Soluciones a cuestionarios y pruebas Capítulo 9 Probabilidad con Procesos Estocásticos, Resúmenes de Estadística

¡Descarga Soluciones a cuestionarios y pruebas Capítulo 9 Probabilidad con Procesos Estocásticos y más Resúmenes en PDF de Estadística solo en Docsity! SOLUTIONS TO SELF-QUIZZES AND SELF-TESTS CHAPTER 9 FUNDAMENTALS OF PROBABILITY wIiTH STOCHASTIC PROCESSES FOURTH EDITION SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA CRC Press Taylor 8 Francis Group Boca Raton London New York (CRC Press is an imprint of the Taylor $ Francis Group, an informa business A CHAPMAN 8. HALL BOOK Chapter 9 — Solutions to Self-Quiz Problems 2 SOLUTIONS TO SELF-QUIZ PROBLEMS Section 9.1 1. Le K = ((2,y,2,u0,0):0<3 <15,0<y<10,0<2<6,0<0u<4, 0<S0<5, T4Yy+H2 HU 0= 20). p(=, y, z, u, v), the joint probability mass function of X, Y, Z, U, and V is given by 15) (10) (6 (AN (5 )= z yz) Xu) o AN 20 The marginal probability mass function of U and V is given by (DC) lo.) MINVEADUTI O <u<40<uS5 40 20 2. We have P(X<Y<Z)= P [f ( P 2y e Ortu+a) dz) dy] de 40 Ji vy = P (fuer dy) de Jo Mz 1/2 o 3 =3 204 1)e Y de == =0.19. 3/ (Qu+1)e la 16 0.19 p(z,y,z,u, v (2,y,z,u,v) € K. Pu, v(u, v) = Section 9.2 1. Let X and X be the points selected randomly and independently from the interval (0, 1). We are interested in E [X (2, — X(1)]. Note that, by Theorem 9.7, f2(21, 22), the joint probability density function of X (,) and Xy, is given by 2 f(21)f (xo) 0<w% <x2<1 Ha(u1, 32) = 0 otherwise, where, 1 0<x<1 0 otherwise. Chapter 9 Solutions to Self-Test Problems 5 SOLUTIONS TO SELF-TEST PROBLEMS 4 1. (a) The volume of a sphere with radius R is re. Since the volume of the unit 4 sphere is 3” by (9.10), 2 ifad+y+22<1 fay=3 0 otherwise. MA a, (b) Ixrt)=] HIdi= 1-24, a+ys<l. Ia dr 27 VEZ (o) fzlz)= F Pa — dy dx. However, it is much easier if we calculate VEZ this double integral by changing it to polar coordinates. Note that, in polar coordinates, dx dy = rdrd0, 0 < 0 < 27, and r changes from 0 to V1 — 22. Therefore, 21 ¡VEZ 3 mt) = f| J/ gn" ard0 27 -/ > (1-2?) d0 = 1-2), -1<2<1. o Sm 2. Let f be the joint probability density function of X 1, Xz, ...., X,. We have Fur, 02... Tn) = fx, (21) fx,1x, (v2l21) £x31x,,x, (23|71, 72) Exa Xi Xo0 Xp 1 (Un | 1) 72, > >> Un 1): Therefore, for 0 < 2 < Ip-1<-"""<wm»z<wu1<l, Harta 0) = 1 tt = and f(11,22,..., Tn) =0, otherwise. 3. We need to find c so that it satisfies So po poo pos 1 o ————————_ didzdydx = 1. e / / / / Me Chapter 9 Solutions to Self-Test Problems 6 Now So po yoo pos 1 dt dz dy de / / / / (Fay E e d -f" aa” dydz -/” 60( a = +5 Therefore, c - (1/120) = 1 implies that c = 120. Now we calculate P(X<Y<Z<T). We have o po 1 PX <Y<Z<T)= 120 / / / J Uraryri aa drdyda auf" f IN Arras 00 poo 1 =3 —————, Qy due / / (1+2+3y)* Ye 1 =5/ Tra 00M, 4. (a) Let Xi, X2, ..., X7 be the lifetimes of the components of the system. For 1<:1<7, f, the probability density function of X ¿, and F, its distribution function are given by fa)=e7?, 2>0; Fla)=1-e* 2>0. Note that the system is still functional two years from now if and only if X (7, > 2. By Remark 9.2, the probability density function of X (7, is given by fala) =T 1, 220. Therefore, o P(Xm > 2) = / Te"(1— e") de = 0.64. 2 (b) At least three components of the system are functional two years from now if and only if X(;, > 2. By Theorem 9.5, the probability density function of X (s, is (1 er [1 (1 e77)]?= 10501079 2>0. Chapter 9 Solutions to Self-Test Problems 7 Therefore, P(X (5) > 2) = 105 / (1 — e da 0.057. 2 5. Let X1, X2, and X3 be, respectively, the number of physicians among the group who retired at ages 64 and below, 65-69, and 70 or later. Let p(11,w2,13) be the joint probability mass function of X1, X2, and Xz. The median age of the five randomly selected retired physiciansis below 65 if and only if the ages of at least three of them are below 65. Therefore, the desired probability is p(5,0,0) + p(4, 1,0) + p(4, 0, 1) + p(3, 2,0) + p(3,0,2)+ p(3,1,1) = OA OO OOO 5! 31111 5! (0.2)%(0.3)?(0.5)"+ (0.23%(0.3)(0.5)?+ 3!