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Midterm Exam
Fall, 2022
Date:
October 15, at 15:00-16:30
Instructions to Students:
All rights reserved. No part of this document
may be reproduced, stored in a retrieval system or transmitted in any form
or by any means, electronic, mechanical, photocopying, recording or otherwise
without prior written permission from all contributing instructors.
1. Do not turn over cover page until instructed.
2. Ink in your name, signature, and Section in the spaces below.
3. Once the exam begins, verify that your exam copy is complete.
Alert
Prof if not.
4. Answer all questions on your scantron sheet, using the exam booklet to
show your work.
5. Formula sheet provided (back of booklet).
6. Books and notes are
not
permitted.
7.
No
Communication Devices allowed on your person or near your desk.
8. Graphing calculators, iPads/Pods/iWatches, notebooks, computers are
not
allowed.
Duration of examination is:
1.5 hours
Statement of Academic Integrity
The Telfer School of Management does
not condone academic fraud. Without limiting the generality of this definition,
academic fraud occurs when a student commits any of the following offences:
plagiarism or cheating of any kind; use of books, notes, mathematical tables,
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Restaurant Service Satisfaction (2 Questions)
In order to improve the quality of the service, a restaurant manager regularly conducts surveys of its
customers. Respondents are asked to assess food quality, service, and price. The responses are (1)
Excellent, (2) Good, and (3) Fair. Surveyed customers are also asked whether they would come back. After
analyzing the responses, an expert in probability determined that 87% of customers say they will return.
Of those who so indicate, 57% rate the restaurant as excellent, 36% rate it as good, and the remainder rate
it as fair. Of those who say they won't return, the probabilities are 14%, 32%, and 54%, respectively.
1. What proportion of customers rate the restaurant as good?
a) 0.754
b) 0.654
c) 0.119
d) 0.355
e) 0. 522
2. If a customer rates the service as "Fair", what is the probability he/she will return?
a) 0.221
b) 0.145
c) 0.466
d) 0.802
e) 0. 396
Theory Questions (4 Questions)
3. Which of the following best describes the concept of marginal probability?
a) A de-minimis probability that can be treated as effectively zero
b) It is a measure of the likelihood that a particular event will occur, if another event has already
occurred.
c) It is a measure of the likelihood of the simultaneous occurrence of two or more events.
d) None of these choices.
e) It is a measure of the likelihood that a particular event will occur, regardless of whether another
event occurs.
4. If A and B are disjoint events with P(A) = 0.70, then P(B):
a) can be any value between 0 and 1
b) can be any value between 0 and 0.70
c) cannot be larger than 0.30
d) cannot be determined with the information given
e) Is properly obtained by way of rectangular subtraction
ADM2303
Page 2 of 11
Fall, 2022
Page 2
5. Suppose P(A) = 0.60, P(B) = 0.85, and A and B are independent. The probability of the complement of
the event (A and B) is:
a) 0.06
b) 0.55
c) 0.45
d) 0.49
e) 0.90
6. Which statements are correct if events A and B have non-zero probabilities?
a) A and B cannot be both independent and disjoint
b) A and B can be both independent and disjoint
c) A and B are always independent
d) A and B are always disjoint
e) The probability of both events happening must also be nonzero
ADM2303
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Fall, 2022
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PREDICTED RETURN:
For the upcoming year, an analyst has constructed the following discrete probability distribution for firm
X's predicted return.
Return
Probability
−
5
0.20
0
0.30
5
0.40
Answer the following four questions based on the above information:
7.
The expected value of this distribution is
a)
1
b)
2
c)
2.5
d)
3
e)
3.5
8.
The variance of this distribution is
a)
14
b)
4.89
c)
17
f)
21
d)
4.58
9.
The coefficient of variation for this distribution is
a)
14.3
b)
7.4
c)
2.29
d)
11.5
e)
8.5
10.
Assuming all returns are doubled, and probabilities remain the same what would happen to the
coefficient of variation? It would be:
a)
doubled
b)
tripled
c)
quadrupled
d)
unchanged
e)
None above
ADM2303
Page 4 of 11
Fall, 2022
Page 4
11.
