Question

Starting at 9 a.m., students arrive to class according to a Poisson process with

parameter λ = 2 (units are minutes). Class begins at 9:15 a.m. There are 30

students.

(a) What is the expectation and variance of the number of students in class by

9:15 a.m.?

(b) Find the probability there will be at least 10 students in class by 9:05 a.m.

(c) Find the probability that the last student who arrives is late.

(d) Suppose exactly six students are late. Find the probability that exactly 15

students arrived by 9:10 a.m.

(e) What is the expected time of arrival of the seventh student who gets

to class?

Answer 1

Here λ = 2

number of students = 30

(a) expected number of students in class by 9:15 am = 2 * 15 = 30

variance of the number of students in class by 9:15 am = sqrt(30) = 5.477

(b) Expected number of students in class by 09:05 AM = 2 * 5 = 10

Pr(x >= 10) = 1 - POISSON(x < 10 ; 10) = 1 - 0.4579 = 0.5421

(c) Pr(Last student arrive is late) = 1- Pr(all students arrives on time)

= 1 - Pr(all 30 students come in time) =

= 1- POISSON(x <= 30 ; 30 ; true)

= 1 - 0.5484 = 0.4516

(d) Exactly six students are late, that means there are 24 students arrive before 9:15 AM. now, we have to find the probability that exactly 15 student arrived by 9:10 AM, that means there are (24 - 15 = 9) students arrived by 9:15 AM.

Expected number of students arrive in between 9:10 am to 9: 15 am = 5 * 2= 10

Pr(x = 9 ; λ = 10) = e^-1010⁹/9! = 0.1251

(e) Expected time taken for seventh student arrival = 7/2 = 3.5

so expected time of arrival = 9:03: 30