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Thread: Poisson Distribution Question

  1. #1
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    Poisson Distribution Question

    So I got a summer homework package from school , and I need to teach myself Poisson Distribution before school starts. So here's the question I've been stuck on for a while. (I need help on b.)


    The number of faults in glass sheets occur at a rate of 2.1 per sq. metre. If a 1x1 sq. m glass sheet contains at least 3 faults it is returned to the manufacturer.

    a) Find the probability that a 1x1 sq. m sheet is returned to the manufacturer.

    My answer:

    $\displaystyle P(X > 2) = 1 - P(0) - P(1) - P(2)$
    $\displaystyle P(X > 2) = 1 - e^{-2.1}(1+2.1+\frac{2.1^{2}}{2})$
    $\displaystyle P(X > 2) = 0.3504$

    b) Six such glass sheets are inspected. What is the probability that at least half of them are returned to the manufacturer? (Here's the answer provided: 0.6817)

    It would be nice if you can show me the work and proper notation, because I barely have a clue.

    Thanks.
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  2. #2
    Senior Member Danneedshelp's Avatar
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    Let $\displaystyle Y$ be a random varibale that represents the number defective glass sheets. Then, $\displaystyle Y$ is $\displaystyle Binomial(n=6, p=0.3504)$, since the probability of a success is equal to the probability of a sheet of glass being defective. Thus, the probability that at least half the glass sheets will be defective out of a batch of 6 glass sheets ought to be

    $\displaystyle P(Y\geq\\2)=\sum_{y=3}^{6}{{6}\choose{y}}(0.3504)^ {y}(1-0.3504)^{6-y}$.

    I think that's correct.
    Last edited by Danneedshelp; Aug 2nd 2010 at 08:17 PM.
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  3. #3
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    Thanks for your reply.

    I see what you are doing using binomial distribution. However, the answer that your equation got wasn't the answer that was provided. The answer that you got was about 0.353.

    I also used the Poisson Distribution way to solve the question based on your logic, and it got the same (or approx.) answer:

    $\displaystyle \lambda = (0.3504)(6) = 2.124$

    so,

    $\displaystyle P(X > 2) = 1 - P(0) - P(1) - P(2)$
    $\displaystyle P(X > 2) = 1 - e^{-2.124}(1 + 2.124 + \frac{2.124^{2}}{2})$
    $\displaystyle P(X > 2) = 0.3568$

    Hm.. are we doing something wrong or the answer provided is just wrong?
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  4. #4
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    Danneedshelp is correct,

    $\displaystyle Y \sim Binomial(n=6, p=0.3504)$

    $\displaystyle P(Y \geq 2) = 0.681689$

    You can use Binomial Distribution: Probability Calculator to check your answers when calculating binomial probabilities.
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  5. #5
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    Quote Originally Posted by timmcan View Post
    So I got a summer homework package from school , and I need to teach myself Poisson Distribution before school starts. So here's the question I've been stuck on for a while. (I need help on b.)


    The number of faults in glass sheets occur at a rate of 2.1 per sq. metre. If a 1x1 sq. m glass sheet contains at least 3 faults it is returned to the manufacturer.

    a) Find the probability that a 1x1 sq. m sheet is returned to the manufacturer.

    My answer:

    $\displaystyle P(X > 2) = 1 - P(0) - P(1) - P(2)$
    $\displaystyle P(X > 2) = 1 - e^{-2.1}(1+2.1+\frac{2.1^{2}}{2})$
    $\displaystyle P(X > 2) = 0.3504$

    b) Six such glass sheets are inspected. What is the probability that at least half of them are returned to the manufacturer? (Here's the answer provided: 0.6817)

    It would be nice if you can show me the work and proper notation, because I barely have a clue.

    Thanks.
    For (b), it looks that the answer provided is the probability when only 4 glass sheets are inspected.
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  6. #6
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    Thanks for the link SpringFan, but Danneedshelp used $\displaystyle P(X \geq 3)$, not $\displaystyle P(X \geq 2)$. Still, it seems that what you did came up as the correct answer according to the answer sheet. I guess there's just something wrong with the wording of the question.

    Thanks for your help.
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