Homework 3 Solutions-2 (1)
pdf
keyboard_arrow_up
School
Iowa State University *
*We aren’t endorsed by this school
Course
305
Subject
Statistics
Date
Feb 20, 2024
Type
Pages
7
Uploaded by EarlCrown13355
Stat 305 Homework 3
Due: March 12th, 2021 by 11:59pm
Show all of your work, and upload this homework to Canvas.
This homework is worth 116 points total
.
Lecture Notes 19 - 20 Material
1. Textbook Problems:
Section 3.1:
4, 7
Section 3.2:
12, 14, 18
,
Section 3.3:
29, 32
Answer:
3.1.4
(3 points)
Im
(
X
) =
{
0
,
1
,
2
,
3
,
4
}
Answers will vary for outcomes, but should be like the following:
Outcome = 1065
→ {
X
= 3
}
, Outcome = 0700
→ {
X
= 1
}
, Outcome = 3681
→ {
X
= 0
}
3.1.7
Not Graded, see solutions in textbook
3.2.12
(5 points)
(a)
P
(
Y
≤
50) =
.
83
(b) This is the compliment of part a)
→
1
-
.
83 =
.
17
(c) 1st standby: you get a seat if less 49 or fewer ticketed passengers show up.
P
(
Y
≤
49) =
.
66
3rd standby: you get a seat if less 47 or fewer ticketed passengers show up.
P
(
Y
≤
47) =
.
27.
3.2.14
(8 points)
p
Y
(
y
) =
k
·
y, y
= 1
,
2
, . . . ,
5
Now,
∑
5
y
=1
p
Y
(
y
)
= 1
⇒
k
∑
5
y
=1
y
= 1
⇒
k
=
1
∑
5
y
=1
y
=
1
1+2+3+4+5
=
1
15
b)
P
(
Y
≤
3) = 1
-
p
Y
(4)
-
p
Y
(5) = 1
-
4
15
-
5
15
= 6
/
15 = 2
/
5
1
Stat 305 Homework 3
Due: March 12th, 2021 by 11:59pm
c)
P
(2
≤
Y
≤
4) =
p
Y
(2) +
p
Y
(3) +
p
Y
(4) = 2
/
15 + 3
/
15 + 4
/
15 = 9
/
15 = 3
/
5
d) Let
p
Y
(
y
) =
y
2
/
50. Now,
5
X
y
=1
p
(
y
) = 1
/
50
(
1
2
+ 2
2
+ 3
2
+ 4
2
+ 5
2
)
=
55
50
6
= 1
So, it is not a pmf.
3.2.18
(5 points)
M
= maximum of two tosses
(a)
P
(
M
= 1) = 1
/
36 (namely
M
(1
,
1) = 1)
P
(
M
= 2) = 3
/
36
P
(
M
= 4) = 7
/
36
P
(
M
= 6) = 11
/
36
P
(
M
= 3) = 5
/
36
P
(
M
= 5) = 9
/
36
PMF of
M
is
P
(
M
=
k
) =
2
K
-
1
36
, k
= 1
,
2
,
3
,
4
,
5
,
6 (Or you can present the PMF in table form like we
have done)
(b)
P
(
M
≤
y
) =
0
y <
1
1
/
36
1
≤
y <
2
4
/
36
2
≤
y <
3
9
/
36
3
≤
y <
4
16
/
36
4
≤
y <
5
25
/
36
5
≤
y <
6
1
y
≥
6
3.3.29
Not Graded. See textbook for solutions
2
Stat 305 Homework 3
Due: March 12th, 2021 by 11:59pm
3.3.32
(9 points)
(a)
E
(
X
) =
X
x
xp
X
(
x
) = 16
·
0
.
2 + 18
·
0
.
5 + 20
·
0
.
3 = 18
.
2
E
(
X
2
) =
X
x
x
2
p
X
(
x
) = 16
2
·
0
.
2 + 18
2
·
0
.
5 + 20
2
·
0
.
3 = 333
.
2
V ar
(
X
) =
E
(
X
2
)
-
[
E
(
X
)]
2
= 333
.
2
-
18
.
2
2
= 1
.
96
(b)
E
(70
X
-
650) = 70
E
(
X
)
-
650 = 70
·
18
.
2
-
650 = 624
(c)
V ar
(70
X
-
650) = 70
2
V ar
(
X
) = 4900
·
1
.
96 = 9604
(d)
E
(
h
(
X
)) =
E
(
X
-
0
.
008
X
2
) =
E
(
X
)
-
0
.
008
E
(
X
2
) = 18
.
2
-
0
.
008
·
333
.
2 = 15
.
5344
2. (10 points total) A box contains seven marbles. Four of them are red and three of them are green. You
reach in and choose three at random without replacement.
