2022W2_MATH_101A_ALL_2022W2.QTPX6H182J03.WW11
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School
University of British Columbia *
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Course
101
Subject
Mathematics
Date
Apr 3, 2024
Type
Pages
9
Uploaded by PresidentPartridgeMaster810
Gyan Edbert Zesiro
2022W2
MATH
101A
ALL
2022W2
Assignment WW11 due 04/13/2023 at 11:59pm PDT
Problem 1.
(1 point)
This assignment is not for marks. It is intended to help you
practice topics appearing at the end of the course (more Tay-
lor series and Probabilty) as you study for the final exam.
Evaluate the given limit.
lim
x
→
0
log
(
1
-
x
)+
x
+
x
2
2
4
x
3
=
Answer(s) submitted:
•
-
1
12
submitted: (correct)
recorded: (correct)
Correct Answers:
• -
0
.
0833333333333333
Problem 2.
(1 point)
Decide if the following series converges. If it does, enter the ex-
act value for its sum (not a decimal approximation); if not, enter
either ”Diverges” or ”D”.
S
=
∞
∑
n
=
0
(
-
1
)
n
2
n
+
1
1
3
2
n
+
1
Answer:
Answer(s) submitted:
•
tan
-
1
1
3
submitted: (correct)
recorded: (correct)
Correct Answers:
•
tan
-
1
1
3
Problem 3.
(1 point)
Let
F
(
x
) =
Z
x
0
e
-
2
t
4
dt
.
(a)
Find the Maclaurin series for
F
(
x
)
. Then enter the polynomial
obtained by discarding all terms involving
x
n
for
n
>
9.
Answer:
T
9
(
x
) =
(b)
Find the indicated derivative:
F
(
29
)
(
0
) =
Answer(s) submitted:
•
x
-
2
5
x
5
+
2
9
x
9
•
(
-
2
)
7
(
29!
)
7!
·
29
submitted: (correct)
recorded: (correct)
Correct Answers:
•
x
-
2
x
5
5
+
2
2
x
9
18
•
(
-
1
)
7
·
2
7
((
4
·
7
)
!
)
7!
1
Problem 4.
(1 point)
Find the infinite series representation, centred at
x
=
0, of the im-
proper integral
f
(
x
) =
Z
x
0
sin
(
5
t
)
2
t
dt
.
Enter the first five non-zero terms, in order of increasing degree.
Answer:
f
(
x
) =
+
+
+
+
+
···
What is the radius of convergence?
Answer:
R
=
Answer(s) submitted:
•
5
x
2
•
-
5
3
x
3
2
·
(
3!
)
·
3
•
5
5
x
5
2
·
(
5!
)
·
5
•
-
5
7
x
7
2
·
(
7!
)
·
7
•
5
9
x
9
2
·
(
9!
)
·
9
•
∞
submitted: (correct)
recorded: (correct)
Correct Answers:
•
1
2
·
5
x
• -
1
2
1
3
·
(
3!
)
·
5
3
x
3
•
1
2
1
5
·
(
5!
)
·
5
5
x
5
• -
1
2
1
7
·
(
7!
)
·
5
7
x
7
•
1
2
1
9
·
(
9!
)
·
5
9
x
9
•
∞
Problem 5.
(1 point)
Let
f
(
x
) =
6
x
sin
(
2
x
)+
px
2
24
-
24cos
(
5
x
)
-
300
x
2
.
(a)
Find the one and only value of the constant
p
for which
lim
x
→
0
f
(
x
)
exists.
Answer:
p
=
(b)
Using the value of
p
found in part
(a)
, evaluate the limit.
Answer:
lim
x
→
0
f
(
x
) =
Answer(s) submitted:
• -
12
•
8
625
submitted: (correct)
recorded: (correct)
Correct Answers:
• -
6
·
2
•
2
3
5
4
Problem 6.
(1 point)
Evaluate the indicated limit.
lim
x
→
0
(
1
+
4
x
+
4
x
2
)
1
/
x
=
Answer(s) submitted:
•
e
4
submitted: (correct)
recorded: (correct)
Correct Answers:
•
exp
(
1
·
4
)
Problem 7.
(1 point)
Find the value of
C
so that the function
f
(
x
) =
(
0
if
x
<
0
Ce
-
2
x
if
x
≥
0
.
is a probability density function.
Answer(s) submitted:
•
2
submitted: (correct)
recorded: (correct)
Correct Answers:
•
2
2
Problem 8.
