Batch Inspection: Binomial, Normal Approximation and a Conditional Check
A production line makes electronic components and, historically, 8% are defective. A batch of 150 components is inspected and rejected if more than 15 defectives are found. The quality manager proposes to use the normal approximation to the binomial distribution instead of summing the exact binomial probabilities, and wants the probability that a batch is rejected. A colleague objects that with only 150 items the approximation will be badly wrong and that a continuity correction is pointless. You are asked to judge both claims. Separately, a sample of 20 components is checked and it is found that at most two are defective; the manager wants the conditional probability that none of the 20 is defective, given that at most two are. Take the defect rate as constant and independent between components.
1. Carry out the normal approximation, state whether the continuity correction is required, and assess the colleague's objection.
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Parameters. X ~ B(150, 0.08): μ = np = 12, σ² = np(1 − p) = 150(0.08)(0.92) = 11.04, σ = 3.32.
Validity. Both np = 12 and n(1 − p) = 138 exceed 5, so the approximation is usable; it is least reliable in the tail, which is where we are working, so the answer is approximate.
Continuity correction. X is discrete and the event is X > 15, i.e. X ≥ 16. On the normal scale this is the region beyond 15.5:
z = (15.5 − 12)/3.32 = 1.05 → P(Z > 1.05) = 1 − 0.8531 = 0.147.
The rejection probability is about 15%.
The colleague is wrong on the correction. Without it, z = (16 − 12)/3.32 = 1.20 and P = 0.115, a large understatement of the tail probability. A correction matters more, not less, when estimating a tail with a discrete variable. The concern about accuracy is fair but does not justify dropping the correction.
2. Given that at most two of a sample of 20 are defective, find the conditional probability that none is defective.
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Work with p = 0.08, n = 20.
P(0) = 0.92²⁰ = 0.1887.
P(1) = C(20,1)(0.08)(0.92¹⁹) = 20(0.08)(0.2051) = 0.3282.
P(2) = C(20,2)(0.08²)(0.92¹⁸) = 190(0.0064)(0.2229) = 0.2711.
P(at most 2) = 0.1887 + 0.3282 + 0.2711 = 0.7880.
Conditional probability = P(0)/P(at most 2) = 0.1887/0.7880 = 0.239.
So there is roughly a 24% chance the sample is completely clean, given that it contains no more than two defectives. This is larger than the unconditional P(0) = 0.189, which is reasonable: knowing the sample is not badly contaminated shifts probability towards the low-count outcomes.
3. A supplier is actually running at 10% defective. Estimate the probability that such a batch passes inspection under the same rule, and comment on the discriminating power of the plan.
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Now p = 0.10, n = 150: μ = 15, σ² = 150(0.10)(0.90) = 13.5, σ = 3.67.
The batch is accepted when X ≤ 15. With the continuity correction use 15.5:
z = (15.5 − 15)/3.67 = 0.136 → P(Z < 0.136) = 0.554.
So a supplier whose defect rate is 10% still has roughly a 55% chance of passing inspection. The plan is not very discriminating near the threshold: the expected number of defectives (15) sits exactly on the acceptance boundary, so about half of such batches pass by chance alone.
Judgement. Tightening the rule - for example rejecting at more than 12 defectives - would cut this escape probability, but it would also reject more genuinely good batches from the 8% line. The choice of rule is a trade-off between the two error rates, not something the binomial calculation alone can settle.