Sample_Prelim_Exam2_Solutions1
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Binghamton University *
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Course
147
Subject
Mathematics
Date
Jan 9, 2024
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Pages
7
Uploaded by GeneralOctopus3438
Math 147 B - Sample Prelim Exam 2, Fall 2023
Honor Code:
I have neither given nor received aid on this exam. I understand that violation
of this statement or the Binghamton code of academic integrity is punished at minimum with
failure of the course.
Printed Name:
Signature:
Student ID:
Instructions:
You may bring in a single sheet of paper with formulas etc., and a basic
calculator. No cell phones, computers or other devices with a communication capability.
The exam consists of
x
sides (?? sheets), which includes this side of instructions, ?? sides
of questions, ??
sides with the standard Normal tables.
For numeric answers give either
decimals with three significant figures or simplified fractions unless otherwise specified in the
question. Hand in all pages of the exam, but not your formula sheet or scrap paper.
Question
Points
Out of
1
20
2
20
3
20
4
20
5
20
1
1. An airline offers discounted “advance-purchase” fares to customers who buy tickets
more than 30 days before travel and charges “regular” fares for tickets purchased during
those last 30 days. The company has noticed that 60% of its customers take advantage
of the advance-purchase fares.
The “no-show” rate among people who paid regular
fares is 30%, but only 5% of customers with advance-purchase tickets are no-shows.
i) What percent of all ticket holders are no-shows?
ii) What is the probability that a customer who didn’t show had an advance-purchase
ticket?
iii) Is being a no-show independent of the type of ticket a passenger holds? Explain.
i) To find this, we compute a tree diagram:
We can conclude that
P
(no show) =
P
(regular and no show) +
P
(Advance and
no show)=0
.
4
·
0
.
3 + 0
.
6
·
0
.
05 = 0
.
15 = 15% of all ticket holders are no-shows.
ii) This is just
P
(advance purchase
|
no show) =
P
(advance purchase and no show)
/P
(
no show) = 0
.
03
/
(0
.
15) = 0
.
20
iii) No, because
P
(advance purchase
|
no show) = 0
.
2
̸
= 0
.
6 =
P
(advance purchase),
which is a necessary condition for independence of these two.
P
(advance purchase
|
no show) =
P
(advance purchase) is false, but it must hold
for independence
2. The price of a particular asset today is 100. There is a 50/50 chance that its price the
next day will either go up or down by 10.
i) Create a probability model for the price of the asset in two days.
ii) Find the expected price of the model in two days.
iii) Do you expect to gain, lose, or stay the same, on average?
iv) Find the standard deviation of the price of the model in two days.
i)
x
P(X=x)
120
.25
80
.25
100
.5
2
ii)
E
[
X
] = 120(
.
25) + 80(
.
25) + 100(
.
5) = 100
iii) We expect the price to stay the same on average
iv)
V
[
X
] = (120
−
100)
2
(
.
25) + (80
−
100)
2
(
.
25) + (100
−
100)
2
(
.
5) = 200
SD
(
X
) =
√
200 = 14
.
14
3. Neurological research has shown that in about 80% of people, language abilities reside
in the brains left side.
Another 10% display right-brain language centers, and the
remaining 10% have two-sided language control.
(The latter two groups are mainly
left-handers; Science News, 161 no. 24 [2002].)
i) In a randomly assigned group of 5 of these students, what’s the probability that
no one has two-sided language control?
ii) In the entire freshman class of 1200 students, how many would you expect to find
of each type?
iii) What are the mean and standard deviation of the number of these freshmen who
might be left-brained in language abilities?
iv) In the entire freshman class of 1200 students, what is the probability that no more
than 720 of them have left-brain language control?
Justify and use a Normal
model.
i) Let
T
be the number of people with two-sided language control from
n
= 5 people.
T
follows a Binom(5
,
0
.
10) distribution and, hence
P
(none have two-sided language control)
=
P
(
T
= 0)
=
5
C
0(0
.
1)
0
(0
.
9)
5
=
(0
.
9)
5
≈
0
.
590
ii) Use Binomial models:
E
(left) =
np
L
= 1200(0
.
80) = 960 people
E
(right) =
np
R
= 1200(0
.
10) = 120 people
E
(two-sided) =
np
T
= 1200(0
.
10) = 120 people.
iii) Let
L
be the number of people with left-brain language control.
