Question
A factory purchases bolts in lots of 800. Acceptance is decided using a single-sampling plan with sample size n = 20 and acceptance number c = 3. Assume that 2% of defective items are considered acceptable quality and 7% defective items are considered unacceptable quality.
Find:
(i) The probability of accepting a lot when the incoming quality level is 5% defective.
(ii) The Average Outgoing Quality (AOQ), assuming rejected lots are completely screened and defectives are replaced.
(iii)The Average Total Inspection (ATI)
Answer :
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Skip in 10
Let the lot size be (N = 800), sample size (n = 20), and acceptance number (c = 3). The probability of accepting a lot with defective fraction (p) is given by the binomial formula: [ P_a = \sum_{k=0}^{c} \binom{n}{k} p^k (1-p)^{n-k} ] (i) Probability of accepting a lot when (p = 0.05) Here (p = 0.05), (n = 20), (c = 3). Compute each term: [ \binom{20}{0} (0.05)^0 (0.95)^{20} = 1 \cdot 1 \cdot 0.95^{20} ] First, approximate (0.95^{20}): [ \ln 0.95 = -0.051293 \quad \Rightarrow \quad \ln 0.95^{20} = -1.02586 \quad \Rightarrow \quad 0.95^{20} = e^{-1.02586} \approx 0.358 ] [ \binom{20}{1} (0.05)^1
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