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  1. #1
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    Question on Poisson Distribution

    A production process outputs items in lots of 50. Sampling plan exist in which lots are pulled aside periodically and exposed to certain type of inspection. It is usually assumed that the proportion defective is very small. It is important to the company that lots containing defectives be a rare event. The current inspection plan is to periodically sample randomly 10 out of the 50 items in a lot and, if none are defective, to perform no intervention.

    a. Suppose in a lot chosen at random, 2 out of 50 are defective. Calculate the probability that at least 1 in the sample of 10 from the lot is defective.

    b. Calculate the mean number of defects found out of 10 items sampled.

    May i know that how can I solve the above questions?
    I appreciate your guides, thank you!
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  2. #2
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    Re: Question on Poisson Distribution

    Quote Originally Posted by HTHVampire View Post
    A production process outputs items in lots of 50. Sampling plan exist in which lots are pulled aside periodically and exposed to certain type of inspection. It is usually assumed that the proportion defective is very small. It is important to the company that lots containing defectives be a rare event. The current inspection plan is to periodically sample randomly 10 out of the 50 items in a lot and, if none are defective, to perform no intervention.

    a. Suppose in a lot chosen at random, 2 out of 50 are defective. Calculate the probability that at least 1 in the sample of 10 from the lot is defective.

    b. Calculate the mean number of defects found out of 10 items sampled.

    May i know that how can I solve the above questions?
    I appreciate your guides, thank you!
    This has nothing to do with the Poisson distribution. It has to do with the binomial distribution.


    The probability $p$ of an item being defective is $p=\frac 2 {50} = \frac 1 {25}$

    The probability that at least 1 out of a sample of 10 is defective is given by

    1-Pr[no defects in 10 samples] = $1 - (1-p)^{10} = 1 - \left(\frac {24} {25} \right)^{10} \approx .335$



    The probability of $k$ defects out of 10 is given by

    $Pr[k] = \left(\begin{array}{r}10 \\ k \end{array}\right) p^k (1-p)^{10-k}$

    The expected value of the number of defects out of 10 is given by

    $\displaystyle{\sum_{k=0}^{10}} k Pr[k] = \displaystyle{\sum_{k=0}^{10}} \left(\begin{array}{r}10 \\ k \end{array}\right) p^k (1-p)^{10-k}$

    you can work out the sum.
    Last edited by romsek; March 30th 2014 at 08:00 PM.
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  3. #3
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    Re: Question on Poisson Distribution

    Hi romsek,

    Thanks for the correction. May I know that for part (b), since the question is binomial distribution where X~B(10, 2/50),
    can I use the formula for mean of X = np ?

    np= 10 * (2/50)

    Is it correct to do so? Thank you.
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  4. #4
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    Re: Question on Poisson Distribution

    yes, it is.
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  5. #5
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    Re: Question on Poisson Distribution

    Hi Romsek,

    Alright, I appreciate your guides in this question! Thank you!
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