2379C-midterm1-23A Solution
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MAT 2379C
Midterm Examination
Version A
October 4, 2023
Instructor: 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).
•
Some statistical tables are included in the last pages of the booklet.
•
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
C
2
D
3
C
4
B
5
A
6
E
1. Consider an experiment in which we study the ability of snails to hold firmly onto a smooth
substrate.
The snail is attached to a smooth rock which is exposed to water, and we record if
the snail has the ability to resist the current. It was found that 38% of snails are resistant to the
current. Suppose that 50% of snails are of species A. Moreover,
30%
of the snails are of species
A and are resistant to the current.
What is the probability that a randomly chosen snail of species A is resistant to the current? Is
resistance to the current independent of the species?
A)
0
.
30
. Yes, resistance to the current is independent of the species.
B)
0
.
6
. Yes, resistance to the current is independent of the species.
C)
0
.
6
. No, resistance to the current is not independent of the species.
D)
0
.
38
. No, resistance to the current is not independent of the species.
E)
0
.
38
. Yes, resistance to the current is independent of the species.
Solution:
Let
A
be the event that a randomly chosen snail belongs to species A and
R
the
event that the snail is resistant to the current.
We know that
P
(
R
) = 0
.
38
,
P
(
A
) = 0
.
5
and
P
(
A
∩
R
) = 0
.
30
. The probability that a randomly chosen snail of species A is resistant to the
current is:
P
(
R
|
A
) =
P
(
R
∩
A
)
P
(
A
)
=
0
.
3
0
.
5
= 0
.
6
Note that
P
(
A
∩
R
) = 0
.
30
6
=
P
(
R
)
P
(
A
) = (0
.
38)(0
.
5) = 0
.
19
Hence
A
and
R
are not independent. The answer is C.
2. In testing the water supply for various cities for two kinds of impurities commonly found in water,
it was found that 40% of cities have water which contains an impurity of type A, 50% have water
which contains an impurity of type B, and 20% have water which contains neither one of the two
impurities. If a city is chosen at random, what is the probability that its water supply has
exactly
one
impurity of type A or B?
A) 0.3
B) 0.8
C) 0.1
D) 0.7
E) 0.2
2
Solution:
Let
A
be the event that the water in this city contains an impurity of type A and
B
be
the event that the water in this city contains an impurity of type B. We know that
P
(
A
) = 0
.
4
,
P
(
B
) = 0
.
5
and
P
(
A
0
∩
B
0
) = 0
.
2
. Hence
P
(
A
∪
B
) = 1
-
P
(
A
0
∩
B
0
) = 1
-
0
.
2 = 0
.
8
. By the
addition rule,
P
(
A
∪
B
) =
P
(
A
) +
P
(
B
)
-
P
(
A
∩
B
)
. Hence
P
(
A
∩
B
) =
P
(
A
) +
P
(
B
)
-
P
(
A
∪
B
) = 0
.
4 + 0
.
5
-
0
.
8 = 0
.
1
The desired probability is
P
(
A
∩
B
0
) +
P
(
A
0
∩
B
) =
P
(
A
∪
B
)
-
P
(
A
∩
B
) = 0
.
8
-
0
.
1 = 0
.
7
The answer is D.
3. Consider the following
R
output:
> pbinom(12,25,0.3)
[1] 0.9825303
> pbinom(3,25,0.3)
[1] 0.03324052
> pbinom(2,25,0.3)
[1] 0.008960528
> pbinom(7,25,0.3)
[1] 0.5118485
> pbinom(8,25,0.3)
[1] 0.6769281
> dbinom(6,25,0.3)
[1] 0.1471665
Let
X
be a binomial random variable wil
n
= 25
trials and
p
= 0
.
3
probability of success. Find
P
(3
≤
X
≤
12)
.
A) 0.9492
B) 0.8324
C) 0.9736
D) 0.8223
E) 0.8017
Solution:
P
(3
≤
X
≤
12) =
P
(
X
≤
12)
-
P
(
X
≤
2) = 0
.
9825303
-
0
.
008960528 = 0
.
9735698
The answer is C.
