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516hw12 - Section 5.4 1 a r 1 2 X2 1 r1 1 1 dx if 0 < z < 1...

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Section 5.4 1. a) 1 , r λ b) 1 2 1 2 ( ) ( ) ( ) X X X X f z f x f z x dx + = - 0 1 0 1;0 2 0 1; 2 0 1 2 1 if 0 1 2 1 1 1 if 1 2 2 2 2 1 if 2 3 2 z x z x x z x z z dx z dx dx dx z dx z < < < - < < < - < < - < < = = = < < < < ° if 0 1 2 1 if 1 2 2 3- if 2 3 2 z z z z z X < < = < < < < c) 1 2 1 2 ( ) ( ) z X X X X F z f x dx + + - = 2 0 1 1 1 , r λ 2 X
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2 0 1 z 0 1 2 1 2 0 1 2 0 if 0 0 if 0 if 0 1 if 0 1 2 4 1 1 1 if 1 2 + if 1 2 2 2 4 2 (3 ) 1 3- 1 if 2 3 if 2 3 4 2 2 2 1 if 3 1 if 3 z z z z x z dx z z x z dx dx z z z x x z dx dx dx z z z < < < < < < - = + < < = < < - - < < + + < < ° 2. a) 0.875 b) By symmetry, this probability would be ½. c) Use the formula for the density of 3 S derived in this chapter. 1 1.1 2 2 3 0 1 3 ( 1.1) 3 0.2213 2 2 t P S dt t t dt ( = + - + - = d) 3 3 (1.0 1.001) (1) 0.001 0.0005 S P S f Z = 5. a) It is clear that the range of 10 10 Z X Y = + is [10, 70]. Take any point z in that range. Then there is a unique integer x between 1 and 6 and a unique number in the interval [0, 1) such that 10 10 z x y = + . So, ( ) (10 10 10 10 ) P Z z P X Y x y = + + ( ) ( 1) ... ( 1) ( ,0 ) 1 1 10 6 6 6 60 P X Y x y P X P X x P X x Y y x y x y z = + + = = + + = - + = < < - + - - = + = = This gives the uniform distribution on [10, 70].
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