ISM_chapter5 - Chapter 5 Multivariate Probability...

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93 Chapter 5: Multivariate Probability Distributions 5.1 a. The sample space S gives the possible values for Y 1 and Y 2 : S AA AB AC BA BB BC CA CB CC ( y 1 , y 2 ) (2, 0) (1, 1) (1, 0) (1, 1) (0, 2) (1, 0) (1, 0) (0, 1) (0, 0) Since each sample point is equally likely with probably 1/9, the joint distribution for Y 1 and Y 2 is given by y 1 0 1 2 01 / 92 / 91 / 9 y 2 1 2/9 2/9 0 2 1/9 0 0 b. F (1, 0) = p (0, 0) + p (1, 0) = 1/9 + 2/9 = 3/9 = 1/3. 5.2 a. The sample space for the toss of three balanced coins w/ probabilities are below: Outcome HHH HHT HTH HTT THH THT TTH TTT ( y 1 , y 2 ) (3, 1) (3, 1) (2, 1) (1, 1) (2, 2) (1, 2) (1, 3) (0, –1) probability 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 y 1 0 1 2 3 –1 1/8 0 0 0 y 2 1 0 1/8 2/8 1/8 2 0 1/8 1/8 0 3 0 1/8 0 0 b. F (2, 1) = p (0, –1) + p (1, 1) + p (2, 1) = 1/2. 5.3 Note that using material from Chapter 3, the joint probability function is given by p ( y 1 , y 2 ) = P ( Y 1 = y 1 , Y 2 = y 2 ) = 3 9 3 2 3 4 2 1 2 1 y y y y , where 0 y 1 , 0 y 2 , and y 1 + y 2 3. In table format, this is y 1 0 1 2 3 0 0 3/84 6/84 1/84 y 2 1 4/84 24/84 12/84 0 2 12/84 18/84 0 0 3 4/84 0 0 0
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94 Chapter 5: Multivariate Probability Distributions Instructor’s Solutions Manual 5.4 a. All of the probabilities are at least 0 and sum to 1. b. F (1, 2) = P ( Y 1 1, Y 2 2) = 1. Every child in the experiment either survived or didn’t and used either 0, 1, or 2 seatbelts. 5.5 a. 1065 . 3 ) 3 / 1 , 2 / 1 ( 2 / 1 0 3 / 1 0 2 1 1 2 1 = = ∫∫ dy dy y Y Y P . b. 5 . 3 ) 2 / ( 1 0 2 / 0 2 1 1 1 2 1 = = y dy dy y Y Y P . 5.6 a. [] . 125 . ) 5 (. 1 ) 5 . ( ) 5 . ( 5 . 0 2 2 2 5 . 0 5 . 0 1 5 . 1 1 5 . 2 1 2 1 2 1 2 2 = = = = + > = > + + dy y dy y dy dy Y Y P Y Y P y y b. = = > = > = < 1 5 . 2 2 1 5 . 1 / 5 . 2 1 2 1 2 1 2 1 ) / 5 . 1 ( 1 1 1 ) / 5 . ( 1 ) 5 . ( 1 ) 5 . ( 2 dy y dy dy Y Y P Y Y P Y Y P y = 1 – [.5 + .5ln(.5)] = .8466. 5.7 a. . 00426 . 1 ) 5 , 1 ( 5 1 2 5 1 0 1 1 0 2 1 5 ) ( 2 1 2 1 2 1 = = = = > < + e e dy e dy e dy dy e Y Y P y y y y b. . 8009 . 4 1 ) 3 ( ) 3 ( 3 3 0 2 1 3 0 ) ( 2 1 2 1 2 2 1 = = = < = < + + e dy dy e Y Y P Y Y P y y y 5.8 a. Since the density must integrate to 1, evaluate 1 4 / 2 1 1 0 1 0 2 1 = = k dy dy y ky , so k = 4. b. 2 2 2 1 2 1 00 2 1 2 2 1 1 2 1 21 4 ) , ( ) , ( y y dt dt t t y Y y Y P y y F yy = = = , 0 y 1 1, 0 y 2 1. c. P ( Y 1 1/2, Y 2 3/4) = (1/2) 2 (3/4) 2 = 9/64. 5.9 a. Since the density must integrate to 1, evaluate 1 6 / ) 1 ( 2 1 1 2 2 = = k dy dy y k y , so k = 6. b. Note that since Y 1 Y 2 , the probability must be found in two parts (drawing a picture is useful): P ( Y 1 3/4, Y 2 1/2) = + 4 / 3 2 / 1 1 1 2 2 2 1 1 2 / 1 1 2 / 1 2 1 ) 1 ( 6 ) 1 ( 6 y dy dy y dy dy y =24/64 + 7/64 = 31/64. 5.10 a. Geometrically, since Y 1 and Y 2 are distributed uniformly over the triangular region, using the area formula for a triangle k = 1. b. This probability can also be calculated using geometric considerations. The area of the triangle specified by Y 1 3 Y 2 is 2/3, so this is the probability.
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Chapter 5: Multivariate Probability Distributions 95 Instructor’s Solutions Manual 5.11 The area of the triangular region is 1, so with a uniform distribution this is the value of the density function.
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This note was uploaded on 03/18/2009 for the course STA 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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ISM_chapter5 - Chapter 5 Multivariate Probability...

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