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Student Solutions Manual Applied Statistics and Probability for Engineers, Fifth Edition by Douglas C. Montgomery, George C. Runger

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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) = ...

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