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2379
Subject
Mathematics
Date
Jan 9, 2024
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MAT 2379
Sample Midterm 1 (with solutions)
Date:
Professor Raluca Balan
Time: 80 minutes
Student Number:
Family Name:
First Name:
This is a closed book examination.
You can bring your own formula sheet (one page, one-sided).
Only Faculty standard calculators are permitted: TI30, TI34, Casio fx-260, Casio fx-300.
You are not allowed to use any electronic device during the exam. Cell phones should be put away.
The exam consists of 6 multiple choice questions and 4 long answer questions.
Each multiple choice question is worth 5 marks and each long answer question is worth 10 marks.
The total number of marks is 70.
NOTE: At the end of the examination, hand in the entire booklet.
.*****************************************************.
For professor’s use:
Number of marks
Total for all MC Questions
Long Answer Question 1
Long Answer Question 2
Long Answer Question 3
Long Answer Question 4
Total
1
Part 1: Multiple Choice Questions
Record your answer to the multiple choice questions in the table below:
Question
Answer
1
2
3
4
5
6
1. Enterococci are bacteria that cause blood infections in hospitalized patients. One antibiotic used
to battle enterocossi is vancomycin. A study revealed that in Canada, the enterocossi bacteria is
resistent to vancomycin in 22% of all hospitalized patients who have this kind of infection. Con-
sider a random sample of three patients with blood infections caused by the enterocossi bacteria.
Assume that all three patients are treated with the antibiotic vancomycin. What is the probability
that the enterocossi bacteria is resistent to the antibiotic for at least one patient?
A) 0.041
B) 0.476
C) 0.062
D) 0.525
E) 0.224
Solution:
Let
A
be the event “the bacteria is resistant to the antibiotic for the first patient”,
B
the event “the bacteria is resistant to the antibiotic for the second patient” and
C
the event “the
bacteria is resistant to the antibiotic for the third patient”. Events
A, B
and
C
are independent
with
P
(
A
) =
P
(
B
) =
P
(
C
) = 0
.
22
. Hence
P
(
A
′
) =
P
(
B
′
) =
P
(
C
′
) = 0
.
78
. The event “the
bacteria is resistent to the antibiotic for at least one patient” is the complement of ”the bacteria
is not resistent to the antibiotic for all three patients”. Therefore, the probability that the bacteria
is resistent to the antibiotic for at least one patient is
P
(
A
∪
B
∪
C
) = 1
−
P
(
A
′
∩
B
′
∩
C
′
) = 1
−
(0
.
78)
3
= 0
.
525
The answer is D.
2. The eye colour of any member of a group of 1770 German men is either blue or brown, and the
hair colour is either blond or brown. In this group, there are 320 men who have brown hair and
brown eyes, and there are 250 men who have brown hair and blue eyes. Finally, 400 have blond
hair and brown eyes. What is the probability that a randomly chosen member of the group has
blond hair and blue eyes?
A) 0.5
B) 0.169
C) 0.226
D) 0.452
E) 0.1
Solution:
Let
A
be the event that a randomly selected man in this group has blond hair, and
B
the event that he has blue eyes. Then
A
′
is the event that the man has brown hair, and
B
′
is the
event that he has brown eyes. We know that
P
(
A
′
∩
B
′
) =
320
1770
,
P
(
A
′
∩
B
) =
250
1770
, P
(
A
∩
B
′
) =
400
1770
.
2
The above three events are disjoint, and their union is the complement of the event
A
∩
B
. We
conclude:
P
(
A
∩
B
) = 1
−
320
1770
+
250
1770
+
400
1770
=
800
1770
= 0
.
452
.
The answer is D.
3. This morning, there were 15 persons who donated blood at a clinic of the Canadian Blood Society.
Here is the distribution of their blood types:
Blood type
O
A
B
AB
Number of donors
8
3
3
1
We select at random 2 persons from these 15 donors. What is the probability that we select exactly
one person with blood type O?
A) 0.2667
B) 0.2489
C) 0.5333
D) 0.4978
E) 0.7156
Solution:
Let
O
i
be the event that the
i
-th selection is a donor with blood type O, for
i
= 1
,
2
.
