chap-5 - Chapter 5 Problems 1. (a) c (1 - x 2 )dx = 1 c = 3...

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64 Chapter 5 Chapter 5 Problems 1. (a) 4 / 3 1 ) 1 ( 1 1 2 = = c dx x c (b) F ( x ) = + = 3 2 3 4 3 ) 1 ( 4 3 3 1 2 x x dx x x , 1 < x < 1 2. = 2 / 2 / 2 / 4 2 x x x e xe dx xe . Hence, 4 / 1 1 0 2 / = = c dx xe c x P { X > 5} = ] 4 10 [ 4 1 4 1 2 / 5 2 / 5 5 2 / + = e e dx xe x 2 / 5 4 14 = e 3. No. f (5/2) < 0 4. (a) = = 20 20 2 2 / 1 10 10 x dx x . (b) F ( y ) = y dx x y 10 1 10 10 2 = , y > 10. F ( y ) = 0 for y < 10. (c) = 6 3 6 3 1 3 2 6 i i i i since 15 10 ) 15 ( = F . Assuming independence of the events that the devices exceed 15 hours. 5. Must choose c so that .01 = 5 1 4 ) 1 ( ) 1 ( 5 c dx x c = so c = 1 (.01) 1/.5 .
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Chapter 5 65 6. (a) E [ X ] = = 0 2 0 2 / 2 2 4 1 dx e y dx e x y x = 2 Γ (3) = 4 (b) By symmetry of f ( x ) about x = 0, E [ X ] = 0 (c) E [ X ] = = 5 5 dx x 7. + 1 0 2 ) ( dx bx a = 1 or 1 3 = + b a 5 3 ) ( 1 0 2 = + dx bx a x or 5 / 3 4 2 = + b a . Hence, a = 5 3 , b = 5 6 8. E [ X ] = 2 ) 3 ( 0 2 = Γ = dx e x x 9. If s units are stocked and the demand is X , then the profit, P ( s ), is given by P ( s ) = bX ( s X ) Ρ if X s = sb i f X > s Hence E [ P ( s )] = + s s dx x sbf dx x f x s bx ) ( ) ( ) ) ( ( 0 A = + + s s s dx x f sb dx x f s dx x xf b 0 0 0 ) ( 1 ) ( ) ( ) ( A A = sb + + s dx x f s x b 0 ) ( ) ( ) ( A Differentiation yields )] ( [ s P E ds d = + + s s dx x f s dx x xf ds d b b 0 0 ) ( ) ( ) ( A = b + + s dx s f s sf s sf b 0 ) ( ) ( ) ( ) ( A = b + s dx x f b 0 ) ( ) ( A
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66 Chapter 5 Equating to zero shows that the maximal expected profit is obtained when s is chosen so that F ( s ) = A + b b where F ( s ) = s dx x f 0 ) ( is the cumulative distribution of demand. 10. (a) P {goes to A } = P {5 < X < 15 or 20 < X < 30 or 35 < X < 45 or 50 < X < 60}. = 2/3 since X is uniform (0, 60). (b) same answer as in (a). 11. X is uniform on (0, L ). < 4 / 1 , min X X L X L X P = 1 > 4 / 1 , min X X L X L X P = 1 > > 4 / 1 , 4 / 1 X X L X L X P = 1 P { X > L /5, X < 4 L /5} = < < 5 / 4 5 1 L X L P = 1 5 2 5 3 = . 13. P { X > 10} = 3 2 , P { X > 25 X > 15} = 30 / 15 30 / 5 } 15 { 25 { = > > X P X P = 1/3 where X is uniform (0, 30). 14. E [ X n ] = 1 1 1 0 + = n dx x n P { X n x } = P { X x 1/ n } = x 1/ n E [ X n ] = 1 1 1 1 1 0 / 1 1 1 1 0 + = = n dx x n dx x n x n n 15. (a) Φ (.8333) = .7977 (b) 2 Φ (1) 1 = .6827 (c) 1 Φ (.3333) = .3695 (d) Φ (1.6667) = .9522 (e) 1 Φ (1) = .1587
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Chapter 5 67 16. P { X > 50} = > 4 10 4 40 X P = 1 Φ (2.5) = 1 .9938 Hence, ( P { X < 50}) 10 = (.9938) 10 17. E [Points] = 10(1/10) + 5(2/10) + 3(2/10) = 2.6 18. .2 = > σ 5 9 5 X P = P { Z > 4/ } where Z is a standard normal. But from the normal table P { Z < .84) .80 and so .84 4/ or 4.76 That is, the variance is approximately (4.76) 2 = 22.66.
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This note was uploaded on 03/31/2008 for the course STAT 418 taught by Professor G.jogeshbabu during the Spring '08 term at Pennsylvania State University, University Park.

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chap-5 - Chapter 5 Problems 1. (a) c (1 - x 2 )dx = 1 c = 3...

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