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23 October, 09:14

Customers arrive at a suburban ticket outlet at the rate of 2 per hour on monday mornings. this can be described by a poisson distribution. selling the tickets and providing general information takes an average of 20 minutes per customer, and varies exponentially. there is 1 ticket agent on duty on mondays. what is the probability that an arriving customer will not have to wait for service

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  1. 23 October, 09:23
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    Answer: 0.513246

    Explanation:

    Given the following;

    Customer arrival rate = 2 per hour

    selling the tickets and providing general information takes an average of 20 minutes per customer.

    1 ticket agent

    Since it is related by the poison distribution;

    λ = 2/hour

    μ = 20 minutes per customer

    μ = (60 : 20) minutes

    μ = 3 per hour

    Average waiting time is calculated by;

    λ : [μ (μ - λ) ]

    Wait time = 2 : [3 (3-2) ]

    Wait time = 2 : 3

    = 0.667hours

    Therefore, probability that customer will not have to wait for service, that is wait time, x = 0

    P (x; μ) = (e^-μ) (μ^x) / x!

    P (0; 0.667) = (0.513246) (1) / 0!

    P (0; 0.667) = 0.513246 / 1

    P (0; 0.667) = 0.513246
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