Define a random variable
X
as:
X
= the
number of red marbles selected.
(a) (2 pts) What are the possible values
X
can take on? (i.e. give
Im
(
X
))
Answer:
Im
(
X
) =
{
0
,
1
,
2
,
3
}
(b) (4pts) Find
P
(
X
=
x
) for all
x
in
Im
(
X
).
Answer:
P
(
X
= 0) =
(
4
0
)
·
(
3
3
)
(
7
3
)
=
1
35
P
(
X
= 1) =
(
4
1
)
·
(
3
2
)
(
7
3
)
=
12
35
P
(
X
= 2) =
(
4
2
)
·
(
3
1
)
(
7
3
)
=
18
35
P
(
X
= 3) =
(
4
3
)
·
(
3
0
)
(
7
3
)
=
4
35
(c) (4 pts) Make a table for the probability distribution of
X
as shown in lecture. (Leave probabilities
as fractions)
Answer:
x
0
1
2
3
p
X
(
x
)
1
35
12
35
18
35
4
35
3. (13 points total) Let
X
be a random variable with image
Im
(
X
) =
{-
2
,
-
1
,
0
,
1
,
2
}
.
(a) (1 pt) Fill in the blank in the table below to make it a valid probability mass function:
x
-
2
-
1
0
1
2
p
X
(
x
)
0
.
1
0
.
3
0
.
3
0
.
1
(b) (2 pts) Add the cumulative distribution function,
F
X
(
x
) to the table.
(c) (3 pts) Using
p
X
(
x
), determine the probabilities that...
i.
X
is at least 1.
3
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Stat 305 Homework 3
Due: March 12th, 2021 by 11:59pm
ii.
X
is greater than -1 and at most 1
iii.
X
is a negative value
(d) (3 pts) Using
F
X
(
x
), find...
i.
F
X
(1)
ii.
F
X
(
.
5)
iii.
P
(
X
≥
0) (
rewrite this first in terms of
F
X
(
x
))
(e) (4 pts) Find the expected value and variance of
X
.
Answer:
(a) Since the sum of the probabilities has to be 1 for a probability mass function,
p
X
(2) = 1
-
0
.
1
-
0
.
3
-
0
.
3
-
0
.
1 = 0
.
2.
(b)
x
-
2
-
1
0
1
2
p
X
(
x
)
0
.
1
0
.
3
0
.
3
0
.
1
0
.
2
F
X
(
x
)
0
.
1
0
.
4
0
.
7
0
.
8
1
(c)
i.
P
(
X
≥
1) =
P
(
X
= 1) +
P
(
X
= 2) = 0
.
1 + 0
.
2 = 0
.
3
ii.
P
(
-
1
< X
≤
1) =
P
(
X
= 0) +
P
(
X
= 1) = 0
.
3 + 0
.
1 = 0
.
4
iii.
P
(
X
≤ -
1) =
P
(
X
=
-
1) +
P
(
X
=
-
2) = 0
.
3 + 0
.
1 = 0
.
4
(d)
i.
F
X
(1) = 0
.
8
ii.
F
X
(
.
5) =
P
(
X
≤
.
5) =
P
(
X
≤
0) =
F
X
(0) =
.
7
iii.
P
(
X
≥
0) = 1
-
P
(
X <
0) = 1
-
P
(
X
≤ -
1) = 1
-
F
X
(
-
1) = 1
-
.
4 =
.
6
(e)
E
[
X
] = (
-
2)
P
(
X
=
-
2) + (
-
1)
P
(
X
=
-
1) + (0)
P
(
X
= 0) + (1)
P
(
X
= 1) + (2)
P
(
X
= 2)
= (
-
2)(0
.
1) + (
-
1)(0
.
3) + (0)(0
.
3) + (1)(0
.
1) + (2)(0
.
2) = 0
To find the variance, we will use the formula
V ar
[
X
] =
E
[
X
2
]
-
(
E
[
X
])
2
.
E
[
X
2
] = (
-
2)
2
P
(
X
=
-
2) + (
-
1)
2
P
(
X
=
-
1) + (0)
2
P
(
X
= 0) + (1)
2
P
(
X
= 1) + (2)
2
P
(
X
= 2)
= (
-
2)
2
(0
.
1) + (
-
1)
2
(0
.
3) + (0)
2
(0
.
3) + (1)
2
(0
.
1) + (2)
2
(0
.
2) = 1
.
6
V ar
[
X
] =
E
[
X
2
]
-
(
E
[
X
])
2
= 1
.
6
-
(0)
2
= 1
.
6.
4
Stat 305 Homework 3
Due: March 12th, 2021 by 11:59pm
Lecture Notes 21 - 24 Material
4. Textbook Problems:
Section 3.4:
48
Section 3.6:
82
(Note: the textbook uses
μ
instead of
λ
for the poisson rate parameter)
Answer:
3.4.48
(12 points)
(a)
P
(
X
≤
3) = 0
.
