STA 401/501
Fall 2012, Miami University
HW4 Solutions
3.2
3.6
We know that P(HH) = P(TT) = P(HT) = P(TH) = 0.25. So, P(Y = -1) = .5, P(Y = 1) =
.25 = P(Y = 2).
5
There are = 10 sample points, and all are equally likely: (1,2), (1,3), (1,4), (1,5),
2
(2,3
STA 401/501, Miami University
Exam 1
September 19, 2011
Write your name on this sheet, and at the end of the exam staple it to the paper
used to work the problems. Show all work and try all parts (even if you cant do
a previous part)! Calculators and a fo
STA 401/501, Miami University
Exam 1
September 13, 2012
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cant do a previous part)!
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This exam has 5 problems and 100 points. Good luck!
1. (a) [6 points] How many ve-letter str
STA 401/501, Miami University
Exam 2
October 18, 2012
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This exam has 6 problems and 100 points. Good luck!
1. Suppose that scores on an exam can be a
STA 401/501, Miami University
Exam 3
November 19, 2012
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This exam has 6 problems and 100 points. Good luck!
1. A certain market has both an express c
STA 401/501
Fall 2012, Miami University
HW8 Solutions
1. WMS 3.145
n
n
Using the binomial theorem, m(t ) E (e tY ) ( pe t ) y q n y ( pe t q ) n .
y 0 y
2. WMS 3.146
d
dt
m(t ) n( pet q) n1 pet . At t = 0, this is np = E(Y).
m(t ) n(n 1)( pet q) n1 ( pet
STA 401/501
Fall 2012, Miami University
HW10 Solutions
1. WMS 5.92
1 y2
From Ex. 5.77, E(Y1) = 1/4 and E(Y2) = 1/2. E(Y1Y2) =
6y
1
y 2 (1 y 2 )dy1 dy 2 = 3/20. So, Cov(Y1, Y2)
00
= E(Y1Y2) E(Y1)E(Y2) = 3/20 1/8 = 1/40 as expected since Y1 and Y2 are depe
STA 401/501
Fall 2012, Miami University
HW11 Solutions
1. WMS 6.42
The probability of interest is P(Y2 > Y1) = P(Y2 Y1 > 0). By Theorem 6.3, the distribution of Y2 Y1 is
normal with = 4000 5000 = 1000 and 2 = 4002 + 3002 = 250,000. Thus, P(Y2 Y1 > 0) = P(
STA 401/501
Fall 2012, Miami University
HW12 Solutions
1. WMS 7.13
P(| Y | .5) = P(2.5 Z 2.5) = .9876.
2. WMS 7.14
We require P(| Y | .5) = P(
.5
.4
n Z
.5
.4
n ) = .95. Thus,
.5
.4
n = 1.96 so that n = 6.15.
Hence, run 7 tests.
3. WMS 7.19
Given that s2
STA 401/501, Miami University
Exam 2
October 11, 2011
Show all workremember that not just the nal answer but how you arrived at
the answer counts! Calculators and a formula sheet (both sides) may be used, but
no other notes and no neighbors.
Name:
This ex
STA 401/501, HW# 1 Solutions
2.2 a. A B, b. A B, c. (A B) = A B, d. (A B) (A B).
2.5
b
B (A B)
= (B A) (B B)
by distributive law
= (B A) S
= (B A)
= A
because B A
d B (A B) = A (B B) = A = . Thus, B and A B are
mutually exclusive.
2.7 Thear are five appli
STA 401/501
Fall 2012, Miami University
HW7 Solutions
1. WMS 4.38
0
y0
y
a. F ( y ) 1dy y 0 y 1
0
1
y 1
b. P(a Y a + b) = F(a + b) F(a) = a + b a = b.
2. WMS 4.44a
a. Y has a uniform distribution (constant density function), so k = 1/4.
3. WMS 4.48
Let Y
STA 401/501
Fall 2012, Miami University
HW6 Solutions
1. WMS 3.138
E[Y (Y 1)] y ( y 1y)! e 2 y ( y 1)y!
y
y 0
y 2
. Using the substitution z = y 2, it is found that E[Y(Y 1)]
e
y 0
= 2. Use this with V(Y) = E[Y(Y1)] + E(Y) [E(Y)]2 = .
2. WMS 3.199
a. Us
STA 401/501, Miami University
Exam 3
November 21, 2011
Show all workremember that not just the nal answer but how you arrived at
the answer counts! Calculators and a formula sheet (both sides) may be used, but
no other notes and no neighbors.
Name:
This e
STA 401/501, Miami University
Exam 1
February 2, 2012
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cant do a previous part)!
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This exam has 5 problems and 100 points. Good luck!
1. In the dice game of Farkle, players may
STA 401/501, Miami University
Exam 2
March 15, 2012
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This exam has 6 problems and 100 points. Good luck!
1. The length of time required by students t
STA 401/501, Miami University
Exam 3
April 19, 2012
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1. Let Y1 and Y2 denote the proportions of t
The Lottery and Lost
STA 401/501
Extra Credit 4, Due Wednesday, October 19, beginning of class
On March 2, 2005, an episode of Lost (the TV series) aired in which a character used the
numbers (4, 8, 15, 16, 23, 42) to win the lottery. On January 4, 2011 t
STA 401/501, Miami University
Final Exam
December 14, 2011
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cant do a previous part)!
Name:
This exam has 7 problems and 100 points (plus 4 bonus points). Good luck!
1. Suppose is a
STA 401/501, Miami University
Final Exam
May 1, 2012
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cant do a previous part)!
Name:
This exam has 6 problems and 100 points (+3 bonus points). Good luck!
1. When the health depart
STA 401/501
Spring 2011, Miami University
HW1 Solutions
b. A B
c. A B
d. ( A B ) ( A B )
2.2
a. AB
2.5
c. ( A B) ( A B ) A ( B B ) 0 . The result follows from part a.
d. B ( A B ) A ( B B ) = 0 . The result follows from part b.
2.6
A = cfw_(1,2), (2,2), (
STA 401/501
Spring 2011, Miami University
HW2 Solutions
2.43
2.44
2.53
9 6 1
= 504 ways.
3 5 1
8 5
a. The number of ways the taxi needing repair can be sent to airport C is = 56.
5 5
So, the probability is 56/504 = 1/9.
6 4
b. 3! = 90, so the
STA 401/501
Fall 2012, Miami University
HW3 Solutions
2.90
a. 1-(49/50)(49/50) = 0.0396.
b. P(at least one injury) = 1 P(no injuries in 50 jumps) = 1 = (49/50)50 = 0.636. Your
friend is not correct.
2.99
Given that P( A B ) = a, P(B) = b, and that A and B
STA 401/501
Fall 2012, Miami University
HW5 Solutions
3.70
Let Y = # of holes drilled until a productive well is found.
a. P(Y = 3) = (.8)2(.2) = .128
b. P(Y > 10) = P(first 10 are not productive) = (.8)10 = .107.
3.71
a.
P (Y a )
q
y 1
pq
y a 1
b. From
STA 401/501, Sols to HW2
2.37
a There are 6 cities, we need to visit all of them in a certain order.
P66 = 6! = 720.
b There are following six types of itineraries.
D
4D
44D
4 4 4D
4444D
4 4 4 4 4D
where San Francisco can be assigned in , but not in 4.