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24 January, 12:09

Messages arrive to a computer server according to a poisson distribution with a mean rate of 10 per hour. determine the length of an interval of time such that the probability that no messages arrive during this interval is 0.90.

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  1. 24 January, 12:20
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    Assuming that the Poisson distribution is applicable:

    The Poisson distribution is f (k) = λ^k * e^ (-λ) / k!

    Where:λ is the expected number of circumstances of the occurrence during interval t;

    In the problem, r = 10 calls/hour. Where it is also = 1/360 calls/sec, therefore λ = t/360 calls during interval t;

    f (k) is the probability of exactly "k" occurrences;

    So based from the problem: f = 0.9; k = 0; λ = t/360

    0.9 = (t/360) ^0 * e^ (-t/360) / 0!

    0.9 = e^ (-t/360)

    t = - 360 ln 0.9 = 37.93 s

    it is approximated: 38 s is the interval with 0.9 probability of no calls received
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