p chapter 6 solutions 3rd ed fall 2007

# P chapter 6 - Probability Third Edition By David J Carr Michael A Gauger Published by BPP Professional Education Solutions to practice questions

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Sheet1 Page 1 Probability, Third Edition By David J Carr & Michael A Gauger Published by BPP Professional Education Solution 6.1 Let the losses due to storm, fire, and theft be denoted X1, X2 and X3 respectively. ({ 23}) ({13}) Prmax 1, , >2.5 =1 -Pr max XXX , 2, =2.5 XXX 1 -Pr (X=2.5 )( Pr X=2.5 ) Pr (X=2.5 ) #NAME? 123 The distribution function for a uniform distribution with range [, ] ab is: xa - () = Fx ba - So, we can calculate the probability as: { ,, }>2.5 )=1 -Pr X=2.5 ) Pr (X=2.5 ) Pr (X=2.5 ) Pr max ( XXX ( 123 123 .2.5 . .2.5 . Err:509 .. .=-1 0.521 Solutions to practice questions h Chapter 6

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Sheet1 Page 2 1 =0.479 .3 . .4 . Solution 6.2 We have: + []= ab =1,879 EX 2 ( -)2 var ()= ba =507 .ba = 12 X - 507 =78 12 Hence: ab+= 1,879 = 2 3,758 . a+(78 +a) =3,758 . a=1,840 . b=1,918 b Solution 6.3 Since we have formulas for [ and EX2], it is convenient to first calculate the second moment: EX][ 2 22 EX. Probability Solutions to practice questions h Chapter 6
Sheet1 Page 3 .=var ()+([]=18.75 +2.5 .. X EX )=25 Hence: 2.5 =[] . EX = a-1 2 2 .2! .. . . (a-1)(a-2) We solve these simultaneous equations to find the parameters: 22 2 25 =EX = 25 EX .a- 2 a [] .2! 1 . 1 Err:520 = .. 2 ([]2 2.5 EX ) (a-1)(a-2). .. a-2 .a=3, .=5 Now it is easy to calculate the 90th percentile: a 3 ... . 5

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Sheet1 Page 4 0.9 =Fx =- 1 () 0.9 1 . .=- . . x +. x +5 .0.9 . .0.9 . 1 .- . 3 .x =50.1 -1 =5.772 0.9 . . .. Solution 6.4 If we notice that X follows a Pareto distribution with parameters a=3 and .=2,000 , the expected return is easily calculated using the general result as: . 2000 []== 1000 EX a-13 -1 Solution 6.5 We want to calculate the following: Pr (XX 1.5 ) Pr 1.5 X 2)- 2 << F 1.5 <n> (
Page 5 F (2) () Pr (X <2 X >1.5 )= Err:520 Pr (X >1.5 ) Pr (X >1.5 ) 1 -( ) F 1.5 Note that the pdf is that of a one-parameter Pareto distribution with a=3 and .=1 , so the cdf is: ()=-x-3 for x >1 Fx 1 Finally: -3 -3 12 -- 1 - 1.5 )F ()

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## This note was uploaded on 09/08/2008 for the course STT 490 taught by Professor Dudziak during the Spring '08 term at Michigan State University.

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P chapter 6 - Probability Third Edition By David J Carr Michael A Gauger Published by BPP Professional Education Solutions to practice questions

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