Twelve customers arrive at the lunch counter at George's Happy Hog per hour and select the hickory smoked meat du jour along with a side and soda for $5. The server doubles as the cashier and can take and fill an order in three minutes on average. What is the likelihood that there is one customer in the system at the George's Happy Hog if the times can be assumed as exponentially distributed?
A) 0.24
B) 0.19
C) 0.81
D) 0.75

Question

Quizplus · Accepted Answer

First, we need to calculate the arrival rate (λ) and the service rate (μ) to find the utilization factor (ρ).
λ = 12 customers/hour
μ = 20 customers/hour (since the server can serve 20 customers in 60 minutes)
ρ = λ/μ = 12/20 = 0.6

Using the formula for the steady-state probability of having n customers in the system (Pn), we can calculate the probability of having one customer in the system (P1).
P0 = 1 - ρ = 1 - 0.6 = 0.4
P1 = ρ * P0 = 0.6 * 0.4 = 0.24

Therefore, the likelihood of there being one customer in the system at George's Happy Hog is 0.24 or 24%. Hence the answer is A) 0.24.

Twelve Customers Arrive at the Lunch Counter at George's Happy