Question
An office supply company ordered a lot of 4,000 pens. When the lot arrives the company inspector will randomly inspect 15 pens. If more than three pens in the sample are non- conforming, the lot will be rejected. If fewer than two pens are non-conforming, the lot will be accepted. If two, three pens are non-conforming, a second sample of size 10 will be taken. The value of c2 is 3. Suppose the inspector finds two nonconforming items in the first sample and two in the second sample. Also AQL and LTPD are 0.05 and 0.10 respectively. Let incoming quality be 4%.
i) What type of acceptance plan is used here?
ii) What is the probability of accepting the lot at the first sample?
iii) What is the probability of accepting the lot at the second sample?
Answer :
Word Count : 558
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Skip in 10
Let's analyze the given problem step by step and solve it numerically. --- ### Given Data: - Lot Size (N) = 4,000 pens - First Sample Size (n₁) = 15 - First Acceptance Criteria: - If more than 3 non-conforming pens → Reject Lot - If 0 or 1 non-conforming pens → Accept Lot - If 2 or 3 non-conforming pens → Take second sample - Second Sample Size (n₂) = 10 - Second Acceptance Criteria: - Accept the lot if total non-conforming pens in both samples ≤ c₂ = 3 - Reject otherwise - Acceptance Quality Limit (AQL) = 0.05 (5%) - Lot Tolerance Percent Defective (LTPD) = 0.10 (10%) - Incoming Quality = 4% (0.04 defect rate) --- ### i) Type of Acceptance Plan The given plan involves: 1. Two-stage sampling: The decision is made based on a second sample if the first sample is inconclusive. 2. Attributes Sampling Plan: Since the lot is being inspected based on the number of defective items (non-conforming pens).
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