966
,
P
(
X <
3) =
P
(
X
≤
2) = 0
.
873
(b)
P
(
X
≥
4) = 1
-
P
(
X
≤
3) = 1
-
0
.
966 = 0
.
034
(c)
P
(1
≤
X
≤
3)
-
P
(
X
≤
3)
-
P
(
X <
1) =
P
(
X
≤
3)
-
P
(
X
≤
0) = 0
.
966
-
0
.
277 = 0
.
689
(d)
E
(
X
) =
np
= 25
·
0
.
05 = 1
.
25
And
V ar
(
X
) =
np
(1
-
p
) = 25
·
0
.
05
·
0
.
95 = 1
.
1875
So
σ
X
=
√
1
.
1875 = 1
.
0897
(e) Let
X
be the number in the sample who have a food allergy, then
X
∼
Bin(50
,
0
.
05), so
P
(No food allergy) =
P
(
X
= 0) = 0
.
05
0
0
.
96
50
= 0
.
0769
3.6.82
(6 points)
(a)
X
∼
Pois(0
.
2), so
P
(
X
= 1) =
e
-
0
.
2
0
.
2
1
1!
= 0
.
1637
(b)
P
(
X
≥
2) = 1
-
P
(
X <
2) = 1
-
P
(
X
≤
1) = 1
-
0
.
982 = 0
.
018
(c)
P
(No missing pulse) =
P
(
X
= 0) =
e
-
0
.
2
0
.
2
0
0!
= 0
.
8187
So
P
(Neither contains a missing pulse) = 0
.
8187
2
= 0
.
6703
5. (10 points total) In the board game
Monopoly
, a player in jail gets out by rolling “doubles” on their
turn, i.e., by rolling the same number on the two dice thrown. Since the only outcome that allows the
player to move his piece is doubles, the player will be less interested in the sum of the two dice and more
interested in whether or not the he or she rolled doubles.
Let
X
be the outcome of a single roll of the two dice, with “success” considered to be rolling doubles
and “failure” rolling anything else. Then
X
∼
Bern(
p
).
(a) (1 pt) What is
p
?
(b) (2 pts) What is
E
(
X
) and
V ar
(
X
)?
5
Stat 305 Homework 3
Due: March 12th, 2021 by 11:59pm
Now, suppose we roll two dice 5 times. Define
Y
to be the number of times doubles are rolled in the 5
trials.
(c) (3 pts) What is the distribution of
Y
? Give its name and parameter values.
(d) (2 pts) What is
E
(
Y
) and
V ar
(
Y
)?
(e) (2 pts) Find
P
(
Y
≤
2)
Answer:
(a) Using the probability under equally likely outcomes, there are 6 doubles outcomes in 36 possibilities,
so
p
=
1
6
.
(b) The expected value is
p
=
1
6
and variance is
p
(1
-
p
) =
5
36
(c)
Y
∼
Bin(
n, p
), where
n
= 5 and
p
=
1
6
(d)
E
(
Y
) =
np
=
5
6
, V ar
(
Y
) =
np
(1
-
p
) = 5(
1
6
)(
5
6
) =
.
694
(e)
P
(
Y
≤
2) =
P
(
Y
= 0) +
P
(
Y
= 1) +
P
(
Y
= 2) =
.
965
6. (10 points total) Suppose that jobs are sent to a printer at an average rate of 10 per hour.
(a) (2 pts) Let
X
= the number of jobs sent in an hour. What is the distribution of
X
? Give the name
and parameter values.
(b) (2 pts) What is the probability that exactly 8 jobs are sent to the printer in an hour?
(c) (2 pts) Let
X
= the number of jobs sent in a 12 min period. What is the distribution of
X
? Give
the name and parameter values.
(d) (2 pts) What is the probability that at least 3 jobs will be sent to the printer in a 12 min period?
(e) (2 pts) How many jobs do you expect to be sent to the printer in a 12 min period?
Answer:
(a)
X
∼
pois(10)
(b)
P
(
X
= 8) =
e
-
10
10
8
8!
=
.
113
(c) 12 mins is 1/5 of an hour, so if we get 10 per hour on average, in 12 mins we should expect 10/5 =
2 jobs. Thus
X
∼
pois(2)
(d)
P
(
X
≥
3) = 1
-
P
(
X
≤
2) = 1
-
.
677 =
.
323 (Using the poisson cdf table with
λ
= 2 and x = 2)
(e)
E
(
X
) =
λ
= 2 (Where
X
∼
pois(2)).
7. (13 points total) Consider the following joint distribution for the weather in two consecutive days. Let
X
and
Y
be the random variables for the weather in the first and the second days, with the weather
coded as 0 for sunny, 1 for cloudy, and 2 for rainy.