We know that
P
(
O
1
) = 8
/
15
. If
O
1
happened, then there are 7 people with blood type O among
the 14 who remain for the second selection.
This means that
P
(
O
2
|
O
1
) = 7
/
14
, and hence
P
(
O
′
2
|
O
1
) = 1
−
7
/
14 = 7
/
14
. If
O
′
1
happened, then there are 8 people with blood type O among
the 14 who remain for the second selection. This means that
P
(
O
2
|
O
′
1
) = 8
/
14
. The probability
that there is exactly 1 person with blood type O is:
P
(
O
1
∩
O
′
2
) +
P
(
O
′
1
∩
O
2
)
=
P
(
O
′
2
|
O
1
)
P
(
O
1
) +
P
(
O
2
|
O
′
1
)
P
(
O
′
1
)
=
7
14
·
8
15
+
8
14
·
7
15
=
8
15
=
0
.
5333
The answer is C.
4. Let
X
be the final mark of a randomly selected student who took the course MAT 2379 in the
fall 2022. Assume that this mark is rounded to the closest integer value (on a scale from 0 to
100).
The following table gives the values of the cumulative distribution function
F
of
X
, for
some particular marks
x
:
mark
x
30
40
60
69
74
79
84
89
100
F
(
x
)
0
0.067
0.105
0.133
0.246
0.333
0.467
0.8
1
What is the probability that a randomly selected student in this class obtained the grade B or B+?
(Recall that the grade B corresponds to a final mark in the range 70-74, and the grade B+ corre-
sponds to a final mark in the range 75-79.)
A) 0.333
B) 0.133
C) 0.466
D) 0.379
E) 0.2
Solution:
X
is a discrete random variable with values
0
,
1
,
2
, . . . ,
100
. The probability that the
student has a final mark in the range 70-79 is:
P
(70
≤
X
≤
79)
=
P
(
X
≤
79)
−
P
(
X
≤
69) =
F
(79)
−
F
(69)
=
0
.
333
−
0
.
133 = 0
.
2
3
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The answer is
E
.
5. Let
X
be a random variable with a binomial distribution with
n
= 20
trials and probability of
success
p
= 0
.
3
. We would like to compute
P
(10
≤
X <
13)
using
R
. Which one of the following
commands gives this probability?
A)
pbinom(13,20,0.3)-pbinom(10,20,0.3)
B)
dbinom(10,20,0.3)+dbinom(11,20,0.3)+dbinom(12,20,0.3)
C)
pbinom(13,20,0.3)-pbinom(9,20,0.3)
D)
pbinom(10,20,0.3)+pbinom(11,20,0.3)+pbinom(12,20,0.3)
E)
dbinom(12,20,0.3)-dbinom(9,20,0.3)
Solution:
We would like to compute
P
(10
≤
X <
13) =
P
(
X
= 10) +
P
(
X
= 11) +
P
(
X
= 12)
.
Answer A gives
P
(
X
≤
13)
−
P
(
X
≤
10) =
P
(
X
= 11) +
P
(
X
= 12) +
P
(
X
= 13)
, which is
not what we want.
Answer B gives
P
(
X
= 10) +
P
(
X
= 11) +
P
(
X
= 12)
, which is the probability that we want.
Answer C gives
P
(
X
≤
13)
−
P
(
X
≤
9) =
P
(
X
= 10)+
P
(
X
= 11)+
P
(
X
= 12)+
P
(
X
= 13)
,
which is not what we want.
Answer D gives
P
(
X
≤
10) +
P
(
X
≤
11) +
P
(
X
≤
13)
, which is not what we want.
Answer E gives
P
(
X
= 12)
−
P
(
X
= 9)
, which is not what we want.
The answer is B.
6. Nearsightedness, or myopia, has become more prevalent in recent years, especially in children.
Although the cause for myopia is unknown, many eye doctors think that this is related to eye
fatigue from computer use, coupled with a genetic predisposition.
