04 STAT HWA a

# 04 STAT HWA a - Dr Valerie R Bencivenga Economics 329...

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Dr. Valerie R. Bencivenga Economics 329 PRACTICE HOMEWORK #4A: ANSWERS 1.a. S = defect in the shaft, B = defect(s) in bushing(s) 02 . ) B S ( P b. 08 . ) B ( P ) B S ( P ) B ( P 06 . 1 . ) S ( P ) B S ( P ) S ( P 08 . 16 . 02 . 08 . 1 . ) B S ( P ) S ( P ) S ( P ) B S ( P c. 14 . 08 . 06 . )] B S ( P ) B ( P [ )] B S ( P ) S ( P [ d. 84 . ) B S ( P 1 ) B S ( P 2.a. HHH e 1 HMH e 4 } e ,..., e { S 8 1 HHM e 2 MHM e 5 MHH e 3 MMH e 6 HMM e 7 MMM e 8 147 . ) 3 )(. 7 (. ) 7 )(. 3 )(. 7 (. ) R with H ( P ) L with H ( P ) R with H ) P ) R with H L with H R with H ( P ) e ( P 2 1 063 . ) 3 )(. 7 (. ) e ( P 2 2 063 . ) 3 )(. 7 (. ) e ( P ) e ( P 2 2 3 343 . 7 . ) e ( P 3 4 027 . 3 . ) e ( P 3 5 147 . ) 3 )(. 7 (. ) e ( P ) e ( P 2 7 6 063 . ) 3 )(. 7 (. ) e ( P 2 8 b. 273 . ) 063 (. 2 147 . ) e ( P ) e ( P ) e ( P ) win ( P 3 2 1 3. Let L i be the length of the i th fiber, i = 1,…,5. Let e i be the basic outcome where the i th fiber breaks first. We are told that the probability of a fiber breaking first is proportional to its length. Therefore, i i L ) e ( P . Since the probabilities of the basic outcomes must sum to one, we have 1 L ) 5 ( ) 4 ( ) 3 ( ) 2 ( ) 1 ( ) e ( P .... ) e ( P 5 1 i i 5 1 Since 5 1 i i 15 5 4 3 2 1 L , it must be that = 1/15. The sample space is } e ,..., e { S 5 1 , and the basic outcomes occur with probabilities 15 / 1 ) e ( P 1 , 15 / 2 ) e ( P 2 ,…, 15 / 5 ) e ( P 5 .

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4. a. 2 6 =64 b. 22 1 6 15 ! 0 ! 6 ! 6 ! 1 ! 5 ! 6 ! 2 ! 4 ! 6 C C C 6 6 5 6 4 6 c. 720 ! 0 ! 6 P 6 6 d. 120 ! 3 ! 6 P 3 6 e. 20 ! 3 ! 3 ! 6 C 3 6 f. 10 ! 2 ! 3 ! 5 ! 1 ! 0 ! 1 C C 2 5 1 1 g. 6 ) 2 )( 3 )( 1 ( ! 1 ! 1 ! 2 ! 1 ! 2 ! 3 ! 1 ! 0 ! 1 C C C 1 2 1 3 1 1 h. 5 . 20 10 i. 8 . 20 ) 1 )( 4 ( ) 2 )( 6 ( C C C C C 3 6 2 2 1 4 1 2 2 4 j. .5 4 =.0625 5.a. L is “terms of lease acceptable” T is “profitable transportation arrangements available” R is “reserves large enough” e 1 = L L e 2 = T L L T e 3 = R T L T R R e 4 = R T L S = {e 1 , e 2 , e 3 , e 4 } b. P(L) = .8 P(T|L) = .7 P(R | T L) = .9 .2 P(L) 1 ) P(e 1 .24 (.3)(.8) L)]P(L) | P(T 1 [ L)P(L) | T P( ) P(e 2 .056 8) (.1)(.7)(. L) L)]P(T T | P(R 1 [ L) L)P(T T | R P( ) P(e 3 .504 8) (.9)(.7)(. L)P(L) | L)P(T T | P(R ) P(e 4 c. 1 P(e 4 ) = 1 .504 = .496
6.a. For each proposal, there are four possible outcomes: AO = proposal accepted outright RO = proposal rejected outright MA = refinement submitted, outcome is acceptance MR = refinement submitted, outcome is rejection Each outcome for the first proposal can occur together with any of the outcomes for the second proposal, so by the product rule, there are 16 basic outcomes in the sample space: Proposal #1 AO RO MA MR AO e 1 e 2 e 3 e 4 RO e 5 e 6 e

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## This note was uploaded on 02/26/2012 for the course ECONOMICS 329 taught by Professor Bencivenga during the Spring '12 term at University of Texas.

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04 STAT HWA a - Dr Valerie R Bencivenga Economics 329...

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