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University of Ottawa *
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Course
2379
Subject
Mathematics
Date
Apr 3, 2024
Type
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13
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1
MAT 2379
Sample Midterm 1 (with solutions)
Date:
Xiao Liang
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
2
Part 1: Multiple Choice Questions
Record your answer to the multiple choice questions in the table below:
Questio
n
Answe
r
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
J
)
=
P
(
B
J
)
=
P
(
C
J
)
=
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
J
∩
B
J
∩
C
J
)
=
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
ey
es.
Then
A
J
is
the
event
that
the
man
has
bro
wn
hair,
and
B
J
is
the event that he has brown eyes. We know that
1770
1770
1770
3
P
(
A
J
∩
B
J
)
=
320
,
P
(
A
J
∩
B
)
=
250
,
P
(
A
∩
B
J
)
=
400
.
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∩
|
4
Blood type
Number of donors
OABAB
8
3
3
1
The above three events are disjoint, and their union is the complement of the event A
B
. We
conclude:
P (
A ∩ B
) = 1 − 320 + 250 + 400 = 800
= 0
.
452
.
The answer is D.
1770
1770
1770
1770
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:
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
m
ea
ns
that
P
(
O
2
|
O
1
)
=
7
/
14
,
and
hence
P
(
O
2
J
|
O
1
)
=
1
−
7
/
14
=
7
/
14
.
If
O
1
J
happ
ened,
then
there
a
re
8
p
eople
with
blo
o
d
t
yp
e
O
among
the 14 who remain for the second selection. This means that P (
O
2
O
1
J
) = 8
/
14
. The probability
that there is exactly 1 person with blood type O is:
P
(
O
1
∩
O
2
J
)
+
P
(
O
1
J
∩
O
2
)
=
P
(
O
2
J
|
O
1
)
P
(
O
1
)
+
P
(
O
2
|
O
1
J
)
P
(
O
1
J
)
7
8
8
7
8
=
14 · 15 + 14 · 15 = 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
10
0
F (
x
)
0
0.06
7
0.10
5
0.13
3
0.24
6
0.33
3
0.46
7
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
5
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
≤
−
≤
−
≤
≤ −
≤
≤
≤
≤
x
frequency
less than -6.00[-6.00, -4.00)[-4.00, -2.00)[-2.00, 0.00)
3%
7%
36%
54%
6
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
:
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)
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10
7
= 1 − (0
.
97)
− 10(0
.
03)(0
.
97) = 0
.
0345
The answer is E.
8
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:
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
9
genotype yyhh
) is
1
/
8 = 0
.
125
.
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≤
2
0
1
1
0
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
.
125)
0
(0
.
875)
10
−
0
+ 10 ! (0
.
125)
1
(0
.
875)
10
−
1
+
10 ! (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.
W
e
kno
w
that
P
(
C
)
=
0
.
05
,
P
(
L
|
C
)
=
0
.
33
and
P
(
L
|
C
J
)
=
0
.
27
.
a)
By the total probability rule,
P
(
L
)
=
P
(
L
|
C
)
P
(
C
)
+
P
(
L
|
C
J
)
P
(
C
J
)
= (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
|
C
)
P
(
C
)
P (
L
)
(0
.
33)(0
.
05)
=
=
0
.
273
P (
L
)
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 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
+)
1
1
=
0
.
14
. W
e
also
kno
w
t
ha
t
the
sensitivit
y
is
P
(
T
est
+
|
T
rue
+)
=
0
.
78
,
and
the
false-p
ositive-rate
is P
(
T
est
+
|
T
rue
−
)
=
0
.
21
.
|
— |
−
/
|
1
2
a)
PPV = P (
True + Test
+) = P
(
True
+
∩
Test
+)
P (
Test
+)
=
P
(
T
es
t
+
|
T
rue
+)
P
(
T
rue
+
)
P
(
T
est
+
|
T
rue
+)
P
(
T
rue
+)
+
P
(
T
est
+
|
T
rue
−
)
P
(
T
rue
−
)
(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
(
T
es
t
−
|
T
rue
−
)
P
(
T
rue
−
)
P
(
T
est
−
|
T
rue
+)
P
(
T
rue
+)
+
P
(
T
est
−
|
T
rue
−
)
P
(
T
rue
−
)
(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?
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 and F )
23
/
1000
23
P
(
D
|
F
)
=
=
=
= 0
.
042
P (
F )
545
/
1000
545
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.
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1
3
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 and N )
70
/
1000
70
P
(
B
|
N
)
=
=
=
= 0
.
27
.
P (
N )
(188 + 70)
/
1000
258