HW4 - × 1 × . 02 + 200000 × . 5 × . 01 + 200000 × . 25...

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Section 4.1 Expectation 4.2 Solution: f ( x ) = ( 3 x )( 1 4 ) x ( 3 4 ) 3 - x ,x = 0 , 1 , 2 , 3 E ( X ) = X x xf ( x ) = 0 ± 3 0 ²± 1 4 ² 0 ± 3 4 ² 3 + 1 ± 3 1 ²± 1 4 ² 1 ± 3 4 ² 2 + 2 ± 3 2 ²± 1 4 ² 2 ± 3 4 ² 1 + 3 ± 3 3 ²± 1 4 ² 3 ± 3 4 ² 0 = 0 + 1 × 3! 2!1! × 1 4 × 9 16 + 2 × 3! 1!2! × 1 16 × 3 4 + 3 × 3! 3!0! × 1 64 × 1 = 3 4 4.9 Solution: The random variable of interest is Y, the amount the gambler can win. The possible values of Y are $3 if event E 1 = { a jack or a queen } occurs and $5 if event E 2 = { a king or an ace } occurs. E ( Y ) = 3 × ( 4 1 ) + ( 4 1 ) ( 52 1 ) + 5 × ( 4 1 ) + ( 4 1 ) ( 52 1 ) = 3 × (4 + 4) 52 + 5 × (4 + 4) 52 = 64 52 = 1 . 23 Therefore, she should pay $1.23 to play if the game is fair. 4.10 Solution: μ X = E ( X ) = X x X y xf ( x,y ) = 1 { f (1 , 1) + f (1 , 2) + f (1 , 3) } + 2 { f (2 , 1) + f (2 , 2) + f (2 , 3) } + 3 { f (3 , 1) + f (3 , 2) + f (3 , 3) } = 1 { 0 . 10 + 0 . 05 + 0 . 02 } + 2 { 0 . 10 + 0 . 35 + 0 . 05 } + 3 { 0 . 03 + 0 . 10 + 0 . 20 } = 2 . 16 μ Y = E ( Y ) = X y X x yf ( x,y ) = 1 { f (1 , 1) + f (2 , 1) + f (3 , 1) } + 2 { f (1 , 2) + f (2 , 2) + f (3 , 2) } + 3 { f (1 , 3) + f (2 , 3) + f (3 , 3) } = 1 { 0 . 10 + 0 . 10 + 0 . 03 } + 2 { 0 . 05 + 0 . 35 + 0 . 10 } + 3 { 0 . 02 + 0 . 05 + 0 . 20 } = 2 . 04 1
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4.11 Solution: The price of the airplane is $200,000. P(total loss may occur)=0.002 P(50% loss may occur)=0.01 P(25% loss may occur)=0.1 Let X be an event that the amount of losses. E ( X ) = X x xf ( x ) = 200000
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Unformatted text preview: × 1 × . 02 + 200000 × . 5 × . 01 + 200000 × . 25 × . 1 = 200000(0 . 002 + 0 . 005 + 0 . 025) = 6400 Therefore, the insurance company should charge $6900(= $6400+$500) in order to obtain an average profit of $500. 4.14 Solution: E ( X ) = Z ∞-∞ xf ( x ) dx = Z 1 x 2( x + 2) 5 dx = Z 1 ± 2 5 x 2 + 4 5 x ² dx = ³ 2 15 x 3 + 2 5 x 2 ´ 1 = 8 15 4.17 Solution: μ g ( X ) = E [ g ( X )] = X x g ( x ) f ( x ) = g (-3) f (-3) + g (6) f (6) + g (9) f (9) = 25 × 1 6 + 169 × 1 2 + 361 × 1 3 = 209 2 4.20 Solution: μ g ( X ) = E [ g ( X )] = Z ∞-∞ g ( x ) f ( x ) dx = Z ∞ g ( x ) f ( x ) dx = Z ∞ e 2 3 x e-x dx = Z ∞ e-1 3 x dx = h-3 e-1 3 x i ∞ = 3 3...
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This note was uploaded on 02/04/2009 for the course STATISTIC 3011 taught by Professor Sally during the Fall '08 term at University of Minnesota Crookston.

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HW4 - × 1 × . 02 + 200000 × . 5 × . 01 + 200000 × . 25...

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