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Math Help - Quality Control?

  1. #1
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    Quality Control?

    Hey, I'm new, and I need to understand how to answer this question asap.

    I'm sorry if this isn't an advanced level problem (I think it's pretty advanced)
    I am in a university level class called Decision Sciences << so I think it's advanced

    Anyway, I would really appreatiate the help:

    The quality control department of a company has decided to select a sample of 20 items from each shipment of goods it receives and inspect them for defects. It has been decided that if the sample contains no defective parts, the entire lot will be accepted.
    a. what is the probability of accepting a lot that contains 10% defective items?
    b. what is the probability of accepting a lot that contains 5% defective items?
    c. what is the probability of rejecting a lot that contains 15% defective items?

    I've tried this so many times and can't figure out how to get these answers provided by my instructer:
    a. .1216
    b. .3584
    c. .9612

    Thank you
    Last edited by StatsStar; May 2nd 2011 at 01:15 PM. Reason: typo
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  2. #2
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    Answer these quick questions...

    If you pick one (1), what is the probability it is defective?
    If you pick one (2), what is the probability both are defective?
    If you pick one (3), what is the probability both are defective?

    If you pick one (1), what is the probability it is NOT defective?
    If you pick one (2), what is the probability NEITHER is defective?
    If you pick one (3), what is the probability NONE is defective?

    Now answer the three questions on the assignment.

    a) Given that 10% are defective, p(pick 20 and get no defectives) = ??
    b) Given that 5% are defective, p(pick 20 and get no defectives) = ??
    c) Given that 15% are defective, p(pick 20 and get ANY defectives) = 1 - p(pick 20 and get ALL NOT defectives) = ??
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  3. #3
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    Is this the only information you have been given? Do you have a sample mean or standard deviation for these 20 items?
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  4. #4
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    Are you saying there is missing information?

    Cause this is all the info I was given...
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  5. #5
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    You didn't answer my questions. Trust me on this. Answer the six questions.
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  6. #6
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    I don't understand what your asking; I really want to answer your questions.

    I can only guess that if I pick one out of the sample of 20 (given that the lot is 10% defective) then the odds of it being defective is 10%

    and then I don't know what you mean when you say one (2) is that picking 1 or 2?

    I don't understand how this helps me. :[
    I really need to solve this by tomorrow morning.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by StatsStar View Post
    Hey, I'm new, and I need to understand how to answer this question asap.

    I'm sorry if this isn't an advanced level problem (I think it's pretty advanced)
    I am in a university level class called Decision Sciences << so I think it's advanced

    Anyway, I would really appreatiate the help:

    The quality control department of a company has decided to select a sample of 20 items from each shipment of goods it receives and inspect them for defects. It has been decided that if the sample contains no defective parts, the entire lot will be accepted.
    a. what is the probability of accepting a lot that contains 10% defective items?
    b. what is the probability of accepting a lot that contains 5% defective items?
    c. what is the probability of rejecting a lot that contains 15% defective items?

    I've tried this so many times and can't figure out how to get these answers provided by my instructer:
    a. .1216
    b. .3584
    c. .9612

    Thank you
    a) If the true defect rate is 10% (or 0.1) then the number of defectives in a batch of size N has a binomial distribution B(20,0.1). For it to have no defectives (and so be accepted) the probability is b(0;20,0.1)

    The others are just the same.

    CB
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  8. #8
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    Have we estblished that this is binomial? Each item is either defective or not.

    Picking only one item, p(defective) = 0.10, p(not defective) = 0.90
    Picking 2 items, p(both defective) = 0.10^2 and p(neither defective) = 0.90^2
    Picking 3 items, p(all defective) = 0.10^3 and p(none defective) = 0.90^3
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  9. #9
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    Hehe, sorry, I ment to post sooner but I actually managed to solve it
    (locked out cause I forgot my pass. xD)

    I used: f(x) = n!/(x!(n-x)!)p^x(1-p)^(n-x)

    thanks :]
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