Problemas do Capítulo 5 - Exercícios de Cálculo e Probabilidade

68 Chapter 5 Problems 1 a 1 2 1 1 1 3 4 c x dx c b Fx 3 2 1 3 3 2 1 4 4 3 3 x x x dx x 1 x 1 2 2 2 2 2 4 x x x xe dx xe e Hence 2 0 1 1 4 c xe x dx c PX 5 2 5 2 5 2 5 1 110 4 4 4 xe x dx e e 5 2 14 4 e 3 No f52 0 4 a 2 20 20 10 10 1 2 dx x x b Fy 2 10 10 10 1 y dx y x y 10 Fy 0 for y 10 c 6 6 3 6 2 1 3 3 i i i i since 10 15 15 F Assuming independence of the events that the devices exceed 15 hours 5 Must choose c so that 01 1 4 5 51 1 c x dx c so c 1 0115 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 5 69 1p xn fay 6 a EIX 7 xe dx2 ye dx 2T34 0 0 b By symmetry of fx about x 0 EX 0 f5 c EIX dr 5 x b 7 ab ax lora1 3 1 xc bx dx 3 or a 35 Hence 5 2 4 a p 5 5 8 ELX xe dv 132 0 9 If s units are stocked and the demand is X then the profit Ps is given by Ps bX s XP ifXs sb ifXs Hence EPs J bx s x0 f xax sbf xdx b of af dx stf fodx so I fdr sbh b Ofc sf ode Differentiation yields d d s s ELP b b olf af dx sf Fade b b ol sf sf s fj fords bb 0 fxdx Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 70 Chapter 5 Equating to zero shows that the maximal expected profit is obtained when s is chosen so that b Fs s bé where Fs J f xdx is the cumulative distribution of demand 10 a Pgoes toAP5X 15 or 20X 30 0r35X45 or50X 60 23 since X is uniform 0 60 b same answer as in a 11 X is uniform on 0 L P min 2 cra LX xX 1P min 2 54 LX xX 1 r 422 5 14 LX xX 1PXL5 X 4L5 1 P E xX 4L3 132 5 5 13 PIX 10 2 Px25 X 15 PIX 2 920 i 3 PX 15 1530 where X is uniform 0 30 1 14 EX xdx nl PX x PXx x 1 1 1 1 1 1 EX fot Me f xa nl n nl 15 a 8333 7977 b 21 1 6827 c 1 3333 3695 d 16667 9522 e 11 1587 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 5 71 16 PX 50 40 10 4 4 P X 1 Φ25 1 9938 Hence PX 5010 993810 17 EPoints 10110 5210 3210 26 18 2 5 9 5 P X σ σ PZ 4σ where Z is a standard normal But from the normal table PZ 84 80 and so 84 4σ or σ 476 That is the variance is approximately 4762 2266 19 Letting Z X 122 then Z is a standard normal Now 10 PZ c 122 But from Table 51 PZ 128 90 and so c 122 128 or c 1456 20 Let X denote the number in favor Then X is binomial with mean 65 and standard deviation 6535 477 Also let Z be a standard normal random variable a PX 50 PX 495 PX 65477 155477 PZ 325 9994 b P595 X 705 P55477 Z 55477 2PZ 115 1 75 c PX 745 PZ 95477 977 22 a P9000 005 X 9000 005 005 005 003 003 P Z P167 Z 167 2Φ167 1 9050 Hence 95 percent will be defective that is each will be defective with probability 1 9050 0950 b 005 005 005 2 P Z σ σ σ Φ 1 99 when 005 005 995 2575 0019 σ σ σ Φ Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 72 Chapter 5 23 a P1495 X 2005 1000 1000 1495 2005 6 6 1 5 1 5 1000 1000 6 6 6 6 P Z 2005 1667 1495 1667 500036 500036 Φ Φ Φ287 Φ146 1 9258 b PX 1495 1495 80015 1 4 800 5 5 P Z PZ 93 1 Φ93 1762 24 With C denoting the life of a chip and φ the standard normal distribution function we have PC 18 106 6 