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p chapter 9 solutions 3rd ed fall 2007

# p chapter 9 solutions 3rd ed fall 2007 - Probability Third...

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Sheet1 Page 1 Probability, Third Edition By David J Carr & Michael A Gauger Published by BPP Professional Education Solution 9.1 Let X be the number of demands made each day, and let Y be the number of demands handled. Then we have: Y=min{X,3 } and: Pr (Y=0)=Pr (X=0)=e-2 =0.13534 Pr (Y=1)=Pr (X=1)=2e-2 =0.27067 02/02/09 Pr (Y=2)=Pr (X=2)= 2 e =0.27067 2! Pr (Y=3)=Pr (X=3)=1 -0.13534 -0.27067 -0.27067 =0.32332 Hence, the expected number of demands handled is: [] 0 0.13534 +1 0.27067 +2 0.27067 +3 0.32332 EY Err:501 1.78 Solution 9.2 We can calculate [][2 in terms of EX and EX2]: EY and EY ][][ 88 8 [] =SyPr( =y) =SyPr( =y) =(1 -a)SyPr( X=y) =(1 -aEX ] =(1 - ) y=0 y=1 y=1 Solutions to practice questions h Chapter 9 EYY Y )[ ah

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Sheet1 Page 2 88 8 222 222 EY[ ] =Sy Pr( Y=y) =Sy Pr( Y=y) =(1 -a)Sy Pr( X=y) =(1 -a)[ ] =(1 -a h EX )( y=0 y=1 y=1 Hence, the variance of Y is: Y =EY [ 2]-([] )2 var( ) EY 2 ) ) Err:511 Err:509 E Solution 9.3 Note that since the pdf is defined for x >0 , the transformation YX2 is 1-1, with: = X =+Y So, using the method of transformations: ( dy ) -y 1 fy =fy ) #NAME? +h) a)( h +h) -((1 - 2 ah Probability Solutions to practice questions h Chapter 9
Sheet1 Page 3 for y >0 Y () X ( dy 2 y Solution 9.4 The pdf of X is: a--1 x /. xe f ()x =a for x >0 X .G()a Now, if Y = ()=cX , we have a 1-1 differentiable transformation. gX The inverse transformation is: XY /c= So, using the method of transformations: y .. fy =f Y () X .. c .. (/) dyc dy

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Sheet1 Page 4 1 yc/( . a-- ) ye Err:520 ().aG() c a a- - 1 ( /)/ yc . ( /) e 1 yc Err:501 .aG()a c for y >0 Hence Y follows a gamma distribution with parameters aand c.. Solution 9.5 We have: .1 .. 05/01/09 1 1 Y() =Pr (Y =y)=Pr . 1 y .=Pr .X = .=1 -Pr .X < .=1 - Fy -= .X . 1 +y 1 +y 1 +y Leths use the method of distribution functions.
Sheet1 Page 5 .

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