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Viewing 6 posts - 1 through 6 (of 6 total)
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  • #43414

    Ruddy
    Participant

    Hello!
    I am looking for some probability guidance!
    I am sampling a lot of 300 units from one of my vendors.
    The vendors past quality has been approx. 2% defective.
    I am taking a sample of 40 pieces and will be rejecting the entire lot if two or more parts are found defective.
    How can I determine the probability of finding two or more defective parts?
    Thanks for everyone’s comments!

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    #137692

    Brit
    Participant

    email me at [email protected] give me your email address
    The symbols don’t wat to seem to come on the screen.

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    #137705

    dangwal
    Participant

    Michael – in all probability this seems to be an examination question……..
    You can use binomial distribution in minitab calc menu to find cumulative prob for <2 (0 & 1) defects. Number of trials is 40 and probability of suuccess is 0.02.
     

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    #137715

    Brit
    Participant

    If you don’t have minitab, let me know. It is easy to calculate by hand.

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    #137946

    Gomes
    Participant

    x  P( X <= x )0 0.445700
    1 0.809537
    2 0.954330

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    #137947

    Gomes
    Participant

    I did run binomila dist on Minitab and got the following results for the problem posted by Michael.For the number of trails = 40For the probability of success of finding atleast 2 defective parts = 6/300 = 0.02 (2% of part are defective). The prob values genertated are as follows:x  P(X<=x)0  0.4457001  0.8095372  0.954330
    The interpretation here is:The prob of finding 0 defective parts = 0.44The prob of finding 1 defective parts = 0.80The prob of finding 2 defective parts = 0.95I would like to know the following:1. Should I use simply the Binomial Prob dist or the Cumulative Prob Dist? Why?2. Are the interpretations right or not?3. Any other answers, suggestions…
    Thanks and Regards,Gomes

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Viewing 6 posts - 1 through 6 (of 6 total)

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