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29 October, 09:44

The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that on the average there are 10 calls per hour. (a) What is the probability that there are three or fewer calls in one hour

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  1. 29 October, 10:19
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    Answer: the probability that there are three or fewer calls in one hour is 0.011

    Step-by-step explanation:

    The formula for poisson distribution is expressed as

    P (x = r) = (e^ - µ * µ^r) / r!

    Where

    µ represents the mean of the theoretical distribution.

    r represents the number of successes of the event.

    From the information given,

    µ = 10

    For the probability that there are three or fewer calls in one hour, it is expressed as

    P (x ≤ 3) = P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3)

    Therefore,

    P (x = 0) = (e^ - 10 * 10^0) / 0! = 0.000045

    P (x = 1) = (e^ - 10 * 10^1) / 1! = 0.00045

    P (x = 2) = (e^ - 10 * 10^2) / 2! = 0.0023

    P (x = 3) = (e^ - 10 * 10^3) / 3! = 0.0077

    Therefore,

    P (x ≤ 3) = 0.000045 + 0.00045 + 0.0023 + 0.0077 = 0.011
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