Let
X
be the number of
diopters of nearsightedness of a randomly selected child with myopia, of age 2-16. (We record the
value with the most severe form of myopia between the left and the right eye. For instance, for
a child with -1.25 in the left eye and -2.50 in the right eye, the value of
X
is -2.5.) Using data
from a large network of optometry offices specializing in pediatric care, we obtain the following
frequency table for the values of
X
:
x
less than -6.00
[-6.00, -4.00)
[-4.00, -2.00)
[-2.00, 0.00)
frequency
3%
7%
36%
54%
A value less than -6.00 is called high myopia. If a sample of 10 children with myopia is randomly
selected, what is the probability that at least two of them have high myopia?
A)
0
.
0016
B)
0
.
2626
C)
0
.
0642
D)
0
.
1316
E) 0.0345
Solution:
Let
Y
be the variable which gives the number of children with high myopia in the selected
sample.
Y
has a binomial distribution with
n
= 10
trials and probability of success
p
= 0
.
03
. The
desired probability is:
P
(
Y
≥
2) = 1
−
P
(
Y
≤
1) = 1
−
P
(
Y
= 0)
−
P
(
Y
= 1)
= 1
−
(0
.
97)
10
−
10(0
.
03)(0
.
97)
9
= 0
.
0345
The answer is E.
4
Long answer questions are included on the following pages.
Part 2: Long Answer Questions
Record your answer to the long answer questions in the space provided below, specifying clearly your
notation and including a proper justification. Show the details of your calculations.
1. IR8 rice (also called miracle rice) is a genetically modified rice which was introduced in Asia in the
1960’s and marked the beginning of the Green Revolution. Since then, more than 400 improved
rice varieties have been created by the International Rice Research Institute. Dwarfism and high
yields are two desirable traits of a rice plant.
Assume that both traits are recessive.
Consider
crossing two plants which do not give high yields, but are heterozygous for this trait. Suppose that
one of the plants is a dwarf, and the other one is a regular size plant which is heterozygous for
this trait.
a) (5 marks) What is the probability that the offspring is a dwarf plant with high yields?
b( (5 marks) Consider 10 offsprings of this pair of plants. What is the probability that at least 3
are dwarf plants with high yields?
Solution:
a) We let
Y
the allele for regular yields, and
y
the allele for high yields. We denote by
H
the allele for normal height and
h
for dwarfism. The dwarf plant has genotype
Y yhh
and the
other plant has genotype
Y yHh
. We draw a tree diagram:
5
We see that the possible genotypes for the offspring are:
Y Y Hh,
Y Y hh,
Y yHh,
Y yhh,
Y yHh,
Y yhh,
yyHh
yyhh
The probability that the offspring is a dwarf plant with high yields (i.e.
has genotype
yyhh
) is
1
/
8 = 0
.
125
.
b) Let
X
be the number of offsprings which are dwarf plants with high yields.
Then
X
has
a binomial distribution with
n
= 10
trials and probability of success
p
= 0
.
125
.
The desired
probability is
P
(
X
≥
3) = 1
−
P
(
X
≤
2)
. We compute first
P
(
X
≤
2)
:
P
(
X
≤
2)
=
P
(
X
= 0) +
P
(
X
= 1) +
P
(
X
= 2)
=
10
0
!
(0
.
125)
0
(0
.
875)
10
−
0
+
10
1
!
(0
.
125)
1
(0
.
875)
10
−
1
+
10
2
!
(0
.
125)
2
(0
.
875)
10
−
2
= 0
.
8805
The desired probability is 1-0.8805=0.1195.
2. Recent studies suggest that there is a link between the use of alcohol-containing mouthwashes
(like Listerine) and oral cancer. Data from several dentist offices show that 33% of the patients
with oral cancer have used Listerine on a regular basis for more than 5 years, while among the
patients who do not have oral cancer, 27% have used Listerine on a regular basis for more than 5
years. It is estimate that approximately 5% of the general population has oral cancer.
a) (5 marks) What is the probability that a randomly chosen patient uses Listerine on a regular
basis?
b) (5 marks) What is the probability that a patient will develop oral cancer, given that he/she has
used Listerine on a regular basis for more than 5 years?
Solution:
We denote by
C
the event that the patient will develop oral cancer and
L
the event that
the patient used Listerine. We know that
P
(
C
) = 0
.
