CHAPTER 5

Section 5-1

5-1.    First, f(x,y) ≥ 0. Let R denote the range of (X,Y).

Then, Image

a)   P(X < 2.5, Y < 3) = f(1.5,2) + f(1,1) = 1/60;

b)   P(X < 2.5) = f (1.5, 2) + f (1.5, 3) + f(1,1) = 1/8 + 1/4 + 1/4 = 5/8

c)   P(Y < 3) = f (1.5, 2) + f(1,1) = 1/8 + 1/4 = 3/8

d)   P(X > 1.8, Y > 4.7) = f ( 3, 5) = 1/8

e)   E(X) = 1(1/4) + 1.5(3/8) + 2.5(1/4) + 3(1/8) = 1.8125E(Y) = 1(1/4) + 2(1/8) + 3(1/4) + 4(1/4) + 5(1/8) = 2.875V(X) = E(X2) − [E(X)]2 = [12(1/4) + 1.52(3/8) + 2.52(1/4) + 32(1/8)] − 1.81252 = 0.4961V(Y) = E(Y2) − [E(Y)]2 = [12(1/4) + 22(1/8) + 32(1/4) + 42(1/4) + 52(1/8)] − 2.8752 = 1.8594

f)   marginal distribution of XImage

g)   Image and fX (1.5) = 3/8. Then,Image

h)   Image and fY(2) = 1/8. Then,Image

i)   E(Y|X = 1.5) = 2(1/3) + 3(2/3) = 2 1/3

j)   Since fY|1.5(y) ≠ fY(y), X and Y are not independent

5-3.    f(x, y) ≥ 0 and

a)   P(X < 0.5,Y < 1.5) = fXY(−1,−2) + fXY(−0.5,−1) =

b)   P(X < 0.5) = ...

Get Student Solutions Manual Applied Statistics and Probability for Engineers, Fifth Edition now with O’Reilly online learning.

O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers.