Answer :
There is 0.17% of the time is Judy idle.
It will take 25 minutes on average, does a student spend waiting in line.
The waiting line on average is 4.1 customers.
The probability that an arriving student will find at least one other student waiting in line is 0.68.
Given students arrive at the Administrative Services Office at an average of one every 6 minutes, and their requests take on average 5 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.
Probability
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 = 60/6 = 10customers/hr
μ = average of 5 minutes = 60/5 = 12customers/hr
1) What percentage of time is Judy idle?
The percentage when Judy was idle = 1- λ/μ = 1- 0.83= 0.17
%Service time = 0.83
%idle time = 0.17
There is 0.17% of time is Judy idle.
2) How much time, on average, does a student spend waiting in line?
A student spends waiting in line then we make use of the formula below:
λ /μ(λ-μ )
= 0.41 × 60
= 25 minutes
It will take 25 minutes on average, does a student spend waiting in line.
3) How long is the waiting line on average?
To calculate How long the waiting line is on average;
average waiting time x arrival rate = 0.41hrs x 10 customers/hr = 4.1 customers
The waiting line on average is 4.1 customers.
What is the probability that an arriving student will find at least one other student waiting in line?
P1( Probability of having a customer to attend to) = 0.17 x 0.83 = 0.141
P2( Probability of having 2 customer to attend to) = 0.17 x 0.83 x 0.83 = 0.117
The probability of finding at least one customer = 1 -[ po + p1]
= 1 - 0.17 - 0.141 = 0.68
The probability that an arriving student will find at least one other student waiting in line is 0.68.
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