Test 5 Solutions

# Test 5 Solutions - STAT 230 Test 5 Short Solutions Circle...

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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 “ So-So ”, with respective probabilities 0 . 5, 0 . 3 and 0 . 2. Marks are independent among different tests....
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## This note was uploaded on 10/21/2010 for the course STAT 230 taught by Professor Various during the Fall '06 term at Waterloo.

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Test 5 Solutions - STAT 230 Test 5 Short Solutions Circle...

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