ATMS 120: FA2015
Snodgrass
Name and NetID:_BENNY CHUI 661284516_
Seasons
On the map below, label the following city locations using the lat/lon information. Use Google to find
each city and use the letter for each city to identify its position on the map.
ATMS 120: FA2015
Snodgrass
Name and NetID: _Benny Chui bchui2_
Weekly Challenge Problem #8: Tornadoes!
Due Tuesday Nov. 3, 2015
Imagine a tornado passed through a residential neighborhood and its track went right down the middle of a street
that was orien
ATMS 120: FA2015
Snodgrass
Name and NetID: _Benny Chui bchui2_
Weekly Challenge Problem #7: Hail and Thunderstorm Wind Power
Due Tuesday Oct 27, 2015
Hail (Part #1): You must show all of your work!
In 2003, the second largest hailstone on record fell in N
HW 5 Solutions
Chapter 3
Problem 64 Let E be the event that the wife answers correctly, and let F be the
event that the husband answers correctly.
(a) If only one of them answers, then the probability of a correct
answer is P (E) = P (F ) = p.
(b) P (corr
Third Homework Set Solutions
Chapter 3
Problem 1 Let E be the event that at least one die lands on six, and let F be the
event that the dice land of different numbers. Then P (EF ) = 2 16 56 =
5
and P (F ) = 30
= 65 . Hence,
18
36
P (EF )
P (E|F ) =
=
P (
Homework 3 Solutions
Chapter 2
Problem 7 (a) There are 615 outcomes in the sample space.
(b) There are 315 outcomes without any blue-collar workers, so that
there are 615 315 outcomes with at least one blue-collar worker.
(c) If there are no independents,
HW 5 Solutions
Chapter 4
Problem 21 (a) E [X] is larger than E [Y ] because the random selection of students
favors larger busloads.
(b) E [X] =
4040+3333+2525+5050
40+33+25+50
=
5814
148
= 39.3, E [Y ] =
148
4
= 37.
Problem 23 (a) Suppose that that you u
Homework 6 Solutions
Chapter 4
Problem 55
P (no errors) =
=
=
P (no errors|first typist) P (first typist)
+P (no errors|second typist) P (second typist)
1 30 3 4.20 4.2
e +
e
2 0!
0!
1 3
e + e4.2 .
2
Problem 57 X is Poisson with parameter = 3.
(a) P cfw_X
Homework 8 Solutions
Chapter 5
Problem 37 Let X be uniformly distributed over (1, 1).
(a) P |X| > 12 = P X > 12 + P X < 12 =
1
2
(b) Let Y = |X|. If y (0, 1), then FY (y) = P cfw_Y y = P cfw_y Y y =
y, so that
(
1
0<y<1
fY (y) =
0
otherwise
Problem
Ninth Homework Set Solutions
Chapter 6
Problem 13 Let X be uniform on (15, 15), and let Y be uniform on (30, 30).
Nobody waits longer than five minutes if |Y X| < 5.
P cfw_|Y X| < 5 = P cfw_5 < Y X < 5
= P cfw_X 5 < Y < X + 5
Z 15 Z x+5
1
dydx
=
15 x5 30
Homework 1 Solutions
Chapter 1
Problem 4 If each of the boys can play any instrument, then there are 4! = 24
possible arrangements. If Jay and Jack can only play piano and drums,
then there are two ways to assign instruments to Jay and Jack, which
leaves
Homework Set April 3, 2015
Solutions
Graded problems are in red
Chapters 5 and 6
Problem 37 Let X be uniformly distributed over (1, 1).
(a) P |X| > 12 = P X > 12 + P X < 12 =
1
2
(b) Let Y = |X|. If y (0, 1), then FY (y) = P cfw_Y y = P cfw_y Y y =
ATMS 120: FA2015
Snodgrass
Name and NetID:_
Seasons
On the map below, label the following city locations using the lat/lon information. Use Google to find
each city and use the letter for each city to identify its position on the map.
City A = 40.0N, 88.0
Homework Set January 30, 2015
Solutions
Chapter 1
Graded problems are in red
Problem 2 Each roll of one die has 6 possible outcomes. Since there are 4 rolls,
there are 64 = 1, 296 possible outcome sequences.
Problem 4 If each of the boys can play any inst
Homework Set February 13, 2015
Solutions
Graded problems are in red
Chapter 2
Problem 43 (a) There are n! ways to arrange n people in a line. There are 2(n1)!
ways to arrange them such that A and B are next to each other.
Hence, the probability of A and B
Homework Set February 20, 2015
Solutions
Graded problems are in red
Chapter 3
Problem 56
P (new) =
m
X
pi P (new|the n-th coupon is of type i) =
i=1
m
X
pi (1pi )n1 .
i=1
2
1
p(1 p) = 2p(1 p)
(b) P (increase by one after three days) = 32 p2 (1 p) = 3p2 (1
Homework Set February 27, 2015
Solutions
Graded problems are in red
Chapter 4
Problem 21 (a) E cfw_X is larger than E cfw_Y because the random selection of students favors larger busloads.
(b) E cfw_X =
Problem 23
4040+3333+2525+5050
40+33+25+50
=
5814
1
Homework Set March 6, 2015
Solutions
Graded problems are in red
Chapter 4
Problem 55
P (no errors) =
=
=
P (no errors|first typist) P (first typist)
+P (no errors|second typist) P (second typist)
1 30 3 4.20 4.2
e +
e
2 0!
0!
1 3
e + e4.2 .
2
Problem 57 X
Homework Set March 20, 2015
Solutions
Graded problems are in red
Chapter 5
Problem 1 (a) We have 1 =
c=
R1
2
1
c(1 x )dx = cx 1
x2
3
|11 =
4
c,
3
so that
3
.
4
Rx
2
2
(b) We have 1 f (y)dy = 43 y 1 y3 |x1 = 21 + 43 x 1 x3 if 1
x 1. Hence,
x < 1,
0
2
F
Homework Set April 24, 2015
Solutions
Graded problems are in red
Chapters 6 and 7
Problem 48 Let X1 , . . . , X5 be independent exponential random variables with parameter .
(a)
P cfw_min(X1 , . . . , X5 ) a = 1 P cfw_min(X1 , . . . , X5 ) > a
= 1 P cfw_X
Math 461 Fall 2006 Test 2 Solutions
Total points: 100. Do all questions. Explain all answers. No notes,
books, or electronic devices.
1. [10=5+5 points] Assume X Exponential(). Justify the following two
formulas, by directly using the exponential density
Homework Set September 18, 2015
Solutions
Chapter 2
Problem 17 There are 64 63 62 61 60 59 58 57 ways of arranging 8 castles on a
chess board. Of these, there are 64 49 36 25 16 9 4 1 = 8 i2 in
i=1
which none of the rooks can capture any of the others. So