Customers arrive every three minutes governed by a Poisson distribution.A single server is capable of handling twenty-five customers per hour and this service time is exponentially distributed.What is the likelihood that the server is not busy at any given time?
A)less than or equal to .1
B)greater than .1 but less than or equal to .15
C)greater than .15 but less than or equal to .25
D)greater than .25

Question

Quizplus · Accepted Answer

The arrival rate of customers is λ= 1/3 per minute, or 20 per hour. 
The service rate is μ = 25 customers per hour. 
The utilization factor ρ = λ/μ = 20/25 = 0.8. 
Using Little's Law, the average time a customer spends in the system R = L/λ, where L is the average number of customers in the system. 
Since the server is capable of handling all customers, L is equivalent to the number of customers in queue, which is given by the Lq= ρ^2 / (1- ρ) = 3.2 customers.
Thus, the expected time a customer spends in the system is R = L/λ = (3.2+0.8)/20 = 0.2 hours. 
The probability that the server is not busy at any given time is equal to the probability that there are no customers in the system, which is equal to the probability of the server being idle. 
The server is idle when there are no customers in the system. Thus, the probability of the server being idle is equal to the probability that no customers arrive during a service time. 
Thus, the probability that the server is not busy at any given time P0= e^(-μT), where T is the length of time we are interested in. 
If we take T = 1 hour, then P0 = e^(-25*1) = 3.8 x 10^-11. 
The likelihood that the server is not busy at any given time is less than or equal to 0.1 since the value is very small. 
Therefore, the correct answer is C.

Customers Arrive Every Three Minutes Governed by a Poisson Distribution