(1 point)
The probability density function for a certain random variable
X
is given by
p
(
x
) =
2
153
x
if 0
≤
x
≤
9
2
8
-
2
136
x
if 9
≤
x
≤
17
0
otherwise.
Find the probability that
X
is between 8 and 10.
Pr
(
8
≤
X
≤
10
) =
Answer(s) submitted:
•
1
9
+
1
4
-
19
136
submitted: (correct)
recorded: (correct)
Correct Answers:
•
0
.
22140522875817
Problem 9.
(1 point)
The probability density function for a certain random variable
X
is given by
p
(
x
) =
2
253
x
if 0
≤
x
≤
11
2
12
-
2
276
x
if 11
≤
x
≤
23
0
otherwise.
Find the expected value of
X
.
E
(
X
) =
Answer(s) submitted:
•
2
253
·
3
·
11
3
+
1
12
23
2
-
11
2
-
2
276
·
3
23
3
-
11
3
submitted: (correct)
recorded: (correct)
Correct Answers:
•
11
.
3333333333333
Problem 10.
(1 point)
Suppose
that,
after
measuring
the
duration
of
many
tele-
phone calls, a telephone company found their data was well-
approximated by the density function
p
(
x
) =
0
.
8
e
-
0
.
8
x
, where
x
is the duration of a call, in minutes.
(a)
What percentage of calls last between 2 and 3 minutes?
Percent =
percent
(b)
What percentage of calls last 2 minutes or less?
Percent =
percent
(c)
What percentage of calls last 4 minutes or more?
Percent =
percent
Solution:
SOLUTION
(a)
The fraction of calls lasting from 1 to 2 minutes is given by the
integral
Z
3
2
p
(
x
)
dx
=
Z
3
2
0
.
8
e
-
0
.
8
x
dx
=
e
-
1
.
6
-
e
-
2
.
4
≈
0
.
11118
,
or about 11.118 percent.
(b)
A similar calculation (changing the limits of integration) gives
the percentage of calls lasting 2 minutes or less as
Z
2
0
p
(
x
)
dx
=
Z
2
0
0
.
8
e
-
0
.
8
x
dx
=
1
-
e
-
1
.
6
≈
0
.
7981
,
or about 79.81 percent.
(c)
The percentage of calls lasting 4 minutes or more is given by
the improper integral
Z
∞
4
p
(
x
)
dx
=
lim
b
→
∞
Z
b
4
0
.
8
e
-
0
.
8
x
dx
=
lim
b
→
∞
(
e
-
3
.
2
-
e
-
0
.
8
b
) =
e
-
3
.
2
≈
0
.
04076
or about 4.076 percent.
Answer(s) submitted:
•
e
-
1
.
6
-
e
-
2
.
4
·
100
•
e
-
0
-
e
-
2
·
0
.
8
·
100
•
e
-
0
.
8
·
4
·
100
submitted: (correct)
recorded: (correct)
Correct Answers:
•
100
e
-
0
.
8
·
2
-
e
-
0
.
8
·
3
•
100 1
-
e
-
0
.
8
·
2
•
100
e
-
0
.
8
·
4
3
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Problem 11.
(1 point)
The probability density function for the duration of riders’
screams on a roller coaster is given by
f
(
x
) =
(
1
10
π
(
1
-
cos
(
8
x
))
if 0
≤
x
≤
10
π
0
otherwise.
Find the mean duration of riders’ screams over the course of the
ride.
The mean duration of riders’ screams is
seconds.
Answer(s) submitted:
•
5
π
submitted: (correct)
recorded: (correct)
Correct Answers:
•
15
.
707963267949
Problem 12.
(1 point)
The random variables
A
,
B
, and
C
have the probability density
functions (PDFs) shown below.
A=
B=
C=
(Click on graph to enlarge)
(a)
Which random variable has the largest expected value?
Enter A, B, or C:
(b)
Select the correct inequality:
•
A. Var
(
B
)
<
Var
(
C
)
•
B. Var
(
B
) =
Var
(
C
)
•
C. Var
(
B
)
>
Var
(
C
)
Answer(s) submitted:
•
B
•
A
submitted: (correct)
recorded: (correct)
Correct Answers:
•
B
•
A
4
Problem 13.
(1 point)
At a restaurant, the probability density function for
T
, the time a
customer has to wait before being seated, is given by
p
(
t
) =
(
0
if
t
<
0
λ
e
-
λ
t
if
t
≥
0
.
Here
t
is measured in minutes, and
λ
= (
1
/
8
)
min
-
1
.
