This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: IEOR 4106 Midterm Exam. Open text book and class notes; 1.5 hours. 100 Points total 1. (35 points) Voice messages are made from a cell phone according to a Poisson process at rate 8 per hour, and independent of this, text messages are sent from the phone according to a Poisson process at rate 2 per hour. (Time is in hours.) (a) (10 points) Given that 4 text messages were sent during the hours of 10:00AM to 11:30AM, what is the probability that exactly two of these four were made between 11:00AM and 11:30AM? SOLUTION: We can treat the (unordered) four arrival times U 1 ,...U 4 as iid uni formly distributed over the 90 minute interval, which for simplicity we denote by (0 , 90). Each one, independently, would arrive between 11:00AM and 11:30AM with “success” probability p = P ( U ∈ (60 , 90)) = 30 / 90 = 1 / 3. So, letting N denote the number of these 4 that do so yields a binomial ( p, 4) distribution; P ( N = 2) = ( 4 2 ) p 2 (1 p ) 2 = 6(1 / 3) 2 (2 / 3) 2 = 8 / 27. (b) (10 points, 5 each) i. What is the probability that the first two messages are both voice, and the third is text? SOLUTION: lambda 1 = 8 , λ 2 = 2. Letting p = λ 1 / ( λ 1 + λ 2 ) = 0 . 80, we get p 2 (1 p ) = (0 . 8) 2 (0 . 20) = 0 . 128. ii. Starting from now (time 0) what is the expected length of time until at least one voice and one text message will been made?...
View
Full
Document
 Spring '08
 Whitt
 Probability theory, Exponential distribution, Poisson process, Text messages

Click to edit the document details