E
(
L
) =
np
L
= 1200(0
.
80) = 960 people and
SD
(
L
) =
p
np
L
(
q
L
) =
p
1200(0
.
80)(0
.
20) = 13
.
86 people.
iv) Since
np
L
= 960 and
nq
L
= 240 are both greater than 10, the Normal model,
N(960
,
13
.
86), may be used to approximate Binom(1200
,
0
.
80).
We are looking
for
P
(
L
≤
720)
=
P
Z
≤
720
−
960
13
.
86
=
P
Z
≤
720
−
960
13
.
86
=
P
(
Z
≤ −
17
.
32)
≈
0
3
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4. The Centers for Disease Control and Prevention say that about 18% of high-school
students smoke tobacco (down from a high of 38% in 1997). Suppose you randomly
select high-school students to survey them on their attitudes toward scenes of smoking
in the movies. What is the probability that
i) none of the first 4 students you interview is a smoker?
ii) the first smoker is the sixth person you choose?
iii) there are no more than 2 smokers among 10 people you choose?
Note:
Randomly selecting high school students can be considered Bernoulli trials.
There are only two possible outcomes, smoker or nonsmoker. The probability that a
student is a smoker is p = 0.18 =
>
non-smoker
q
= 1
−
p
= 1
−
0
.
18 = 0
.
82. The trials
are not independent, since the population is finite, but we are not sampling more than
10% of all high school students.
i) P(none of the first 4 are smokers) = (0
.
82)
4
≈
0
.
452
ii) Use Geom(0.18).
P(first smoker is the sixth person) = (0
.
82)
5
×
(0
.
18)
≈
0
.
067
iii) Use Binom(10, 0.18). Let X = the number of smokers among n = 10 students.
P
(no more than 2 smokers of 10)
=
P
(
X
≤
2)
=
P
(
X
= 0) +
P
(
X
= 1) +
P
(
X
= 2)
=
10
C
0
(0
.
18)
0
(0
.
82)
10
+
10
C
1
(0
.
18)
1
(0
.
82)
9
+
10
C
2
(0
.
18)
2
(0
.
82)
8
≈
0
.
737
5. In June 2017, Pew Research asked a random sample of 2504 U.S. adults, “Do you
strongly favor, favor, oppose, or strongly oppose allowing gays and lesbians to marry
legally?” (www.people-press.org/2017/06/26/ support-for-same-sex-marriage-grows-even-
among-groups-that -had-been-skeptical/). Of those polled, 62% said they favored mar-
riage equality, the highest percentage to date.
i) Create a 95% confidence interval for the percentage of all U.S. adults who support
marriage equality.
ii) Based on your confidence interval, can you conclude that a majority of U.S. adults
support legally recognizing gay marriages? Explain.
iii) If pollsters wanted to follow up on this poll with another survey that could de-
termine the level of support for gay marriage to within 2% with 98% confidence,
how many people should they poll?
i)
Randomization condition:
Pew Research randomly selected 2504 U.S. adults.
10% condition:
2504 results is less than 10% of all U.S. adults.
Success/Failure Condition:
n
ˆ
p
= (2504)(0
.
62) = 1552
.
5 and
n
ˆ
q
= (2504)(0
.
38) =
951
.
5 are both greater than 10 so the sample is large enough.
4
Since the conditions are met, we can use a one-proportion z-interval to estimate
the percentage of U.S. adults who support marriage equality. Since we are finding
a 95% CI, we find the Z-score that corresponds to p=0.975 on our Z-table. This
gives us
z
∗
= 1
.
96
ˆ
p
±
z
∗
q
ˆ
p
ˆ
q
n
= (0
.
62)
±
1
.
96
q
(0
.
62)(0
.
38)
2504
= (
.
601
, .
639).
We are 95% confident that between 66.1% and 63.9% of U.S. adults support
marriage equality.
ii) Since the interval is entirely above 50%, there is evidence that a majority of U.S.
adults support marriage equality.
iii) Find p=.99 on our Z-table to get
z
∗
= 2
.
33. We do not know the true proportion
of U.S. adults who support marriage equality, so use ˆ
p
= ˆ
q
= 0
.
50, for the most
cautious estimate.
ME
=
z
∗
q
ˆ
p
ˆ
q
n
0
.
02 = 2
.
33
q
(0
.
5)(0
.
5)
n
n
≈
3394 people.
5
6
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7