(a)
Find the probability that a customer will have to wait at least
3 minutes to be seated.
Pr
(
T
≥
3
) =
.
(b)
What is the mean waiting time?
E
(
T
) =
.
(c)
What is the variance in waiting times?
Var
[
T
] =
.
Answer(s) submitted:
•
e
-
3
8
•
8
•
64
submitted: (correct)
recorded: (correct)
Correct Answers:
•
0
.
687289278790972
•
8
•
64
Problem 14.
(1 point)
Four random variables with values in the interval
[
0
,
1
]
, named
A
,
B
,
C
, and
D
, have the probability density functions (PDFs)
sketched below.
A
=
B
=
C
=
D
=
(a)
Which random variable has the smallest mean? [Enter A, B,
C, or D.]
(b)
Select the correct inequality.
•
A. Var
(
B
) =
Var
(
C
)
•
B. Var
(
B
)
<
Var
(
C
)
•
C. Var
(
B
)
>
Var
(
C
)
(c)
Select the correct inequality.
•
A. Var
(
A
)
>
Var
(
D
)
•
B. Var
(
A
) =
Var
(
D
)
•
C. Var
(
A
)
<
Var
(
D
)
Answer(s) submitted:
•
A
•
C
•
C
submitted: (correct)
recorded: (correct)
Correct Answers:
•
A
•
C
•
C
5
Problem 15.
(1 point)
Let
X
be a real-valued random variable with probability density
function
p
(
x
) =
x
2
27
√
2
π
e
-
1
2
·
(
x
3
)
2
a) Let
p
(
x
) =
a
0
+
a
1
x
+
a
2
x
2
+
...
be the Taylor explansion of
p
(
x
)
around
x
=
0. Find
a
0
=
(0.5 points.)
a
1
=
(0.5 points.)
a
2
=
(1 point.)
b) Use the Taylor approximation of degree 2 from part a) to ap-
proximate the probability Pr
(
0
≤
X
≤
3
)
.
Pr
(
0
≤
X
≤
3
)
≈
(1.5 points.)
Answer(s) submitted:
•
0
•
0
•
1
27
√
2
π
•
1
3
√
2
π
submitted: (correct)
recorded: (correct)
Correct Answers:
•
0
•
0
•
0
.
0147756400148679
•
0
.
132980760133811
Problem 16.
(1 point)
Consider a group of people who have received treatment for a dis-
ease such as cancer. Let
t
be the
survival time
, the number of
years a person lives after receiving treatment. The density func-
tion giving the distribution of
t
is
p
(
t
) =
Ce
-
Ct
for some positive
constant
C
.
(a)
The survival function,
S
(
t
)
, is the probability that a randomly
selected person survives for at least
t
years. Find a formula for
S
(
t
)
.
S
(
t
) =
(b)
Suppose that a patient has a 75 percent chance of surviving at
least 5 years. Find
C
.
C
=
(c)
Using the value of
C
you found in (b), find each of the follow-
ing:
•
the probability that the patient survives up to (that is, less than
or equal to) 4 years:
•
the mean survival time for patients with this survival function,
in years:
Solution:
SOLUTION
The cumulative distribution function
P
(
t
) =
R
t
0
p
(
x
)
dx
=
Area un-
der graph of density function
p
(
x
)
for 0
≤
x
≤
t
gives the fraction
of the population who survive
t
years or less after treatment. We
could also say that this is the fraction of the population who sur-
vive up to
t
years after treatment.
(a)
The probability that a randomly selected person survives for at
least
t
years is the probability that she or he lives
t
years or longer,
so
S
(
t
) =
Z
∞
t
p
(
x
)
dx
=
lim
b
→
∞
Z
b
t
Ce
-
Cx
dx
=
lim
b
→
∞
-
e
-
Cb
+
e
-
Ct
=
e
-
Ct
.
Note that this is the same as
S
(
t
) =
1
-
P
(
t
)
.
(b)
The probability of surviving at least 5 years is
S
(
5
) =
e
-
5
C
=
75
100
.
Thus
C
=
-
1
5
ln
(
0
.
75
)
.
(c)
The probability that a patient survives up to 4 years is just
P
(
4
) =
R
4
0
p
(
x
)
dx
=
e
-
Cx
4
0
≈
0
.
205582.
