A shirt manufacturing company supplies shirts in lots of size 250 to the buyer. A single sampling plan with n = 20 and c = 1 is being used for the lot inspection. The company and the buyer decide that AQL = 0.04 and LTPD = 0.10. If there are 15  defective in each lot, compute the
i) probability of accepting the lot. 
ii) producer’s risk and consumer’s risk. 
iii) average outgoing quality (AOQ), if the rejected lots are screened and all defective shirts
are replaced by non-defectives. 
iv) average total inspection (ATI). 

We can use the binomial distribution to calculate the probabilities and risks associated with this sampling plan.

Given:

Lot size = 250
Sample size = 20
Acceptable Quality Level (AQL) = 0.04
Limiting Quality Level (LTPD) = 0.10
Defective items in the lot = 15
i) Probability of accepting the lot:

The lot will be accepted if the number of defects in the sample is less than or equal to the acceptance number, which is c = 1 in this case. We can use the binomial distribution to calculate the probability of observing 0 or 1 defects in a sample of size 20 when the true proportion of defects in the lot is 15/250 = 0.06 (since there are 15 defects in a lot of 250):

P(0 or 1 defects) = P(X = 0) + P(X = 1)
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