A recent survey shows that the probability of a college student drinking alcohol is 0.6. Further, given
that the student is over 21 years old, the probability of drinking alcohol is 0.8. It is also known that
30% of the college students are over 21 years old. The probability of drinking or being over 21 years
old is __________.
a)
0.24
b)
0.42
c)
0.66
d)
0.90
e)
68%
12.
Rumi, a production manager, is trying to improve the efficiency of his assembly line. He knows that
the machine is set up correctly only 60% of the time. He also knows that if the machine is set up
correctly, it will produce good parts 80% of the time, but if set up incorrectly, it will produce good
parts only 20% of the time. Rumi starts the machine and produces one part before he begins the
production run. He finds the first part to be good. What is the revised probability that the machine
was set up correctly?
a)
48%
b)
56%
c)
86%
d)
96%
e)
68%
ADM2303
Page 5 of 11
Fall, 2022
Page 5
Anger and cardiovascular disease.
Studies have shown that people who are angry more frequently and more intensely are at higher risk for all
heart disease events, including heart attack, silent heart attack and bypass surgery. In one particular study,
12,986 participants were recruited to examine the hypothesis that people with high trait anger might be
particularly susceptible to coronary heart disease. The study participants were followed for five years. The
table below shows the resulting data for participants identified as having normal blood pressure.
Anger Score
Total
Low
Moderate
High
Heart
Disease
Event
Yes
53
110
27
190
No
3057
4704
606
8284
Total
3110
4731
633
8474
13.
What percentage of the study participants had moderate anger score?
a) 56
b) 55
c) 45
d) 46
e) 35
14.
What percentage of the study participant had a heart disease event?
a) 3.2
b) 2.2
c) 1.2
d) 4.2
e) 5.2
15.
What percentage of the participant had a heart disease event and high anger score?
a) 0.12
b) 0.22
c) 0.42
d) 0.32
e) 0.52
16.
What percentage of individuals who experienced a heart disease event had moderate anger scores?
a) 58
b) 55
c) 45
d) 46
e) 35
17.
What percentage of participants with high anger scores experienced a head disease event?
a) 9
b) 3
c) 2
d) 5
e) 4
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18.
Which one of the following is correct? Anger score and heart disease event are most likely:
a) Complementary
b) Mutually exclusive
c) Independent
d) Dependent
e) Disjoint
COVID-19 SIR model
A population during the pandemic can be partitioned into three groups: number of susceptible people (S),
number of infected (I), and number of people who have recovered (R). Consider the population of a
community in the midst of the Covid-19 pandemic where 50% of the population is considered susceptible,
20% of the population is infected, and 30% is recovered. The only rapid antigen test that is available to this
community is accurate 90% of the time for susceptible individuals, 95% for infected individuals, and only
70% for recovered individuals.
P(S) = 0.5
P(I) = 0.2
P(R) = 0.3
P(+ | S) = 0.9
P(+ | I) = 0.95
P(- | R) = 0.7
19.
A randomly selected individual from this population is tested with the rapid antigen test. What is
the probability that the result of the test is positive?
a) 0.22
b) 0.45
c) 0.32
d) 0.55
e) 0.73
20.
If a randomly selected individual from this population has tested positive for Covid-19, what is the
probability that they are infected?
a) 0.12
b) 0.34
c) 0.26
d) 0.53
e) 0.53
ADM2303
Page 7 of 11
Fall, 2022
Page 7
Air Travel
During the summer, Canadian airlines across the country had to cancel hundreds of flights as they were not
able to cope with the increased demand after lifting pandemic restrictions. Canadian Transportation Agency
receives on average 400 calls a day to help stranded travellers deal with airlines. Suppose that the hotline
is staffed for 16 hours a day. Assume that the number of calls on a time period can be modelled by Poisson
distribution. (Choose the answer that is closest to your answer.)
21.
What is the average number of calls in a 15-minute interval?
a) 5.00
b) 1.32
c) 6.25
d) 3.67
e) 1.53
22.
What is the probability of exactly 6 calls in a 15-minute interval?
a) 0.74
b) 1.32
c) 0.1598
d) 0.012
e) 0.539
23.
What is the probability of no calls in a 15-minute interval?
a) 0.1239
b) 0.0019
c) 0.5399
d) 0.0129
e) 0.5349
24.
