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Chapter 5 Student Solutions Manual Applied Statistics and Probability for Engineers Fifth Edition
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Chapter 5 Student Solutions Manual Applied Statistics and Probability for Engineers Fifth Edition
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2016/12/23 Chapter 5 - Student Solutions Manual Applied Statistics and Probability for Engineers, Fifth EditionStudent Solutions Manual Applied Statistics and Probability for Engineers, Fifth Edition
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Chapter 4 Chapter 6
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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,
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) = 1.8125
E(Y) = 1(1/4) + 2(1/8) + 3(1/4) + 4(1/4) + 5(1/8) = 2.875
2 2 2 2 2
1 5 V(X) = E(X5 ) − [E(X)] = [ (1/4) + 1. (3/8) + 2. (1/4)
2 2
+ 35 (1/8)] − 1.812 = 0.4961
2 2 2 2 2
Y 1 V(Y) = E( ) − [E(Y)] = [ (1/4) + 2 (1/8) + 3 (1/4) +
2 2 2
5 4
5 (1/4) + (1/8)] − 2.87 = 1.8594
f) marginal distribution of X
g) and fX (1.5) = 3/8. Then,
h) and fY(2) = 1/8. Then,
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) = fXY(−1,−2) + f XY(−0.5,−1) =
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,2016/12/23 Chapter 5 - Student Solutions Manual Applied Statistics and Probability for Engineers, Fifth Edition
c) P(Y < 1.5) = fXY(−1,−2)+ fXY(−0.5,−1)+ fXY(0.5,1) =
d) P(X > 0.25, Y < 4.5) = fXY(0.5,1)+ fXY(1,2) =
e) E(X) =
2 2
V(X) = (− 1 − 1/8) (1/8) + (− 0.5 − 1/8) (1/4) + (0.5 −
2 2
1/8) (1/2) + (1 − 1/8) (1/8) = 0.4219
2 2
V(Y) = (− 2 − 1/4) (1/8) + (− 1 − 1/4) (1/4) + (1 −
2 2
1/4) (1/2) + (2 −1/4) (1/8) = 1.6875
f) marginal distribution of X
g)
h)
i) E(X|Y = 1) = 0.5
j) No, X and Y are not independent
5-5. a) The range of (X,Y) is
The problem needs probabilities to total one. Modify so that the probability
of moderate distortion is 0.04.
b)
c) E(X) = 0(0.970299) + 1(0.029403) + 2(0.000297) + 3*
(0.000001) = 0.03 (or np = 3*0.01) ⬆
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, 2016/12/23 Chapter 5 - Student Solutions Manual Applied Statistics and Probability for Engineers, Fifth Edition
d)
e) E(Y|X = 1) = 0(.920824) + 1(0.077543) + 2(0.001632) =
0.080807
g) No, X and Y are not independent because, for example,
fY(0)≠fY|1(0).
5-7. a) The range of (X,Y) is X ≥ 0, Y ≥ 0 and X + Y ≤ 4.
Here X and Y denote the number of defective items found with inspection
device 1 and 2, respectively.
For x = 1,2,3,4 and y = 1,2,3,4
b)
c) Because X has a binomial distribution E(X) = n(p) = 4*
(0.993) = 3.972
d)
e) E(Y|X = 2) = E(Y) = n(p) = 4(0.997) = 3.988
f) V(Y|X=2) = V(Y) = n(p)(1 − p) = 4(0.997)(0.003) =
0.0120
g) Yes, X and Y are independent.
x y 4−x−y
5-9. (a) fXY(x,y)= (10%) (30%) (60%) , for X + Y <
=4
⬆
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