SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

34 CHAPTER 5. ARRIVAL-COUNTING PROCESSES
Pr{Y 200 >c} = 1− Pr{Y 200 ≤ c}
= 1−
c∑
j=0
e − 4 50 (200) ( 4(200)
50
) j
j!
c∑ e −16 (16) j
= 1−
j!
j=0
≥ 0.95
Using trial and error, or the approximation from Exercise 2, c =9.
Pr{Y 200 > 9} ≈0.957
Pr{Y 200 > 10} ≈0.923
We will declare an “in-control” process “out-of-control” if λ =1/50 and {Y 200 >
9}.
Pr{Y 200 > 9} =1−
9∑
j=0
e − 1 50 (200) ( 200
50 )j
j!
≈ 0.008
7. Suppose we model the arrival of students to the bus stop as a Poisson arrival process
with expected time between arrivals of
̂1/λ =17.6 seconds
giving ̂λ =1/17.6 students/second ≈ 3.4 students/minute.
Now we can compute the probability of more than 60 students arriving during each
proposed time interval
Pr{Y 15 > 60} = 1− Pr{Y 15 ≤ 60}
= 1−
60 ∑
j=0
≈ 0.094
Pr{Y 12 > 60} ≈ 0.002
Pr{Y 10 > 60} ≈ 0.000
e −3.4(15) (3.4(15)) j
Ignoring students left behind on one pick up that add to the next pick up, we see that
there is nearly a 1 in 10 chance of filling the bus when pick up is every 15 minutes.
The carryover will only make the problem worse. Pick up every 12 minutes effectively
eliminates the problem. Every 10 minutes is more often than needed.
j!