Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Students arrive at the Administrative Services Office at an average of one every

ID: 423085 • Letter: S

Question

Students arrive at the Administrative Services Office at an average of one every 24 minutes, and their requests take on average 16 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? (Round your answer to 1 decimal place.)


Percentage of time             %


b. How much time, on average, does a student spend waiting in line? (Do not round intermediate calculations. Round your answer to 1 decimal place.)


Average time              minutes

c. How long is the (waiting) line on average? (Round your answer to 2 decimal places.)


Average length of the waiting line             customers


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? (Do not round intermediate calculations. Round your answer to 2 decimal places.)


Explanation / Answer

Following details are provided :

Arrival rate of students ( @ every 24 minutes ) = a = 60/24 = 2.5 students per hour

Service rate of students ( i.e. processing time of requests @ 16 minutes per student) = S = 3.75 students per hour

= ( 1 – a/s) x 100

= ( 1 – 2.5/3.75) x 100

= 0.3333 x 100

= 33.33%

= a/ s x ( s – a) hour

= 2.5/3.75 x ( 3.75 – 2.5 ) hour

= 2.5 / 3.75 x 1.25 hour

= 0.5333 hour

= 31.998 minute ( 32 rounded of)

AVERAGE TIME STUDENT SPENDS WAITING IN LINE = 32 MINUTES

= a^2/ s x ( s – a )

= ( 2.5 x 2.5)/ 3.75 x ( 3.75 – 2.5 )

= ( 2.5 x 2.5) / ( 3.75 x 1.25 )

= 1.33

AVERAGE LENGTH OF THE WAITING LINE = 1.33 CUSTOMERS

= 1 - probability that there are no students waiting in the line

Probability that there are no students waiting in the line = Po = 1 – a/s = 1 – 2.5/3.75 = 0.333

Therefore ,

Probability that at least one student waiting in the line

= 1 – 0.333

= 0.667( 0.67 rounded to 2 decimal places )

PROBABILITY THAT AN ARRIVING STUDENT WILL FIND AT LEAST ONE OTHER STUDENT WAITING IN THE LINE = 0.667