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% 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
We are given: * Lot size ( N = 800 ) * Sample size ( n = 20 ) * Acceptance number ( c = 3 ) * Acceptable quality ( p_a = 2% = 0.02 ) * Unacceptable quality ( p_u = 7% = 0.07 ) * Question asks for calculations at ( p = 5% = 0.05 ) We solve step by step manually. --- (i) Probability of accepting a lot when defective fraction is 5% The probability of accepting a lot in a single-sampling plan is given by the cumulative probability of observing at most ( c ) defectives in the sample: [ P_a = \sum_{x=0}^{c} \binom{n}{x} p^x (1-p)^{n-x} ] Here, ( n = 20, c = 3, p = 0.05 ). Step 1: Compute individual probabilities. * ( \binom{20}{0} (0.05)^0 (0.95)^{20} = 1 \cdot 1 \cdot 0.95^{20} ) We estimate ( 0.95^{20} ) manually: [ 0.95^2 = 0.9025 ] [ 0.95^4 = 0.9025^2 \approx 0.8145 ] [ 0.95^8 = 0.8145^2 \approx 0.6634 ] [ 0.95^{16} = 0.6634^2 \approx 0.4391 ] [ 0.95^{20} = 0.95^{16} \cdot 0.95^4
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