6 5 18 10 14 10 3 10 φ φ133 9082 Thus if N is the number of the chips whose life is less than 18 106 then N is a binomial random variable with parameters 100 9082 Hence PN 195 1 195 9082 90820918 φ 1 φ247 1 25 Let X denote the number of unacceptable items among the next 150 produced Since X is a binomial random variable with mean 15005 75 and variance 1500595 7125 we obtain that for a standard normal random variable Z PX 10 PX 105 75 105 75 7125 7125 P X PZ 11239 8695 The exact result can be obtained by using the text diskette and to four decimal places is equal to 8678 27 PX 57995 7995 2500 P Z PZ 1599 negligible Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 5 73 28 Let X equal the number of lefthanders Assuming that X is approximately distributed as a binomial random variable with parameters n 200 p 12 then with Z being a standard normal random variable PX 195 20012 195 20012 2001288 2001288 X P PZ 9792 8363 29 Let s be the initial price of the stock Then if X is the number of the 1000 time periods in which the stock increases then its price at the end is suXd1000X sd1000 u X d Hence in order for the price to be at least 13s we would need that d1000 13 u X d or X log13 1000log log d u d 4692 That is the stock would have to rise in at least 470 time periods Because X is binomial with parameters 1000 52 we have PX 4695 100052 4695 100052 10005248 10005248 X P PZ 3196 9993 30 Pin black 5 black 5 black 5 white1 P P P α α α 2 2 2 5 4 8 5 4 8 5 6 18 1 2 2 1 1 1 2 2 3 2 e e e α π α α π π 18 18 18 2 1 2 3 e e e α α α α is the value that makes preceding equal 12 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 74 Chapter 5 31 a E X a 2 0 2 A a a dx dx A a x a a x a A A A 2 1 0 2 d a a A da A b E X a 0 a x x a a x e dx x a e dx λ λ λ λ 1 1 a a a a a a e e a e ae ae ae λ λ λ λ λ λ λ λ λ Differentiation yields that the minimum is attained at a where 1 2 e λa or a log 2λ c Minimizing a median of F 32 a e1 b e12 33 e1 34 a PX 20 e1 b PX 30X 10 30 1 4 10 3 4 P X P X 13 35 a 50 40 exp λ t dt e35 b e121 36 a 1 F2 2 3 0 exp t dt e4 b exp444 exp1444 c 2 3 1 exp t dt e154 37 a PX 12 PX 12 PX 12 12 b PX a Pa X a a 0 a 1 Therefore 1 f X a 0 a 1 That is X is uniform on 0 1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 5 75 38 For both roots to be real the discriminant 4Y2 44Y 2 must be 0 That is we need that Y2 Y 2 Now in the interval 0 Y 5 Y2 Y 2 Y 2 and so PY2 Y 2 PY 2 35 39 FYy Plog X y PX ey FXey fYy fXeyey y y e ee 40 FYy PeX y FXlog y fYy 1 1 Xf log y y y 1 y e Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 76 Chapter 5 Theoretical Exercises 1 The integration by parts formula udv uv vdu with dv 2 2 bxebx u x2b yields that 2 2 2 2 0 0 0 1 2 2 bx bx bx xe x e dx e dx b b 2 2 3 2 0 1 2 e y dy b by y x 2 b 3 2 3 2 2 1 2 2 4 b b π π where the above uses that 2 2 0 1 2 e y dy π 12 Hence a 3 2 4b π 2 0 P Y y dy 0 y Yf x dx dy 0 0 0 x Y Y f x dy dx xf x dx Similarly 