supplier is complaining about. An answer of 'about 19%' with no working does not demonstrate that the candidate knows which tail of the distribution is being measured; note that 0.81 is the acceptance probability, not the rejection probability.
2. Does relaxing the acceptance number to two bring the producer's risk within the manager's 5% limit? If not, what acceptance number is required?
Read the worked answer
At c = 2, add P(2) = C(20,2) × 0.04² × 0.96^18 = 190 × 0.0016 × 0.4796 = 0.1458.
P(accept) = 0.4420 + 0.3683 + 0.1458 = 0.9561, so P(reject) = 0.0439, about 4.4%.
That is below 5%, so an acceptance number of 2 does meet the manager's stated target. The current c = 1 gives 19%, so the relaxation is required, not merely welcome.
Judgement note: 4.4% is uncomfortably close to 5%. If the manager wants genuine headroom, c = 3 gives P(reject) = 0.0439 − P(3) with P(3) = 1140 × 0.000064 × 0.4996 = 0.0365, i.e. 0.7% - but see the next part before recommending it, because producer's risk is only half of the trade-off.
3. Compute the consumer's risk (probability of accepting a 12% defective lot) under c = 1 and under c = 2, and state a recommendation.
Read the worked answer
With p = 0.12 and n = 20:
- P(0) = 0.88^20 = 0.0776
- P(1) = 20 × 0.12 × 0.88^19 = 2.4 × 0.0882 = 0.2117
- P(2) = 190 × 0.0144 × 0.1002 = 0.2742
c = 1: P(accept) = 0.0776 + 0.2117 = 0.2893, about 29%. c = 2: P(accept) = 0.2893 + 0.2742 = 0.5635, about 56%.
Relaxing the acceptance number roughly doubles the chance of waving through a bad lot - 29% to 56%. A 12% defective lot reaching the line implies about 60 defectives in 500, at Rs. 1,800 each, versus a sampling cost of only 20 × Rs. 60 = Rs. 1,200 per lot; the rework exposure dwarfs the inspection saving.
Recommendation: do not simply move to c = 2. Either keep c = 1 and accept the supplier's 19% rejection exposure as the price of protection, or move to c = 2 and raise n (or tighten the supplier's process) so that both risks are controlled. Producer's risk and consumer's risk cannot be tuned independently with a single acceptance number at fixed n - that is the point of the question.