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# S2103_MA340_Q3_sol.pdf

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Wilfrid Laurier University

Mathematics

MA340

Adam Metzler

Spring

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MA340 Quiz: July 9-10, 2013 || Solutions
1. An emergency center (open 24 hours per day) receives phone calls according to a Poisson process with an
average of 2 calls in any 20-minute period. Employees in the center work in shifts of eight hours each (i.e. one
shift is eight hours long and di▯erent shifts do not overlap each other).
(a) What is the probability that the center receives three or fewer phone calls during a given 45-minute period?
The hourly rate is 6 per hour and 45 minutes accounts for3 of one hour. Therefore the number of calls in
3 4
a 45-minute period is Poisson with parameter 6 ▯4 = 4:5 and the probability that there are three or fewer
calls is
X 3 k ▯4:5
(4:5) e
:
k=0 k!
(b) If Jillian begins her shift at 9:00 a.m., what is the probability that she receives her ▯rst call between 9:15
a.m. and 9:45 a.m.?
There are at least two ways to solve this problem. First, if T denotes the time (measured in hours after
9:00 a.m.) at which she receives her ▯rst call then T has an exponential distribution with parameter 4:5
and
P(▯rst call betweeen 9 : 15 a:m: and 9 : 45 a:m:= P(0:25 ▯ T ▯ 0:75)
Z 0:75
▯4:5t
= 4:5e dt
0:25
▯4:5t t=0:75
= [▯e ]t=0:25
= e▯1:125▯ e▯3:375
= 0:2904 :
Alternatively, we could observe that she receives her ▯rst call between 9:15 and 9:45 if and only if she
▯4:5(0:25)
(i) receives no calls between 9:00 and 9:15 (which occurs with probability e ) and (ii) receives at
least one call between 9:15 and 9:45 (which occurs with probability 1 ▯ e:5(0:5. Since these events are
independent we have argue as follows
P(▯rst call betweeen 9 : 15 a:m: and 9 : 45 a:m:= [e▯4:5(0:2][1 ▯ e▯4:5(0:])
= e▯1:125[1 ▯ e▯2:2]
▯1:125 ▯3:375
= e ▯ e
= 0:2904
(c) Paul’s shift runs from 10:00 p.m. to 6:00 a.m. and he receives 50 calls during that time. How likely is it
that 20 of those calls arrived between 3:00 a.m. and 6:00 a.m.?
We have
P(20 between 3 and 6 \ 50 between 10 and 6)
P(20 between 3 and 6 j 50 between 10 and 6) =
P(50 between 10 and 6)
= P(20 between 3 and 6 \ 30 between 10 and 3)
P(50 between 10 and 6)
e▯4:5(3[(4:5)(3)]e▯4:5([(4:5)(5)]
= 20! ▯ 30!
e▯4:5([(4:5)(8)]
50!
20 30
= 50! ▯ 3 5
20! ▯ 30! 850
▯ ▯
50 20 30
= 20 (3=8) (5=8)
= 0:1072 ;
which is a binomial probability. The intuition is as follows - the window from 3 to 6 accounts for 3/8
of the entire shift, and therefore each of the 50 calls that was received has a 3/8 chance of having been
received in that window, independently of the other calls.
(d) Say that a shift is \quiet" if no calls are

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