ChapterSolutions - 5

# ChapterSolutions - 5 - Chapter 5 Problems 1(a c(1 x 2)dx =...

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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
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.

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

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