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Math Help - poisson

  1. #1
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    poisson

    Customers arrive at photocopying machine at an average rate of 2 each 5 minutes. Assume that this arrivals are independent, with a constant arrival rate and hat thi sproblem follows a poisson distributionn, with X denoting the number of arrivals in a 5 minute period and mean lambda=2. Find the probab that more than 2 customers arrive in a 5 minute period.

    This is no problem to solve. I just want to see if I got Poisson well. Like, 5 is my time base, right. So I know, that 2 persons come each 5 minutes. So people are computed on a base(subinterval) of 5 minutes. So if they ask me the prob that more than 2 customers arrive in a 1 minute period I should do

    2: 5= x:1 x=2/5=0.4 and my lambda would be 0.4
    so P=1 - P(0) - P(1) -P(2)

    right?? Thank you so much
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  2. #2
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    Quote Originally Posted by 0123 View Post
    Customers arrive at photocopying machine at an average rate of 2 each 5 minutes. Assume that this arrivals are independent, with a constant arrival rate and hat thi sproblem follows a poisson distributionn, with X denoting the number of arrivals in a 5 minute period and mean lambda=2. Find the probab that more than 2 customers arrive in a 5 minute period.
    The probability that k people arrive in that time interval is P(k) = \frac{2^ke^{-2}}{k!}. Here k\geq 2.
    Thus,
    P(k\geq 2) = 1 - P(0\leq k \leq 1) = 1 - e^{-2} - 2e^{-2} \approx .59
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    The probability that k people arrive in that time interval is P(k) = \frac{2^ke^{-2}}{k!}. Here k\geq 2.
    Thus,
    P(k\geq 2) = 1 - P(0\leq k \leq 1) = 1 - e^{-2} - 2e^{-2} \approx .59
    well actually it's not. Because it asks for "more than 2" so you should exclude also P(X=2) and the result is 0. 3233.

    But this is clear. I was wondering about the case in which the period to be examined changes.Can you help? Thanks
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