This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: STAT 230 Test 5 Short Solutions Circle Your Section Number : 001 (10:30) 002 (12:30) 003 (1:30) 004 (2:30) 1. The joint probability function f ( x, y ) = P ( X = x, Y = y ) of random variables X and Y is tabulated as follows. x f ( x, y ) 2 3 4 0.4 0.1 0.1 y 1 0.1 0.1 0.2 [3] (a) Find the marginal probability function f X ( x ) of X and f Y ( y ) of Y . x 2 3 4 y 1 f X ( x ) 0.5 0.2 0.3 f Y ( y ) 0.6 0.4 [3] (b) Find the conditional distribution of X given Y = 1. P ( X = 2  Y = 1) = 1 / 4, P ( X = 3  Y = 1) = 1 / 4 and P ( X = 4  Y = 1) = 2 / 4 = 1 / 2. [2] (c) Are X and Y independent? Why or why not? No, since P ( X = 2 , Y = 0) = 0 . 4 6 = P ( X = 2) P ( Y = 0) = 0 . 3. [2] (d) Find the probability function of W = XY + 1. Possible values for W are 1, 3, 4 and 5. P ( W = 1) = f (2 , 0) + f (3 , 0) + f (4 , 0) = 0 . 6, P ( W = 3) = f (2 , 1) = 0 . 1, P ( W = 4) = f (3 , 1) = 0 . 1 and P ( W = 5) = f (4 , 1) = 0 . 2. [2] (e) Find the correlation coefficient ρ between X and Y and interpret the value of ρ . ( Note: V ar ( X ) = 0 . 76, V ar ( Y ) = 0 . 24) We need to find Cov ( X, Y ) = E ( XY ) E ( X ) E ( Y ) = 1 . 3 2 . 8 × . 4 = 0 . 18; ρ = Cov ( X, Y ) / p V ar ( X ) V ar ( Y ) = 0 . 18 / √ . 76 × . 24 = 0 . 4215. There is a positive correlation between X and Y but the correlation is not very strong. 2. Joe has to write five tests in a stat course. For each test, Joe’s mark can be “ Excellent ”, “ Good ” or “ SoSo ”, with respective probabilities 0 . 5, 0 . 3 and 0 . 2. Marks are independent among different tests....
View
Full
Document
This note was uploaded on 10/21/2010 for the course STAT 230 taught by Professor Various during the Fall '06 term at Waterloo.
 Fall '06
 various
 Probability

Click to edit the document details