05
,
P
(
L
|
C
) = 0
.
33
and
P
(
L
|
C
′
) = 0
.
27
.
a) By the total probability rule,
P
(
L
)
=
P
(
L
|
C
)
P
(
C
) +
P
(
L
|
C
′
)
P
(
C
′
)
=
(0
.
33)(0
.
05) + (0
.
27)(0
.
95)
=
0
.
0165 + 0
.
2565 = 0
.
273
.
b) Using Bayes’ rule, the desired probability is
P
(
C
|
L
)
=
P
(
C
∩
L
)
P
(
L
)
=
P
(
L
|
C
)
P
(
C
)
P
(
L
)
=
(0
.
33)(0
.
05)
0
.
273
=
0
.
0165
0
.
273
= 0
.
06
.
3. A patient with high blood pressure is called hypertensive, and a patient with normal blood pressure is
called normotensive. Suppose that 78% of hypertensives and 21% of normotensives are classified as
6
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hypertensives by an automated blood-pressure machine. Assume that 14% of the adult population
is hypertensive.
a) (5 marks) What is the positive predictive value (PPV) of this machine?
b) (5 marks) What is the negative predictive value (NPV) of this machine?
Solution:
Let True
+
be the event of being hypertensive.
We know that
P
(
True
+) = 0
.
14
.
We also know that the sensitivity is
P
(
Test
+
|
True
+) = 0
.
78
, and the false-positive-rate is
P
(
Test
+
|
True
−
) = 0
.
21
.
a)
PPV
=
P
(
True
+
|
Test
+) =
P
(
True
+
∩
Test
+)
P
(
Test
+)
=
P
(
Test
+
|
True
+)
P
(
True
+)
P
(
Test
+
|
True
+)
P
(
True
+) +
P
(
Test
+
|
True
−
)
P
(
True
−
)
=
(0
.
78)(0
.
14)
(0
.
78)(0
.
14) + (0
.
21)(0
.
86)
=
0
.
1092
0
.
1092 + 0
.
1806
=
0
.
1092
0
.
2898
=
0
.
38
.
b)
NPV
=
P
(
True
− |
Test
−
) =
P
(
True
− ∩
Test
−
)
P
(
Test
−
)
=
P
(
Test
− |
True
−
)
P
(
True
−
)
P
(
Test
− |
True
+)
P
(
True
+) +
P
(
Test
− |
True
−
)
P
(
True
−
)
=
(0
.
79)(0
.
86)
(0
.
22)(0
.
14) + (0
.
79)(0
.
86)
=
0
.
6794
0
.
0308 + 0
.
6794
=
0
.
6794
0
.
7102
=
0
.
96
.
4. According to the Ontario legislation, passengers aged 13 or older can travel in the front seat of a
motor vehicle. The following table gives the extent of injuries and the passenger position for 1000
accidents.
Extent of injury
Front Seat
Back Seat
None
188
70
Minor
232
295
Major
102
75
Death
23
15
Total
545
455
a) (5 marks) Based on this data, what is the probability of a passenger dying in a motor vehicle
accident, given that the passenger was traveling in the front seat? Is death independent of the
passenger travelling in the front seat? Justify your answer.
b) (5 marks) What is the probability that a passenger traveled in the back seat, given that the
passenger did not have any injuries?
7
Solution
: a) Let
D
be the event that the passenger dies, and
F
be the event that the passenger
travels is the front seat. From the table, we know that:
P
(
D
|
F
) =
P
(
D
and
F
)
P
(
F
)
=
23
/
1000
545
/
1000
=
23
545
= 0
.
042
Since
P
(
D
) = 38
/
1000 = 0
.
038
̸
= 0
.
042 =
P
(
D
|
F
)
, death is not independent of the passenger
travelling in the front seat: the passengers travelling in the front seat have an increased chance of
dying.
b) Let
B
be the event that the passenger traveled in the back seat and
N
be the event that the
passenger did not have any injury. The desired probability is:
P
(
B
|
N
) =
P
(
B
and
N
)
P
(
N
)
=
70
/
1000
(188 + 70)
/
1000
=
70
258
= 0
.
27
.
8