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Math Help - Poisson Distribution Question

  1. #1
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    Poisson Distribution Question

    A cashier passes an average of 7 customers an hour. Using Poisson distribution, what is probability that she will pass 2 customers in 2 different 1 hour time frames (i.e between 1:00-2:00 & 3:00-4:00)?

    Solution:

    since they it's 2 diffent one hour periods, I would think that you would have: 2 \left( \frac{7^1}{1!} e^{-7} + \frac{7^2}{2!} e^{-7} + e^{-7} \right) = 0.05927237 , since she can have 0 in the first interval but 2 in the second one, or one in the first and another in the second. Is this correct?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lllll View Post
    A cashier passes an average of 7 customers an hour. Using Poisson distribution, what is probability that she will pass 2 customers in 2 different 1 hour time frames (i.e between 1:00-2:00 & 3:00-4:00)?

    Solution:

    since they it's 2 diffent one hour periods, I would think that you would have: 2 \left( \frac{7^1}{1!} e^{-7} + \frac{7^2}{2!} e^{-7} + e^{-7} \right) = 0.05927237 , since she can have 0 in the first interval but 2 in the second one, or one in the first and another in the second. Is this correct?
    If this mean what is the probability that she passess 2 customers total in the
    two time slices, then it does not matter that they are disjoint, this is the
    same probability that she passess two customers in a two hour interval.

    RonL
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  3. #3
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    I'm not sure I fully understand. Are you saying that it doesnt make a difference if the intervals are split or joined, implying that the mean would be 14=2*7?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by lllll View Post
    I'm not sure I fully understand. Are you saying that it doesnt make a difference if the intervals are split or joined, implying that the mean would be 14=2*7?

    Yes the mean would be that, and the number has a Poisson distribution
    with that mean.

    RonL
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