Students come to the Stat Department’s Software Help Desk on average at a rate o
ID: 3208466 • Letter: S
Question
Students come to the Stat Department’s Software Help Desk on average at a rate of 1.5 students per hour. Students arrive independently to the Help Desk.
a) Let J be the number of students arriving at the Help Desk between 9:30 am and 11:30 am. Find the probability that at least one student comes to the Help Desk during this period. What are the distribution, parameter(s), and support of J?
b) Given that at least one student came between 9:30 am and 11:30 am, what is the probability that fewer than 4 students came during that period?
c) What is the probability that 2 students come between 9:30 am and 11:30 am and 1 student comes between 11:30 am and 12:15 pm?
d) The Help Desk is open from 10:00 am to 4:00 pm on every weekday. What is the expected number of students that will come to the Help Desk on Monday through Wednesday?
Explanation / Answer
Solution
Back-up Theory
Number of students coming to the Help Desk is Poisson Distributed with parameter , where = average number of students coming to the Help Desk per unit time, say 1 hour……..................................................……….(1)
If number of students coming to the Help Desk per hour is Poisson Distributed with parameter , then number of students coming to the Help Desk per k hours is Poisson Distributed with parameter k,……….............…….(2)
If A and B are two events, P(A/B) = P(A&B)/P(B)…………………………………………...................................(3)
Now, to work out the solution,
Given = 1.5
Note: all probabilities are obtained using Excel Functions directly.
Part (a) Here = 2x1.5 = 3 by (2) under Back-up Theory
Probability that at least one student comes to the Help Desk during 9:30 am and 11:30 am (i.e., a 2 hour period) = P(J 1 with = 2x1.5 = 3) = 1- P(J = 0 with = 3) =1 – 0.0498 = 0.9502 ANSWER
As already given under Back-up Theory, J is supported by Poisson with parameter 1.5.
Part (b) Here also = 2x1.5 = 3 by (2) under Back-up Theory
Let B be the event that at least one student came between 9:30 am and 11:30 am, and A be the event that fewer than 4 students came between 9:30 am and 11:30 am, then event (A&B) would imply number of students who came between 9:30 am and 11:30 is 2 or 3.
Now, given that at least one student came between 9:30 am and 11:30 am, probability that fewer than 4 students came during that period
= P(A/B) = P(A&B)/P(B) = P(2 or 3 students coming)/P(at least one student coming)
= { P(2 students coming) + P(3 students coming)}/0.9502 [from Part (a) answer]
= (0.2240 + 0.2240)/0.9502 = 0.4718 ANSWER
Part (c)
Probability that 2 students come between 9:30 am and 11:30 am (i.e., 2 hour period and hence = 3) and 1 student comes between 11:30 am and 12:15 pm (i.e., ¾ hour period and hence = 1.125)
= P(2 students coming in 2 hour period) x P(1 student coming in ¾ hour period)
= 0.2240 x 0.3652 = 0.0818 ANSWER
Part (d)
10:00 am to 4:00 pm on every weekday for Monday through Wednesday = 6 x 3 = 18 hours. Since = 1.5 is the mean, i.e., expected number of students coming per hour, for 18 hours, expected number of students coming
= 18 x 1.5 = 27 ANSWER