Question

customers arrive at an average of 30 per hour. A single server in the store serves customers, taking 1.5 minutes on average to serve each customer. Inter-arrival times and service times follow the exponential distribution. What is the expected fraction of time that the server will be busy? On average, how many people will there be in the store? On average, how long will someone be in the store? What is the probability that there will be more than 2 people in the store at a given time?

Answer 1

Arrival rate = λ = 30 / hour

Service Rate = μ = 60/1.5 = 40 / hour

Fraction of time that the server is busy = Probability that the server is busy = λ/μ = 30/40 = 0.75 or 75%

The average number of customers in the store = λ/(μ-λ) = 30/(40-30) = 3

The average time a customer spends in the store = 1/(μ-λ) = 1/(40-30) = 0.10 hour = 6 mins

Probability of n people in the store = (λ/μ)ⁿ(1 - λ/μ)

Hence, Probability of more than 2 people in the store = 1 - Probability of 2 people in store - Probability of 1 person in store - Probability of 0 people in store = 1 - (λ/μ)²(1 - λ/μ) - (λ/μ)(1 - λ/μ) - (1 - λ/μ) = 1 - (30/40)²(1 - 30/40) - (30/40)(1 - 30/40) - (1 - 30/40) = 0.4218