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Math Help - Poisson Dist.

  1. #1
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    Poisson Dist.

    The number of people arriving for treatment at an emergency room can be modelled by a Poisson process with a rate parameter of five per hour. What is the probability that more than 12 people arrive during a two hour period?

    I said >12 people arriving in 2 hours would be the same as finding >6 people in one hour using the same parameter \lambda = 5

    Using P(X) = \frac{e^{-\lambda}\lambda^x}{x!}

    P(X>6) = 1- P(X\leq 6) = 1-0.762 = .238

    but this was not correct. I think the problem here is my logic.

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  2. #2
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    Quote Originally Posted by Bushy View Post
    The number of people arriving for treatment at an emergency room can be modelled by a Poisson process with a rate parameter of five per hour. What is the probability that more than 12 people arrive during a two hour period?

    I said >12 people arriving in 2 hours would be the same as finding >6 people in one hour using the same parameter \lambda = 5 Mr F says: This is not correct.

    Using P(X) = \frac{e^{-\lambda}\lambda^x}{x!}

    P(X>6) = 1- P(X\leq 6) = 1-0.762 = .238

    but this was not correct. I think the problem here is my logic.

    Calculate Pr(Y > 12) where Y ~ Poisson(mean = 10).
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  3. #3
    Super Member Random Variable's Avatar
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    I'm fairly certain that those aren't equivalent statements.

    Let Y be the number of people who arrive during any two hour period.

    Then Y follows a Poisson distribution with a mean of 10.

    and  P(Y>12) = 1-P(Y \le 12) = 1 - \sum_{j=0}^{12} \frac{e^{-10} 10^{j}}{j!}
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