0 0 Y P Y y dy xf x dx Subtracting these equalities gives the result 4 EaX b ax b f x dx a xf x dx b f x dx aEX b 5 EXn 0 P X n t dt 1 0 n n n P X x nx dx by t xn dt nxn1dx 1 0 n P X x nx dx 6 Let X be uniform on 0 1 and define Ea to be the event that X is unequal to a Since a Ea is the empty set it must have probability 0 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 5 77 7 SDaX b 2 2 Var aX b a a σ σ 8 Since 0 X c it follows that X2 cX Hence VarX EX2 EX2 EcX EX2 cEX EX2 EXc EX c2α1 α where α EXc c24 where the last inequality first uses the hypothesis that P0 X c 1 to calculate that 0 α 1 and then uses calculus to show that 0 1 maximum α α1 α 14 9 The final step of parts a and b use that Z is also a standard normal random variable a PZ x PZ x PZ x b PZ x PZ x PZ x PZ x PZ x 2PZ x c PZ x 1 PZ x 1 2PZ x by b 1 21 PZ x 10 With c 1 2πσ we have fx 2 2 2 x ce μ σ f x 2 2 2 2 ce x x μ σ μ σ f x 2 2 2 2 4 2 2 2 2 x x c e x c e μ σ μ σ σ μ σ Therefore f μ σ fμ σ cσ2e12 cσ2e12 0 11 a Integrate 1 2 E g Z π 2 2 g x e x dx by parts 2 2 1 2 x dv g x dx u e π to obtain the result b Let gz zn and apply part a c Let n 3 in part b to obtain 4 2 3 3 E Z E Z 12 EX2 2 1 2 0 0 2 2 2 2 x P X x x dx xe dx E X λ λ λ Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 78 Chapter 5 13 a 2 b a b μ c 1 eλm 12 or m 1 λ log 2 14 a all values in a b b μ c 0 15 PcX x PX xc 1 eλxc 16 λt 1 1 f t a F t a t a a t 0 t a 17 If X has distribution function F and density f then for a 0 FaXt PaX t Fta and fax 1 a f t a Thus 1 1 1 aX X a f t a t t a F t a a λ λ 19 EXk 0 k x x e λ dx λ 0 k x k e x dx λ λ λ λ λk Γk1 kλk Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 5 79 20 EXk 1 0 1 k x t x e x dx t λ λ λ Γ 1 0 k x t k e x dx t λ λ λ λ Γ k t k t λ Γ Γ Therefore EX tλ EX2 λ2Γt 2Γt t 1tλ2 and thus VarX t 1tλ2 t2λ2 tλ2 21 Γ12 1 2 0 e x x dx 2 2 0 2 e y dy by x y22 dx ydy 2x dy 2 1 2 2 0 2 2 e y dy π π 2 0 π P Z where Z is a standard normal π 22 1λs 1 1 x t s t x s e x dx e s λ λ λ λ λ λ 1 x s t x s e x s dx λ 1 0 1 y t y e y s dy λ by letting y x s As the above equal to the inverse of the hazard rate function is clearly decreasing in s when t 1 and increasing when t 1 the result follows 23 λs cs vβ1 s v which is clearly increasing when β 1 and decreasing otherwise 24 Fα 1 e1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 80 Chapter 5 25 Suppose X is Weibull with parameters v α β Then 1 X v X v P x P x β β α α PX v αx1β 1 expx 26 We use Equation 63 EX Ba 1 bBA b 1 1 a a b a a b a a b Γ Γ Γ Γ EX2 Ba 2 bBa b 2 1 2 1 a a b a a a b a a b a b Γ Γ Γ Γ Thus VarX 2 2 2 1 1 1 a a a ab a b a b a b a b a b 27 X ab a 29 PFX x PX F1x FF1x x 30 FYx PaX b x x b P X a when a 0 FXx ba when a 0 fYx 1 a fXx ba if a 0 When a 0 FYx 1 X x b x b P X F a a and so fYx 1 X x b a f a 31 FYx PeX x PX log x FXlog x fYx fXlog xx 2 2 log 2 1 2 x e x μ σ πσ Copyright 2010 Pearson Education Inc Publishing as Prentice Hall