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Math Help - Exponential Distributions

  1. #1
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    Exponential Distributions

    I am having a hard time distinguishing the difference between Poisson and Exponential distributions.

    Here is the question I am on.

    The time between calls to a plumbing supply business is exponentially distributed with a mean time between calls of 15 minutes

    a.
    What is the probability that there are no calls within a 30-minute interval?
    b.
    What is the probability that at least one call arrives within a 10-minute interval?
    c.
    What is the probability that the first call arrives within 5 and 10 minutes after opening?


    I am confused about the part when the question asks, for example, "no calls within a 30 minute interval" or "at least one call in a 10 min interval" thanks for any help.
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  2. #2
    MHF Contributor matheagle's Avatar
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    1 The Poisson is discrete, in this case it's the number of calls per unit of time
    2 The Exponential is continuous, it's the time between calls.

    Let X\sim Poisson(\lambda t)

    So P(X=x)={e^{-\lambda t}(\lambda t)^x\over x!}

    Then P(X=0)=e^{-\lambda t} means that we had zero arrivals by time t.

    Let Y be the time of the first arrival (this is the exponential rv).

    P(Y>y) means the same, we have had no arrivals by time y.

    So P(Y>y)=e^{-\lambda y}

    But that is the distribution of an exponential, with parameter \lambda

    So F_Y(y)=P(Y\le y)=1-e^{-\lambda y}

    and f_Y(y)= \lambda e^{-\lambda y}

    So in (a) you want no calls in any 30 minute period which is twice the mean

    giving you P(Y>2)=e^{-2}
    Last edited by matheagle; October 18th 2009 at 09:32 PM.
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  3. #3
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    thanks for the clarification, what if I had the question.

    The distance between major cracks in a highway follows an exponential distribution with a mean of 8 miles.

    and I was asked

    a) What is the probability that there are 2 major cracks in a 10-mile stretch of the highway?

    what would I do about that "2 major cracks" part? Thanks.

    could I just do (e^-1.25 * 1.25^2)/2! or would it be 1 - e^-1.25?
    Last edited by Oblivionwarrior; October 18th 2009 at 08:18 AM.
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  4. #4
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    Quote Originally Posted by Oblivionwarrior View Post
    thanks for the clarification, what if I had the question.

    The distance between major cracks in a highway follows an exponential distribution with a mean of 8 miles.

    and I was asked

    a) What is the probability that there are 2 major cracks in a 10-mile stretch of the highway?

    what would I do about that "2 major cracks" part? Thanks.

    could I just do (e^-1.25 * 1.25^2)/2! or would it be 1 - e^-1.25?
    Let X be the random variable number of cracks in a 10 mile stretch.

    X ~ Poisson (\lambda = 10/8 = 1.25) where the mean is got by understanding the relationship between the Poisson distribution and the exponential distribution.

    You require Pr(X = 2) so yes, your first answer is the correct one.
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