Consider an acceptance sampling plan with n = 20 and c = 0. Compute the producer's...
Consider an acceptance sampling plan with n = 20 and c = 0. Compute the producer's risk for a lot with a defective rate of 2%.
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0.6676 |
||
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0.3324 |
||
|
0.3965 |
||
|
0.6035 |
Homework Answers
Given that, an acceptance sampling plan with n = 20 and c = 0.
Defective rate = q = 2% = 0.02
=> Successive rate = p = 100 - 2 = 98% = 0.098
First we find, P(accept lot)
P(accept lot) = 20C0 * (0.98)20 * (0.02)0 = 1 * (0.98)20 * 1
=> P(accept lot) = 0.6676
Therefore,
P(producer's risk) = 1 - P(accept lot) = 1 - 0.6676 = 0.3324
=> P(producer's risk) = 0.3324
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