Capítulo 8: Problemas e Exercícios

140 Chapter 8 Problems 1 P0 X 40 1 PX 20 20 1 20400 1920 2 a PX 85 EX85 1517 b P65 X 85 1 PX 75 10 1 25100 c 1 25 75 5 25 n i i P X n n so need n 10 3 Let Z be a standard normal random variable Then 1 75 5 n i i P X n PZ n 1 when n 3 4 a 20 1 15 2015 i i P X b 20 20 1 1 15 155 i i i i P X P X 155 20 20 P Z PZ 1006 8428 5 Letting Xi denote the ith roundoff error it follows that 50 1 i i E X 0 Var 50 1 i i X 50 VarX1 5012 where the last equality uses that 5 X is uniform 0 1 and so VarX Var5 X 112 Hence 3 i P X PN0 1 3125012 by the central limit theorem 2PN0 1 147 1416 6 If Xi is the outcome of the ith roll then EXi 72 VarXi 3512 and so 79 79 1 1 300 3005 i i i i P X P X 1 2 3005 797 2 01 01 158 79 3512 P N P N 9429 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 8 141 7 100 1 525 500 525 01 01 5 100 25 i i P X P N P N 3085 where the above uses that an exponential with mean 5 has variance 25 8 If we let Xi denote the life of bulb i and let Ri be the time to replace bulb i then the desired probability is 100 99 1 1 550 i i i i P X R Since i i X R has mean 100 5 99 25 52475 and variance 2500 9948 2502 it follows that the desired probability is approximately equal to PN0 1 550 52475250212 PN0 1 505 693 It should be noted that the above used that VarRi Var 1 Unif01 2 148 9 Use the fact that a gamma n 1 random variable is the sum of n independent exponentials with rate 1 and thus has mean and variance equal to n to obtain 01 X n P n 01 P X n n n 01 01 P N n 2 01 01 P N n Now PN0 1 258 005 and so n 2582 10 If Wn is the total weight of n cars and A is the amount of weight that the bridge can withstand then Wn A is normal with mean 3n 400 and variance 09n 1600 Hence the probability of structural damage is PWn A 0 400 3 09 1600 P Z n n Since PZ 128 1 the probability of damage will exceed 1 when n is such that 400 3n 128 09 n 1600 The above will be satisfied whenever n 117 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 142 Chapter 8 12 Let Li denote the life of component i 100 1 1 1000 50101 10 i i E L 1505 Var 2 100 100 100 2 2 1 1 1 1 10 100 100101 10 100 i i i i i L i Now apply the central limit theorem to approximate 13 a 74 80 157 145 X P X P PPZ 214 0162 b 74 80 247 148 Y P Y P PZ 343 0003 c Using that 19664 196 25 SD Y X 330 we have 22 P Y X 330 22330 P Y X PZ 67 2514 d same as in c 14 Suppose n components are in stock The probability they will last for at least 2000 hours is p 1 2000 100 2000 30 n i i n P X P Z n where Z is a standard normal random variable Since 95 PZ 164 it follows that p 95 if 2000 100 30 n n 164 or equivalently 2000 100n n 492 and this will be the case if n 23 15 10000 1 2700000 i i P X PZ 2700000 2400000800 100 PZ 375 0 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 8 143 16 a Number AJs jobs let Xi be the time it takes to do job i and let XA 20 1 i i X be the time that it takes AJ to finish all 20 jobs Because 2050 1000 E X A VarXA 20100 2000 the central limit theorem gives that 1000 900 1000 900 2000 2000 A A X P X P 2236 1 2236 013 P Z Φ b Similarly if we let XM be the time that it takes MJ to finish all of her 20 jobs then by the central limit theorem XM is approximately normal with mean and variance 2052 1040 E XM VarXM 20225 4500 Thus 1040 900 1040 900 4500 4500 M M X P X P 2087 1 2087 018 P Z Φ c Because the sum of independent normal random variables is also normal M A D X X is approximately normal with mean and variance ED 1040 1000 40 VarD 4500 2000 6500 Hence 40 40 0 6500 6500 D P D P 4961 4961 691 P Z Φ Thus even though AJ is more likely than not to finish earlier than MJ MJ has the better chance to finish within 900 minutes Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 144 Chapter 8 19 Let Yi denote the additional number of fish that need to be caught to obtain a new type when there are at present i distinct types Then Yi is geometric with parameter 4 4 i EY 3 0 4 4 25 1 4 3 2 3 i i E Y VarY Var 3 0 4 130 2 12 9 9 i i Y Hence 25 25 1300 1 3 3 9 10 P Y and so we can take a 25 1300 3 b 25 1300 3 Also 2 25 130 1 3 10 130 9 P Y a a when a 1170 3 Hence 25 1170 3 P Y 1 21 gx xnn1 is convex Hence by Jensens Inequality EY nn1 EYnn1 Now set Y X n1 and so EX n EX n1nn1 or EX n1n EX n11n1 22 No 23 a 2026 769 b 2020 36 514 357 d p PZ 255 20 20 PZ 123 1093 e p 112184 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 8 145 Theoretical Exercises 1 This follows immediately from Chebyshevs inequality 2 PD α PX μ αμ 2 2 2 2 2 1 r ς α μ α 3 a λ λ λ b 1 1 np np p np p c answer 1 d 1 2 3 112 e answer 1 d answer μ σ 4 For ε 0 let δ 0 be such that gx gc ε whenever x c δ Also let B be such that gx B Then EgZn n n x c x c g x dF x g x dF x δ δ ε gcPZn c δ BPZn c δ In addition the same equality yields that EgZn gc εPZn c δ BPZn c δ Upon letting n we obtain that lim sup EgZn gc ε lim inf EgZn gc ε The result now follows since ε is arbitrary 5 Use the notation of the hint The weak law of large numbers yields that 1 lim 0 n n P X X n c ε Since X1 Xn is binomial with parameters n x we have 1 1 1 n k n k n k n X X E f f k n x x k n The result now follows from Exercise 4 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 146 Chapter 8 6 EX 1 1 k i i k i P X i i P X i 1 k i i P X k PX kkk 12 2 2 k P X k 7 Take logs and apply the central limit theorem 8 It is the distribution of the sum of t independent exponentials each having rate λ 9 12 10 Use the Chernoff bound etiMt 1 te ti eλ will obtain its minimal value when t is chosen to satisfy λet i and this value of t is negative provided i λ Hence the Chernoff bound gives PX i eiλλii 11 etiMt pet qneti and differentiation shows that the value of t that minimizes it is such that npet ipet q or et iq n i p Using this value of t the Chernoff bound gives that PX i n i i i iq q n i p iq n i n i i i i n nq n i p i q n i 12 1 EeθX eθEX by Jensens inequality Hence θEX 0 and thus θ 0 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall