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    An engineering company, which supplies car components in batches of size 1000 to an assembly plant, applies the following double sampling scheme to all batches to monitor the quality of its production.
    A random sample of 50 components is taken from a batch. The batch is accepted if no defectives are found and is rejected if more than 2 defectives are found. If one or two defectives are found, a second sample of 100 components is taken from the batch. The batch is then accepted if this second sample contains no more then two defectives; otherwise it is rejected.

    (i) For a randomly chosen batch containing a proportion p=0.025 of defectives, what is the probability on the basis of the first sample that
    (a) a batch is accepted
    (b) a batch is rejected?

    For any randomly chosen batch, what is the probability that a second sample is taken?
    Under the double sampling scheme what is the overall probability that
    (c) a batch is accepted
    (d) a batch is rejected?

    (ii) The policy with rejected batches is to inspect every component, replacing all the defective ones. Any defectives found in samples from accepted batches are also replaced.

    Let the random variable D be the number of defectives remaining in the batch after this policy has been applied to the double sampling scheme.

    What values can D take, and what probabilities are associated with these values, if, on average, batches contain a proportion p=0.025 of defectives?

    Hence, evaluate the average outgoing quality under the double sampling scheme.

    ---------------------------------------------------------------------

    Right this is what I have done so far, so if anyone could correct me or help me it would be appreciated, I really am stuck on this question, and have no idea but I had a go at the first parts

    (i)
    (a) X~Bin(50,0.025)
    therefore X~Po(1.25)

    p(accept)= P(x=0) = 0.287

    (b) p(reject)
    p(x>2) = 1 - p(x less than or equal to 2)
    1- [p(x=0) + p(x=1) + p(x=2)]
    = 1 - 0.869 = 0.131

    p(second sample is taken) = P(one or two defectives are found)

    = p(one defective) + p(two defectives)
    = 0.287 + 0.358 = 0.645

    I'm not too sure about these rest of the parts, and don't even know if the first parts I've done are right. If anyone could help it'd be appreciated. Thanks
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    Agree with 0.287, but the 2nd answer should be 0.132. Hence 3rd answer is 1 - .287 - .132 = 0.581. [You got P(2) wrong I think.]
    Chance it is finally accepted = 0.287 + 0.581 X exp[-2.5] = 0.335.
    Hence chance it is finally rejected = 1 - .335 = 0.665.
    Now let's see your attempts at the rest of the question.
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    For p(x=2) I get 0.224, so therefore p(second sample is taken) = 0.287+0.224 = 0.511. What did you get for p(x=2), I must be putting it into the calculator wrong. Also I'm really not sure about part (ii) of the question concerning the random variable D.

    Wouldn't the values of D just be 1.25 and 2.5?
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    P[2nd sample] = 0.358 + 0.224 = 0.582.
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    Ah I see! Thank you. Any ideas on part (ii) anyone?
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    Anyone please got any tips on what to do for part (ii)?
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    I am guessing here, but this is what I reckon:
    You expect the batch of 1000 to have 1000 X 0.025 = 25 duds.
    If your 1st sample of 50 has no duds, then you accept the batch - which is expected to have 25 duds. [Probability = 0.28765]
    If your 1st sample has 1 dud and your 2nd sample has no duds, then you accept the batch - which is expected to have 25 - 1 = 24 duds.[Probability = 0.02940]
    If your 1st sample has 2 duds and your 2nd sample has no duds, then you accept the batch - which is expected to have 25 - 2 = 23 duds. [Probability = 0.01837]
    In all other cases, the batch is rejected and so it will then be cleared of all duds. [Probability = 1 - sum of probabilities above = 0.6657]
    And so: Expected number of duds per batch: 25 24 23 0
    Probability: 0.2865 0.0294 0.01837 0.6657
    Now work out the mean of this distribution to get an expected value of 8.3 duds per sold batch.
    Any good?
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    Ok, thank you so much for your help. For 24 defectives for example did you get 0.358 x 0.0821 = 0.02938 = 0.0294? (just making sure!)

    Right going to be cheeky and see if I can get help on another question, it's a similar one, but I'm having trouble working out the proportion p of defectives:

    Batches of 400 items each containing 3 defective items are subject to the following double sampling plan. A batch is accepted if a first sample of 5 items contians no defectives and is rejected if this sample contains two or more defectives. If the first sample contains one defective, this defective is replaced by a good item and the first sample is then returned to the batch. A new sample of 7 items is then taken from the batch, The batch is then accepted if this second sample contains no more than one defective; otherwise it is rejected. At any stage if a defective item is found in a sample, it is replaced by a good item. If a batch is rejected, it is then fully inspected and all defective items are replaced.

    (i) What is the probability on the basis of the first sample that:
    (a) the batch is accepted
    (b)a second sample is taken.

    Under the double sampling scheme, what is the overall probability that the batch is accepted.

    (ii) If the random variable D is the number of defective items in a batch after the double sampling scheme has been applied, what values can D take and what probabilities are assoicated with these values?
    Evaluate the average outgoing quality under this double sampling scheme.

    (iii) Repeat your calculations for parts (i) and (ii) above using the appropriate Poisson approximation. Comment on what effect this has on your results, and why?

    -----------------------------------------------------------------------------------------

    Right since part (iii) is asking for a Poisson approximation I'm therefore guessing for parts (i) and (ii) we use a Binomial distribution. I'm having trouble finding what the value of p would be in X~Bin(n=400,p=?). Would it be 3/400 = 0.0075? Therefore using X~Bin(400,0.0075)?

    Thanks again for the help!
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    Anyone?! Got a paper on this stuff tomorrow! Arghhh
 
 
 
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