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# Chapter 2 Answers - Chapter 2 Answers 2 We know from Table...

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Chapter 2 Answers 2. We know from Table 2.2 that = = . Pr( 0) 0 22, Y = = . Pr( 1) 0 78, Y = = . Pr( 0) 0 30, X = = . Pr( 1) 0 70. X So (a) ( ) 0 Pr( 0) 1 Pr( 1) 0 0 22 1 0 78 0 78, ( ) 0 Pr( 0) 1 Pr( 1) 0 0 30 1 0 70 0 70 Y X E Y Y Y E X X X μ μ = = × = + × = = × . + × . = . = = × = + × = = × . + × . = . . (b) 2 2 2 2 2 2 2 2 2 2 2 2 [( ) ] (0 0.70) Pr( 0) (1 0.70) Pr( 1) ( 0 70) 0 30 0 30 0 70 0 21 [( ) ] (0 0.78) Pr( 0) (1 0.78) Pr( 1) ( 0 78) 0 22 0 22 0 78 0 1716 X X Y Y E X X X E Y Y Y σ μ σ μ = - = - × = + - × = = - . × . + . × . = . , = - = - × = + - × = = - . × . + . × . = . . (c)Table 2.2 shows Pr( 0, 0) 0 15, X Y = = = . Pr( 0, 1) 0 15, X Y = = = . Pr( 1, 0) 0 07, X Y = = = . Pr( 1, 1) 0 63. X Y = = = . So cov( , ) [( )( )] (0 - 0.70)(0 - 0.78)Pr( 0, 0) (0 0 70)(1 0 78)Pr( 0 1) (1 0 70)(0 0 78)Pr( 1 0) (1 0 70)(1 0 78)Pr( 1 1) ( 0 70) ( 0 78) 0 15 ( 0 70) 0 22 0 15 0 30 ( 0 78) 0 07 0 XY X Y X Y E X Y X Y X Y X Y X Y σ μ μ = = - - = = = + - . - . = , = + - . - . = , = + - . - . = , = = - . - . . + - . . . + . - . . + . 30 0 22 0 63 0 084, 0 084 ( , ) 0 4425 0 21 0 1716 XY X Y cor X Y σ σ σ . . = . . = = = . . . .

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5. Let X denote temperature in ° F and Y denote temperature in ° C. Recall that Y = 0 when X = 32 and Y = 100 when X = 212; this implies (100/180) ( 32) or 17.78 (5/9) . Y X Y X = × - = - + × Using Key Concept 2.3, = 70 F X μ implies that 17.78 (5/9) 70 21.11 C, Y μ = - + × = ° and = 7 F X σ implies (5/9) 7 3.89 C. Y σ = × = ° 6. The table shows that Pr( 0, 0) 0 045, X Y = = = . Pr( 0, 1) 0 709, X Y = = = . Pr( 1, 0) 0 005, X Y = = = .
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Chapter 2 Answers - Chapter 2 Answers 2 We know from Table...

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