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# S2103_MA340_Q3_sol.pdf

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School
Wilfrid Laurier University
Department
Mathematics
Course
MA340
Professor
Semester
Spring

Description
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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