Sampling Plan Problem

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    Can someone help me to solve a sampling plan problem?

    A single sampling plan calls for a sample size of 80 with an acceptance number of 5 and a rejection number of 6. If the quality of the submitted lots is ten percent defective, then the percent of lots expected to be accepted in the long run is approximately

    A. 6%
    B. 10%
    C. 20%
    D. 30%

    explanation is greatly appreciated.


    Katie Barry

    Tracy — Although the iSixSigma audience is helpful, they are not here to do your work for you. You are far more likely to get a response if you first explain your process. What do you think the answer is? Show your work! If you think you’re wrong say where you think you got off the rails. The more information you share, the better advice you are likely to receive.


    Amit Kumar Ojha

    Hi Tracy,

    Katie is right but since you asked the question, I am mentioning below how you can resolved it.
    See here any lot which has more than or equal to 6 defects will be considered defective. We are concerned about the number of lots which would be accepted (non defective) in long run. Since here we are talking about defectives, the appropriate distribution would be Binomial Distribution.
    No of Trials(N) = 80 , Probability of success(p) 0.1

    Probability of getting a lot which is defective = P(Defects <=5) [Let r=5]
    The formula for calculating the probability in binomial distribution is NCr*(p^N)*(q^N-r).

    Hope it helps..If you find further difficulties, please feel free to let me know.

    All the best!!!

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