The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that on the average ther

Question

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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Eva 5 hours 2021-09-12T03:37:15+00:00 1 Answer 0

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    2021-09-12T03:38:51+00:00

    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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