Ask Question
17 August, 15:24

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

+4
Answers (1)
  1. 17 August, 18:00
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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. ...” in 📘 Business if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers