Exam (elaborations)
Chapter 5 Student Solutions Manual Applied Statistics and Probability for Engineers Fifth Edition
- Course
- Institution
Chapter 5 Student Solutions Manual Applied Statistics and Probability for Engineers Fifth Edition
[Show more]
Content preview
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
PREV NEXT
⏮ ⏭
Chapter 4 Chapter 6
🔎
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) =
Enjoy Safari? Subscribe Today
https://www.safaribooksonline.com/library/view/student-solutions-manual/9780470888445/Chapter05.html 1/19
,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) ⬆
Enjoy Safari? Subscribe Today
https://www.safaribooksonline.com/library/view/student-solutions-manual/9780470888445/Chapter05.html 2/19
, 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
⬆
Enjoy Safari? Subscribe Today
3/19
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller Edumaxsolutions. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $13.69. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
75323 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now
