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Thread: Poisson distribution

  1. #1
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    Poisson distribution

    I am really stuck on this one.
    The number of phone calls received by an operator in 5-minute period follows a Poisson distribution with a mean of lamba (L). Find the probability that the total number of phone calls received in 10 randomly selected 5-minute periods is 10.

    Let x_1 be the number of calls in the 1st 5-min period

    Let x_2 be the number of calls in the 2nd 5-min period

    ...Let x_10 be the number of calls in the 10th 5-min period.

    The way I see it gives too many possibilities.

    I want to compute p(10,0,0,..0) + p(0,10,0,0,...0) + ... = p(0,0,...,10) + p(9,1,0...0) + p(9,0,1,0,...0) + p(9,0,0...,1) +...+ p(1,1,1,1....,1)
    = 10[(e^L*L^10)/10!][e^L]^9 + 9[e^L*L^9)/9!][(e^L*L][e^L]^8 + ....+ [(e^L*L]*10
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  2. #2
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    Re: Poisson distribution

    Hey JaguarXJS.

    Hint - Consider the independence criterion [of distributions] and what happens when you add Poisson distributions together. [Other hint - look at moment generating functions].
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  3. #3
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    Re: Poisson distribution

    Quote Originally Posted by chiro View Post
    Hey JaguarXJS.

    Hint - Consider the independence criterion [of distributions] and what happens when you add Poisson distributions together. [Other hint - look at moment generating functions].
    Professor Chiro,
    Thanks for responding to my post. I truly appreciate it.
    So the answer is [(e^-10L)(10L)^10]10!?I think this is what your hint is saying.
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  4. #4
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    Re: Poisson distribution

    Basically when you have two independent Poisson distributions and you add them together, then you should get a Poisson distribution with the two lambdas added together.

    From that you can look at probabilities.
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