Quality-Control Sampling with Binomial and Geometric Models
A bottled-water company in Limbe samples 20 bottles from each production batch and checks the caps. Historically 4% of bottles have a misaligned cap. The quality-control rule is to reject a batch if at least 2 of the 20 sampled bottles are defective. The supervisor treats the number of defective bottles in the sample as a binomial random variable, with independent bottles and constant defect probability within a batch. A customer complaint has raised concern that the defect rate may have risen to 10%. The supervisor wants to know how often the current rule rejects a batch at the historical 4% rate and how often it would reject a batch if the rate were really 10%. There is also a separate check in which bottles are inspected one by one until the first defective is found; the supervisor asks how likely it is that this first defective is the fifth bottle inspected, and what average inspection position would be expected. Finally, the company considers increasing the sample to 30 bottles while keeping the same rejection rule.
1. Using the binomial model with n = 20 and p = 0.04, calculate the probability that a batch is rejected.
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Let X be the number of defective bottles in the sample. Then X ~ B(20, 0.04). The batch is rejected when X ≥ 2, so P(reject) = P(X ≥ 2) = 1 − P(X = 0) − P(X = 1).
P(X = 0) = (0.96)^20 ≈ 0.4420. P(X = 1) = C(20,1)(0.04)(0.96)^19 = 20(0.04)(0.4604) ≈ 0.3683.
Therefore P(reject) ≈ 1 − 0.4420 − 0.3683 = 0.1897.
So the current rule rejects about 19.0% of batches when the defect rate is 4%.
2. Calculate the probability that the first defective bottle is the fifth inspected, and state the expected number of inspections until the first defective when p = 0.04.
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Let T be the inspection number of the first defective bottle. Then T has a geometric distribution with success probability p = 0.04. For the first defective to be the fifth bottle, the first four bottles must be non-defective and the fifth defective: P(T = 5) = (1 − p)^4 p = (0.96)^4(0.04) ≈ 0.8493 × 0.04 ≈ 0.0340.
So the probability is about 3.4%.
The expected number of inspections until the first defective is E(T) = 1/p = 1/0.04 = 25.
On average, 25 bottles would be inspected before finding the first defective when the defect rate is 4%.
3. If the true defect rate rises to 10%, calculate the rejection probability under the current n = 20 rule and under a proposed n = 30 rule with the same rule (reject if at least 2 defective). Does the change to 30 achieve a rejection probability of at least 0.8? Comment.
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For n = 20 and p = 0.10: P(X = 0) = 0.9^20 ≈ 0.1216, P(X = 1) = 20(0.10)(0.9^19) ≈ 0.2702, P(reject) ≈ 1 − 0.1216 − 0.2702 = 0.6082.
For n = 30 and p = 0.10: P(X = 0) = 0.9^30 ≈ 0.0424, P(X = 1) = 30(0.10)(0.9^29) ≈ 0.1413, P(reject) ≈ 1 − 0.0424 − 0.1413 = 0.8163.
Yes, the n = 30 rule gives a rejection probability of about 0.816, just above 0.8, whereas the n = 20 rule gives only about 0.608. However, increasing the sample size also raises the false-alarm rate at the historical 4% rate: for n = 30 and p = 0.04, P(reject) = 1 − 0.96^30 − 30(0.04)(0.96^29) ≈ 1 − 0.2939 − 0.3673 = 0.3388, compared with 0.1897 for n = 20. The company must balance better detection of a 10% defect rate against more frequent rejection of good batches.