5solution

# 5solution - ECEN 303 Assignment 5 Problems 1 An irregular...

This preview shows pages 1–3. Sign up to view the full content.

ECEN 303: Assignment 5 Problems: 1. An irregular student is taking a difficult class. On any given homework, his score takes value from 1 to 10, with equal probability 0.1, independently of other assignments. Determined to do well, the student decides to take advantage of a “best out of two” grading policy, where only the top score out of two assignments is recorded. His final score can be computed as F = max { X 1 ,X 2 } . (a) Calculate the PMF of F . Consider the probability that F is less than k , Pr( F k ) = Pr( { X 1 k } ∩ { X 2 k } ) = Pr( X 1 k )Pr( X 2 k ) = k 2 100 . The PMF of F is given by p F ( k ) = Pr( F k ) - Pr( F k - 1) = k 2 100 - ( k - 1) 2 100 = 2 k - 1 100 . (b) By how much has his expected score improved as a result of submitting two assignments? The expected score of the student on a single test is 10 summationdisplay k =1 kp X ( k ) = 10 summationdisplay k =1 k 10 = 5 . 5 . When the student takes advantage of the “best out of two” grading policy, his expected score becomes 10 summationdisplay k =1 kp F ( k ) = 10 summationdisplay k =1 k 2 k - 1 100 = 770 - 55 100 = 7 . 15 . 2. Five distinct numbers are randomly distributed to players numbered 1 through 5. Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, player 1 and 2 compare their numbers; the winner then compares with player 3, and so on. Let X denote the number of times player 1 is a winner. Find Pr( X = i ) ,i = 0 , 1 , 2 , 3 , 4. We can use conditioning to solve this problem. Let N be the number held by player 1. Pr( X = i | N = j ) X = 0 X = 1 X = 2 X = 3 X = 4 N = 1 1 0 0 0 0 N = 2 3 / 4 1 / 4 0 0 0 N = 3 1 / 2 1 / 3 1 / 6 0 0 N = 4 1 / 4 1 / 4 1 / 4 1 / 4 0 N = 5 0 0 0 0 1 . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Since player 1 is equally likely to get any of he numbers, we conclude that the probability mass function of X is given by p X (0) = 1 2 , p X (1) = 1 6 , p X (2) = 1 12 , p X (3) = 1 20 , p X (4) = 1 5 . 3. A total of 4 buses carrying 148 students from the same school arrives at a football stadium. The buses carry, respectively, 40, 33, 25, and 50 students. One of the students is randomly selected. Let X denote the number of students that were on the bus carrying this randomly selected student. One of the 4 bus drivers is also randomly selected. Let Y denote the number of students on her bus.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern