Students arrive at the Administrative Services Office at an average of one every 20 minutes, and their requests take on average 15 minutes to be processed. The service counter is staffed by only one clerk, Judy gumshoes, who works eight hours per day. Assume Poisson arrivals and exponential service times.

A. What percentage of time is Judy idle?

B. How much time, on average, does a student spend waiting in line?

C. How long is the waiting line on average?

D. What is the probability that an arriving student will find at least one other student waiting in line?

Step-by-step explanation:

a) percentage when judy was idle = 1- λ/μ = 1- 0.75 = 0.25

%service time = 0.75

%idle time = 0.25

b) How much time, on average, does a student spend waiting in line;

= λ/ μ( μ- λ)

= 0.75hrs = 0.75 x60 = 45minutes

c) How long is the waiting line on average;

= average waiting time x arrival rate = 0.75hrs x 3 customers/hr

= 2.25customers

d) What is the probability that an arriving student will find at least one other student waiting in line ; Po( probability of idle time i.e no customer to attend to) = 0.25

P1( Probability of having a customer to attend to) = 0.25 x 0.75 = 0.1875

P2( Probability of having 2 customer to attend to) = o.25 x 0.75x0.75 = 0.14

Hence, probability of finding at least one customer = 1 -[ po + p1]

= 1 - 0.25 - 0.1875 = 0.5625