What is the probability of at least two calls in a 15-minute interval?
a) 0.123
b) 0.986
c) 0.689
d) 0.012
e) 0.033
Graduate school
40% of undergraduate students of a business school plan on pursuing a graduate degree. Fifteen
undergraduate students are randomly selected from this business school. (Round your answer to 4 decimal
points)
25.
What is the probability that no more than two of these students plan to pursue a graduate degree?
a) 0.1231
b) 0.1398
c) 0.5478
d) 0.0271
e) 0.0843
26.
What is the probability that at least seven but less than nine of the students plan to pursue a graduate
degree?
a) 0.2311
b) 0.1381
c) 0.4781
d) 0.2951
e) 0.8491
ADM2303
Page 8 of 11
Fall, 2022
Page 8
BookWal1
Industries supplies bookshelf kits that pair solid wood with iron-pipe
supporting structure.
We begin by focusing on the wood material demanded per month. This depends on the
number of bookcases demanded per month, which is a
random variable
and denoted
?
.
The probability model for
?
has an expected value of 509.9 (units/month) and a standard
deviation of 147 (units/month).
Determine the monthly wood demand, noting that each bookshelf requires
1.5
square-
meters of wood, and that standard practice involves ordering an additional
55
square-
meters per month (regardless of
?
) as a contingency (spare wood to compensate for
mistakes in carpentry). Consider BookWal1 Industries’ monthly wood order (MWO).
Determine the following for the random variable MWO.
27.
Its expected value
a)
566.40 square-meters
b)
764.85 square-meters
c)
819.85 square-meters
d)
can not be determined
e)
55 square-meters
28.
Its coefficient of variation
a)
Can not be determined without full probability model
b)
3.72
c)
0.29
d)
0.27
e)
Is zero (owing to independence)
The length of iron pipe required (meters) per month to complete the Bookshelf kits is also
a random variable (labeled
?
). The expected value and standard deviation are 0.14
(kilometers per month) and 0.06 (kilometers per month) respectively. In subsequent
questions you may assume that the random variables
MWO
and
Y
are independent (though
admittedly in reality this would not be the case).
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If the price of wood is
25
dollars per square-meter and the price of pipe is
204
dollars per
linear kilometer, determine the following for the total cost of materials (wood and pipe)
per month (label this Z).
29.
The expected value (per month)
a)
21 Thousand dollars.
b)
20 Thousand dollars.
c)
518 Thousand dollars.
d)
Can not be determined.
e)
188 Thousand dollars.
30.
The coefficient of variation
a)
Can not be determined.
b)
0.999
c)
1.001
d)
3.723
e)
0.269
31.
Consider inflation whereby the aforementioned prices are both multiplied by 1.3.
How would the
coefficient of variation
for monthly material (wood and pipe) costs
change? It would be …
a)
Can not be determined.
b)
Increased multiplicatively by 1.3.
c)
Decreased multiplicatively by 0.77.
d)
Reduced by subtracting 1.3 from the original estimate (i.e., addition/complement
rule).
e)
Unchanged.
ADM2303
Page 10 of 11
Fall, 2022
Page 10
ADM2303 formula sheet - Fall 2022 zz
Probability theory
Rule of sum of probabilities:
P(
S
) = 1
Complement rule
(Let
A
c
be complement of
A
, i.e., Not
A
):
P(
A
) = 1
−
P(
A
c
)
Addition rule for two mutually exclusive events (where
∪
connotes “or” aka union):
P(
A
∪
B
) = P(
A
) + P(
B
)
Addition rule for two not mutually exclusive events:
P(
A
∪
B
) = P(
A
) + P(
B
)
−
P(
A
∩
B
)
Multiplication rule for two independent events (where
∩
connotes “and” aka intersection):
P(
A
∩
B
) = P(
A
)
×
P(
B
)
Multiplication rule for
n
independent events:
P(
A
1
∩
A
2
∩
...A
n
) = P(
A
1
)
×
P(
A
2
)
×
...