6
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The average survival time for patients with this survival function
is
Z
∞
0
x
·
p
(
x
)
dx
=
lim
b
→
∞
C
(
-
(
1
+
Cx
)
e
-
Cx
C
2
b
0
=
lim
b
→
∞
-
(
1
+
Cb
)
e
-
Cb
C
+
1
C
=
1
C
.
Thus the average survival time is
T
=
-
5
ln
(
0
.
75
)
≈
17
.
3803.
Answer(s) submitted:
•
e
-
Ct
•
ln
(
0
.
75
)
-
5
•
1
-
e
-
ln
(
0
.
75
)
-
5
·
4
•
1
ln
(
0
.
75
)
-
5
submitted: (correct)
recorded: (correct)
Correct Answers:
•
e
-
Ct
• -
1
5
ln
75
100
•
1
-
75
100
4
5
•
-
5
ln
75
100
Problem 17.
(1 point)
The probability density function for a certain random variable
X
is given by this expression involving a constant
C
:
p
(
x
) =
(
Cxe
-
9
x
,
for
x
≥
0
,
0
,
for
x
<
0
.
Find
C
and the expected value
E
(
X
)
.
Answers:
C
=
,
E
(
X
) =
.
Solution:
We
know
that
p
(
x
)
is
a
probability
density
function,
so
Z
∞
0
p
(
x
)
dx
=
1.
Integration by parts shows that
Z
∞
0
Cxe
-
9
x
dx
=
C
/
81. Matching
this with the required value 1 gives
C
=
81.
Further integration by parts reveals that
E
(
X
) =
Z
∞
0
Cx
2
e
-
9
x
dx
=
2
9
.
Answer(s) submitted:
•
81
•
2
9
submitted: (correct)
recorded: (correct)
Correct Answers:
•
9
2
•
2
9
7
Problem 18.
(1 point)
The probability density function for a certain random variable
X
is given by
p
(
x
) =
(
Cxe
-
kx
,
for
x
≥
0
,
0
,
for
x
<
0
.
Given Var
(
X
) =
1
2
, find the constants
C
and
k
.
Answers:
C
=
,
k
=
.
Solution:
There are three things to know:
1
=
Z
∞
0
p
(
x
)
dx
=
C
k
2
,
E
(
X
) =
Z
∞
0
xp
(
x
)
dx
=
2
C
k
3
,
E
(
X
2
) =
Z
∞
0
x
2
p
(
x
)
dx
=
6
C
k
4
.
The first fact implies
C
=
k
2
, which we can use when enforcing
1
2
=
Var
(
X
) =
E
(
X
2
)
-
(
E
(
X
))
2
=
6
C
k
4
-
4
C
2
k
6
=
2
k
2
.
This
gives
k
=
r
2
(
2
)
1
,
and
back-substitution
gives
C
=
k
2
=
2
(
2
)
1
.
Answer(s) submitted:
•
4
•
2
submitted: (correct)
recorded: (correct)
Correct Answers:
•
2
·
2
1
•
r
2
·
2
1
Problem 19.
(1 point)
Every day an experimentalist acquires a new measurement from
some random variable
X
. The measurement from Day
i
is
x
i
. On
Day
N
, the experimentalist uses all the data gathered so far to cal-
culate the following estimates for the expected value and variance
of
X
:
E
N
=
x
1
+
x
2
+
...
+
x
N
N
,
V
N
=
1
N
N
∑
i
=
1
(
x
i
-
E
N
)
2
.
On Day 27, the experimentalist observes
x
27
=
59
.
81 and looks
up
E
26
=
52
.
41 and
V
26
=
8
.
74. Find the updated estimates of the
expected value and variance.
Answers:
E
27
=
,
V
27
=
.
Notes:
The measurement
x
27
is clearly an outlier. A serious exper-
imentalist would deal with this in ways beyond the scope of this
question.
They would probably also use the “sample variance”
instead of the
V
N
defined here.
A preliminary observation that may help mimics one that was
shown in class: for every
N
,
V
N
=
1
N
N
∑
i
=
1
(
x
i
-
E
N
)
2
=
···
=
1
N
N
∑
i
=
1
x
2
i
!
-
E
2
N
.
Solution:
This is a simple algebra problem wrapped up in ex-
perimental terminology. Rearranging the given definitions gives
N
∑
i
=
1
x
i
=
NE
N
,
N
∑
i
=
1
x
2
i
=
N
(
V
N
+
E
2
N
)
.
This works for every
N
. That makes it possible to find the recur-
rence identities
N
+
1
∑
i
=
1
x
i
=
NE
N
+
x
N
+
1
,
N
+
1
∑
i
=
1
x
2
i
=
N
(
V
N
+
E
2
N
)+
x
2
N
+
1
.
