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Math Help - Uniform random variable question

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    Uniform random variable question

    Suppose that 10^6 people arrive at a service station at times that are independent random variables, each of which is uniformly distributed over (0,10^6). Let N denote the number that arrive in the first hour. Find an approximation for P(N = i).
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    Hello,

    Consider the event : the customer arrives at the first hour, and its complementary : the customer doesn't arrive at the first hour.

    Then we have a binomial distribution.
    You can approximate it by a Poisson distribution (see here : Binomial distribution - Wikipedia, the free encyclopedia )
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    Super Member Anonymous1's Avatar
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    I guess I'm just confused by the wording of the question. Is the probability that the customer arrives at the first hour = \frac{1}{10^6}?
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    Quote Originally Posted by Anonymous1 View Post
    I guess I'm just confused by the wording of the question. Is the probability that the customer arrives at the first hour = \frac{1}{10^6}?
    Yes (depending on what 10^6 represents...minutes or hours)

    And n=10^6
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    Super Member Anonymous1's Avatar
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    Thanks.

    So,

    P(N = i) = e^{-1} \frac{1^i}{i!}

    I'm only worried because my answer seems to simple. I don't need to use a summation from 0 to i, do I?
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    Quote Originally Posted by Anonymous1 View Post
    Thanks.

    So,

    P(N = i) = e^{-1} \frac{1^i}{i!}

    I'm only worried because my answer seems to simple. I don't need to use a summation from 0 to i, do I?
    No you don't. You're asked the probability for a given i

    And it's rather P(N=i)~{\color{red}\approx}~e^{-1}\cdot\frac{1^i}{i!}~\left[=\frac{e^{-1}}{i!}\right]
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