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Math Help - Poisson distribution

  1. #1
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    Poisson distribution

    Calls for dial-in connections to a computer centre arrive at an average rate of four per minute. The calls follow a Poisson distribution. If a call arrives at the beginning of a 1-minute interval, what is the probability that a second call will not arrive in the next 20 seconds?

    I'm having trouble putting this into the Poisson probability distribution formula. It seems different from some of the other problems I've done. How do you solve this?

    Jack
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  2. #2
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    Quote Originally Posted by JackStolerman View Post
    Calls for dial-in connections to a computer centre arrive at an average rate of four per minute. The calls follow a Poisson distribution. If a call arrives at the beginning of a 1-minute interval, what is the probability that a second call will not arrive in the next 20 seconds?

    I'm having trouble putting this into the Poisson probability distribution formula. It seems different from some of the other problems I've done. How do you solve this?

    Jack
    For a Poisson process, if the arrivals per unit of time have a Poisson distribution with mean \lambda, the waiting time between arrivals has an exponential distribution with mean \lambda^{-1} and cumulative distribution function F(x) = 1 - e^{-\lambda x}. Can you take it from here?
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  3. #3
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    Well I'm pretty sure that I have to integrate.

    Also is the formula the same as:
    f(y)=(1/B)*e^(-y/B) where y is greater than or equal to zero and less than infinity.
    I have to integrate this, but what are the bounds and everything?

    Jack
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  4. #4
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    OK. I think I got it, but could someone confirm this:

    I let Y be the time between the arrival of two dial-in connections in minutes. We have P(Y>1/3)
    (lambda)t=4 and (1/lambda)=(1/4)
    Thus, f(y)=(1/0.25)e^(-y/0.25)=4e^(-4y)
    P(Y>1/3)=integration of 4e^(-4y) from (1/3) to infinity. This equals e^(-4/3).

    Is this the probability?
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  5. #5
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    Quote Originally Posted by JakeD View Post
    For a Poisson process, if the arrivals per unit of time have a Poisson distribution with mean \lambda, the waiting time between arrivals has an exponential distribution with mean \lambda^{-1} and cumulative distribution function F(x) = 1 - e^{-\lambda x}. Can you take it from here?
    Quote Originally Posted by JackStolerman View Post
    OK. I think I got it, but could someone confirm this:

    I let Y be the time between the arrival of two dial-in connections in minutes. We have P(Y>1/3)
    (lambda)t=4 and (1/lambda)=(1/4)
    Thus, f(y)=(1/0.25)e^(-y/0.25)=4e^(-4y)
    P(Y>1/3)=integration of 4e^(-4y) from (1/3) to infinity. This equals e^(-4/3).

    Is this the probability?
    Yes. Note the cumulative distribution function is F(x) = P(Y \le x) so the probability P(Y > 1/3) = 1-F(1/3) = e^{-\lambda /3} = e^{-4/3}.
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