Y
X
0
1
2
0
0
.
3
0
.
1
0
.
1
1
0
.
2
0
.
1
0
2
0
.
1
0
.
1
0
(a) (4 pts) Find the marginal probability mass functions for
X
and
Y
.
(b) (4 pts) Calculate the expectation and variance for
X
and
Y
.
(c) (4 pts) Calculate the covariance and correlation between
X
and
Y
.
How strong does the linear
relationship seem?
(d) (1 pts) Are the weather in two consecutive days independent?
Answer:
6
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Stat 305 Homework 3
Due: March 12th, 2021 by 11:59pm
(a) The marginal distributions for X and Y are
x
0
1
2
p
X
(
x
)
0
.
5
0
.
3
0
.
2
y
0
1
2
p
Y
(
y
)
0
.
6
0
.
3
0
.
1
(b) The expectation and variance are
E
(
X
) = (0)(0
.
5) + (1)(0
.
3) + (2)(0
.
2) = 0
.
7
E
(
X
2
) = (0)
2
(0
.
5) + (1)
2
(0
.
3) + (2)
2
(0
.
2) = 1
.
1
V ar
(
X
) =
E
(
X
2
)
-
[
E
(
X
)
2
] = 1
.
1
-
0
.
7
2
= 0
.
61
E
(
Y
) = (0)(0
.
6) + (1)(0
.
3) + (2)(0
.
1) = 0
.
5
E
(
Y
2
) = (0)
2
(0
.
6) + (1)
2
(0
.
3) + (2)
2
(0
.
1) = 0
.
7
V ar
(
Y
) =
E
(
Y
2
)
-
[
E
(
Y
)]
2
= 0
.
7
-
0
.
5
2
= 0
.
45
(c) We have:
Cov
(
X, Y
) =
E
(
XY
)
-
E
(
X
)
E
(
Y
)
E
(
XY
) = (0)(0)(0
.
3) + (0)(1)(0
.
5) + (0)(2)(0
.
1)
+ (1)(0)(0
.
2) + (1)(1)(0
.
1) + (1)(2)(0)
+ (2)(0)(0
.
1) + (2)(1)(0
.
1) + (2)(2)(0)
= 0
.
3
Cov
(
X, Y
) =
.
3
-
(
.
7)(
.
5) =
-
0
.
05
ρ
=
Corr
(
X, Y
) =
Cov
(
X, Y
)
p
V ar
(
X
)
V ar
(
Y
)
=
-
0
.
05
p
(0
.
61)(0
.
45)
=
-
0
.
095
X
and
Y
have a weak, negative linear relationship
(d)
X
and
Y
are not independent since
Cov
(
X, Y
) =
-
0
.
05
6
= 0,
or
X
and
Y
are not independent since
0 =
P
(
X
= 2
, Y
= 2)
6
=
P
(
X
= 2)
P
(
Y
= 2) = 0
.
2
×
0
.
1 = 0
.
02
8. (12 points total) Using the joint distribution table given in problem 7, calculate the following probabilities:
(a) (2 pts)
P
(
X
=
Y
)
(b) (2 pts)
P
(
X < Y
)
(c) (2 pts)
P
(
X > Y
)
(d) (2 pts)
P
(
X
= 2
|
Y
= 1) (Just use the definition of conditional probability)
(e) (2 pts) Probability that the weather is sunny on two consecutive days.
(f) (2 pts) Probability that the weather is cloudy on the first day, and rainy on the second day.
Answer:
(a)
P
(
X
=
Y
) = 0
.
3 + 0
.
1 + 0 = 0
.
4
(b)
P
(
X < Y
) = 0
.
1 + 0
.
1 = 0
.
2
(c)
P
(
X > Y
) = 0
.
2 + 0
.
1 + 0
.
1 + 0 = 0
.
4
(d)
P
(
X
= 2
|
Y
= 1) =
.
1
.
3
=
1
3
(e)
P
(
X
= 0
, Y
= 0) = 0
.
3
(f)
P
(
X
= 1
, Y
= 2) = 0
7
Related Documents
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage
Intermediate Algebra
Algebra
ISBN:9781285195728
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Cengage Learning
Algebra: Structure And Method, Book 1
Algebra
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:McDougal Littell
Recommended textbooks for you
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageIntermediate AlgebraAlgebraISBN:9781285195728Author:Jerome E. Kaufmann, Karen L. SchwittersPublisher:Cengage LearningAlgebra: Structure And Method, Book 1AlgebraISBN:9780395977224Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. ColePublisher:McDougal Littell
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage
Intermediate Algebra
Algebra
ISBN:9781285195728
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Cengage Learning
Algebra: Structure And Method, Book 1
Algebra
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:McDougal Littell