The problem setup gives
N
,
E
N
,
V
N
, and
x
N
+
1
, so we have enough
to finish.
Answer(s) submitted:
•
52
.
41
+
59
.
81
-
52
.
41
27
•
(
8
.
74
+
52
.
41
2
)
·
26
+
59
.
81
2
27
-
52
.
41
+
59
.
81
-
52
.
41
27
2
submitted: (correct)
recorded: (correct)
Correct Answers:
•
26
·
52
.
41
+
59
.
81
27
•
26
·
(
8
.
74
+
52
.
41
2
)
+
59
.
81
2
27
-
26
·
52
.
41
+
59
.
81
27
2
8
Problem 20.
(1 point)
Background:
The probability density function for a normally dis-
tributed random variable with mean
μ
and variance
σ
2
is
p
(
x
) =
1
σ
√
2
π
e
-
(
x
-
μ
)
2
2
σ
2
,
x
∈
R
.
The special choices
μ
=
0 and
σ
2
=
1
2
are used to define the
error
function
, a standard tool in spreadsheets and scientific packages:
erf
(
x
) =
2
√
π
Z
x
0
e
-
t
2
dt
.
So for non-negative
x
, erf
(
x
)
gives the probability that such a ran-
dom variable falls in the interval
[
-
x
,
x
]
.
Problem:
Suppose
*
that the height of Canadian women is a nor-
mally distributed random variable with mean
μ
=
162
.
9cm and
variance
σ
2
=
23
.
9cm
2
.
Find positive numbers
a
and
b
that satisfy both of the following
conditions:
•
1
2
[
erf
(
a
)
-
erf
(
b
)]
equals the probability that a randomly se-
lected Canadian woman is between 153cm and 158cm tall.
•
1
2
[
1
-
erf
(
a
)]
equals the probability that a randomly selected
Canadian woman is not more than 153cm tall.
a
=
,
b
=
.
*
These values for
μ
and
σ
2
may not be reliable outside this prob-
lem.
Solution:
For a random variable
X
whose probability density is
given in the question and any real numbers
α
≤
β
, we have
Pr
(
α
≤
X
≤
β
) =
Z
β
α
1
σ
√
2
π
e
-
(
x
-
μ
)
2
2
σ
2
dx
.
The form of the error function suggests the substitution
t
=
x
-
μ
σ
√
2
.
This calls for
dx
=
σ
√
2
dt
. Keeping the limits of integration syn-
chronized with the integration variable gives
Pr
(
α
≤
X
≤
β
) =
1
σ
√
2
π
Z
(
β
-
μ
)
/
(
σ
√
2
)
(
α
-
μ
)
/
(
σ
√
2
)
e
-
t
2
(
σ
√
2
dt
)
=
1
√
π
Z
(
β
-
μ
)
/
(
σ
√
2
)
(
α
-
μ
)
/
(
σ
√
2
)
e
-
t
2
dt
=
1
2
erf
β
-
μ
σ
√
2
-
1
2
erf
α
-
μ
σ
√
2
.
In our problem we have
α
<
β
<
μ
, so both inputs to the error
function above are negative. It’s helpful to note that
the function
erf is odd
(exercise: prove this!), so we can also write
Pr
(
α
≤
X
≤
β
) =
1
2
erf
μ
-
α
σ
√
2
-
1
2
erf
μ
-
β
σ
√
2
.
The results for our specific situation involve
α
=
153 and
β
=
158,
along with
μ
=
162
.
9 and
σ
2
=
23
.
9, leading to
a
=
μ
-
α
√
2
σ
2
=
1
.
43192821399742
,
b
=
μ
-
β
√
2
σ
2
=
0
.
708732146321956
.
(In the limit as
α
→ -
∞
, the fact that erf
(
z
)
→
1 as
z
→
+
∞
gives
Pr
(
X
≤
β
) =
1
2
(
1
)
-
1
2
erf
μ
-
β
σ
√
2
. Thus the second stated
requirement in the question is compatible with the development
shown above.)
Answer(s) submitted:
•
-
(
153
-
162
.
9
)
√
2
·
23
.
9
•
-
(
158
-
162
.
9
)
√
2
·
23
.
9
submitted: (correct)
recorded: (correct)
Correct Answers:
•
1
.
43193
•
0
.
708732
Generated by ©WeBWorK, http://webwork.maa.org, Mathematical Association of America
9
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