Question

5. Students arrive at a cafeteria according to a Poisson process at a rate of 20 students per hour. With probability of 0.8, a student will dine in (rather than making a to go order) (a) What is the expected number of students to arrive at a cafeteria in 1 hour? (b) What is the expected number of students to arrive at a cafeteria in a 5 hour period? What assumption did you make? (c) What is the probability that at most (less than or equal to) 25 students will arrive in the 1 hour time period? (d) What is the probability that at most 220 students will arrive in a 10 hour time period? (e) In a 4 hour time period, what is the probability that more than 50 students will arrive? (f) Say the students are independent of one another. What is the probability that 20 students will arrive in a 1 hour time period AND all 20 will dine in? Hnt: PAnB)-P(A)P(BA).

Answer 1

Given: The rate of the Poisson process (students arriving at a cafeteria) is, 1 = 20/ hour. The probability that a student who arrived at the cafeteria dine there is, P (BA)= 0.80. A= a student arrives at a cafeteria and B=a student dines at the cafeteria. (a) It is provided that students arrive at a cafeteria at a rate of 20 per hour. Then in 1 hour the expected number of students to arrive at the cafeteria is = it = 20 x1 = 20 Therefore, the expected number of students to arrive at the cafeteria in 1 hour is 20 (b) In 1 hour there are 20 students arriving at the cafeteria. Compute the expected number of students arriving at the cafeteria in a 5 hour period- =ht = 20 x 5 = 100 Therefore, the expected number of students arriving at the cafeteria in a 5 hour period is 100

c) Compute the probability that at most 25 students will arrive in 1hour period: P(X = x) = *** x! P(A< 25) = Vela Use the Excel function"=POISSON (25,20, TRUE)" to determine the probability, fx =POISSON (25,20,TRUE) DE F P(A < 25) = 0.8878 IF EPOISSON(25,20 TRUE Therefore, the probability that at most 25 students will arrive in 1hour period is 0.8878 (d) In 1 hour there are 20 students arriving at the cafeteria. Compute the expected number of students arriving at the cafeteria in a 10 hour period- = At = 20x10 = 200 Compute the probability that at most 220 students will arrive in 10 hour period: P(X=x) = ** P(AS 220) = 5e-200 2004 P AS220)=2 a !

Use the Excel function" =POISSON (220,200, TRUE)" to determine the probability, fax =POISSON(220,200, TRUE) DE F P (A S220) = 0.9247 Therefore, the probability that at most 220 students will arrive in 10 hour period is 0.9247 (e) In 1 hour there are 20 students arriving at the cafeteria. Compute the expected number of students arriving at the cafeteria in a 4 hour period- = it = 20 x 4 = 80 Compute the probability that more than 50 students will arrive in 4 hour period: P(X=x) = ** P(A> 50)=1-P(A350) - x! Use the Excel function" =POISSON(50,80, TRUE)" to determine the probability, fx =1 - POISSON(50,80, TRUE) D P (A > 50) = 1-PCA S 50) P(A > 50) = 0.9998 Therefore, the probability that more than 50 students will arrive in 4 hour period is 0.9998

Compute the probability that 20 students will arrive in a 1 hour period and all 20 will dine in- P(An B)=P(BA)*P(A) = (0.75)mox 479 2018 (For 20 students.) =0.00317x0.08884 = 0.00028 Therefore, the probability that 20 students will arrive in a 1 hour period and all 20 will dine in is 0.00028