×
P(
A
n
)
Multiplication rule for dependent events:
P(
A
∩
B
) = P(
B
|
A
)P(
A
) = P(
A
|
B
)P(
B
)
Partition rule: for a partition
B
1
, B
2
, ..., B
k
:
P(
A
) =
k
i
=1
P(
A
∩
B
i
) =
k
i
=1
P(
A
|
B
i
)P(
B
i
)
Bayes’ formula:
P(
B
i
|
A
) =
P(
A
|
B
i
)P(
B
i
)
P(
A
)
=
P(
A
|
B
i
)P(
B
i
)
∑
k
i
=1
P(
A
|
B
i
)P(
B
i
)
Events A and B are independent if:
P(
A
|
B
) = P(
A
) and P(
B
|
A
) = P(
B
)
or:P(
A
∩
B
) = P(
A
)
×
P(
B
)
Random variables (RV)
Expected value of discrete RV
X
:
E
(
X
) =
µ
=
n
i
=1
x
i
P(
X
=
x
i
)
Variance of discrete RV
X
: Var(X) =
σ
2
=
n
i
=1
(
x
i
−
µ
)
2
P(
X
=
x
i
)
=
n
i
=1
x
2
i
P(
X
=
x
i
)
−
µ
2
Standard deviation of discrete RV
X
:
SD
(
X
) =
σ
=
V ar
(
X
)
Coefficient of variation of discrete RV
X
:
CV
(
X
) =
SD
(
X
)
E
(
X
)
Correlation of two discrete RV
X
and
Y
:
ρ
x,y
=
∑
n
i
=1
(
x
i
−
µ
x
)(
y
i
−
µ
y
))P(
X
=
x
i
∩
Y
=
y
j
)
s
x
s
y
Combining random variables
Adding a constant
c
to random variable
X
:
E
(
X
±
c
) =
E
(
X
)
±
c
V ar
(
X
±
c
) =
V ar
(
X
) =
σ
2
X
Multiplying random variable
X
by a constant
a
:
E
(
aX
) =
aE
(
X
)
V ar
(
aX
) =
a
2
σ
2
X
Expected value of linear combination of RVs:
1
E
(
aX
+
bY
+
c
) =
aE
(
X
) +
bE
(
Y
) +
c
Variance of linear combination of RVs
(1)
:
V ar
(
aX
+
bY
+
c
) =
a
2
σ
2
X
+
b
2
σ
2
Y
+ 2
a b Cov
(
X, Y
)
where
Cov
(
X, Y
) =
ρ
x,y
σ
X
σ
Y
. If
X
and
Y
independent
then
ρ
x,y
= 0 and covariance component drops out.
Discrete and continuous distributions
The Binomial probability distribution:
P(
X
=
x
) =
n
!
(
n
−
x
)!
x
!
p
x
(1
−
p
)
n
−
x
for
x
= 0
,
1
, ..., n
E
(
X
) =
np, V ar
(
X
) =
np
(1
−
p
)
The Poisson probability distribution
(if approx’n of binomial,
λ
=
np
):
P(
X
=
x
) =
e
−
λ
λ
x
x
!
for
x
= 0
,
1
,
2
, ...
E
(
X
) =
λ, V ar
(
X
) =
λ, e
= 2
.
718
The Geometric probability distribution:
P(
X
=
x
) = (1
−
p
)
x
−
1
p
for
x
= 1
,
2
, ...
E
(
X
) =
1
p
, V ar
(
X
) =
1
−
p
p
2
The Normal distribution:
X
∼
N
(
µ, σ
)
⇒
f
(
x
) =
1
σ
√
2
π
exp
−
1
2
(
x
−
µ
σ
)
2
Z
=
X
−
µ
σ
∼
N
(0
,
1)
⇒
P(
Z<z
) = using normal table
The Exponential distribution :
X
∼
Expo
(
λ
)
⇒
f
(
x
) =
λe
−
λx
P(
X
≤
a
) = 1
−
e
−
aλ
E
(
X
) =
1
λ
, V ar
(
X
) = (
1
λ
)
2
The Uniform distribution:
X
∼
Uniform(
a, b
)
⇒
f
(
x
) =
1
b
−
a
for
a
≤
x
≤
b
P(
x
1
<X<x
2
) =
x
2
−
x
1
b
−
a
E
(
X
) =
a
+
b
2
, V ar
(
X
) =
(
b
−
a
)
2
12
1
For cases like 2
X
−
3
Y
the coefficient on
Y
is (
−
3), thus treat accordingly; want to subtract a constant rather than add — put minus sign on
c
.
ADM2303
Page 11 of 11
Fall, 2022
Page 11
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Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
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