Students arrive at the Administrative Services Office at an average of one every 15 minutes, and their requests take on average 10 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 (just before entering the Administrative Services Office) will find at least one other student waiting in line?

a) %idle time = 0.33

b) 40.2minutes

c) 40.02customers

d) 0.5219

Step-by-step explanation:

It was said in the question Students arrive at the Administrative Services Office at an average of one every 15 minutes which means that

λ = 60/15 = 4customers/hr

It was stated that their requests take on average 10 minutes to be processed which means that

μ = average of 10minutes = 60/10 = 6customers/hr

Then let us use these information to solve the given questions

a) percentage when judy was idle = (1 - λ/μ) = 1 - 0.67 = 0.33

%service time = 0.67

%idle time = 0.33

b) To calculate How much time, on average that a student spend waiting in line then we make use of the formula below

= λ / μ (μ - λ)

= 0.67hrs = 0.67 x60 = 40.2minutes

c) To calculate How long the waiting line on average;

= average waiting time x arrival rate = 0.67hrs x 6customers/hr

= 40.02customers

d) the probability that an arriving student will find at least one other student waiting in line is calculated below;

P (idle time i. e no customer to attend to) = 0.33

P1 (Probability of having a customer to attend to) = 0.33 x 0.67 = 0.2211

P (Probability of having 2 customer to attend to) = 0.33 x 0.67x0.67 = 0.1481

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

= 1 - 0.33 - 0.1481 = 0.5219