4. A screening test is applied to 200 patients suffering from a certain disease.
This test is also
applied to 200 persons selected randomly from the community who do not have the disease, called
“controls”. The following table summarizes the test results:
Patients who
Community
have the disease
controls
Total
Positive tests
166
18
184
Negative tests
34
182
216
Total
200
200
400
3
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What is the specificity of this test?
A) 0.73
B) 0.91
C) 0.82
D) 0.89
E) 0.10
Solution:
specificity =
P
(Test
- |
True
-
) =
P
(True
-
and Test
-
)
P
(True
-
)
=
182
/
400
200
/
400
=
182
200
= 0
.
91
.
The answer is B.
5. In all major Canadian cities, hospitals are experiencing an increase in the number of patients ad-
mitted to an intensive care unit (ICU). In a certain hospital, 35% of ICU patients have Covid-19.
What is the probability that in a hospital room of 10 ICU patients, at least 2 patients have Covid-19?
A) 0.9140
B) 0.7384
C) 0.0860
D) 0.6171
E) 0.2616
Solution:
Let
X
be the number of patients with Covid-19 in this room. Then
X
has a binomial
distribution with
n
= 10
trials and probability of success
p
= 0
.
35
. The desired probability is
P
(
X
≥
2)
=
1
-
P
(
X
≤
1) = 1
-
P
(
X
= 0)
-
P
(
X
= 1)
=
1
-
10
0
(0
.
35)
0
(0
.
65)
10
-
10
1
(0
.
35)
1
(0
.
65)
9
=
1
-
(0
.
65)
10
-
10(0
.
35)(0
.
65)
9
= 0
.
9140
The answer is A. The wrong answer B is obtain by calculating
1
-
P
(
X
≤
2) = 0
.
7384
.
6. A new drug has been introduced to bring hypertension under control—that is, to reduce high blood
pressure to normotensive levels. For any clinical practice among 4 patients, let
X
be the number
of patients of four who are brought under control. There is a table that gives the values of
X
and
the corresponding probabilities, according to a previous study with the drug. However, part of the
table is damaged.
x
0
1
2
3
?
P
(
X
=
x
)
0.8%
7.6%
26.5%
?
24.0%
Determine the variance for the random variable
X
.
A) 1.33
B) 1.54
C) 2.80
D) 8.68
E) 0.84
Solution:
Since
X
is a discrete random variable with values
{
0
,
1
,
2
,
3
,
4
}
, and the sum of the
probabilities should be 1, the distribution of
X
is
x
0
1
2
3
4
P
(
X
=
x
)
0.8%
7.6%
26.5%
41.1%
24.0%
4
Thus,
E
(
X
) = 0
×
0
.
8% + 1
×
7
.
6% + 2
×
26
.
5% + 3
×
41
.
1% + 4
×
24
.
0% = 2
.
799
,
E
(
X
2
) = 0
2
×
0
.
8% + 1
2
×
7
.
6% + 2
2
×
26
.
5% + 3
2
×
41
.
1% + 4
2
×
24
.
0% = 8
.
675
.
Therefore,
Var(
X
) =
E
(
X
2
)
-
(
E
(
X
))
2
= 8
.
675
-
2
.
799
2
= 0
.
84
The correct answer is E.
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. Sturgeon is a name used to describe a family of fish. It is one of the oldest fish families in Canada,
dating back 200 million years. They have undergone little change over time and are often described
as living fossils. Typical adult sturgeons are between 2 meters and 6 meters long. Sturgeons are
bottom feeders that eat mainly crustaceans and small shells. As they grow larger they may eat
other fish and depending on their size, they can even swallow a whole salmon. Assume that 45%
of the sturgeons in a particular river are large enough to swallow a whole salmon. If 10 sturgeons
are randomly selected from this river, what is the probability that
(a) (2 marks) none is large enough to swallow a whole salmon;
(b) (3 marks) at least two is large enough to swallow a whole salmon;
(c) (5 marks) between 2 and 4, inclusively, are large enough to swallow a whole salmon.
Solution:
Let
X
be the number of sturgeons large enough to swallow a salmon whole among
10
sturgeons.
X
has a binomial distribution with
n
= 10
and
p
= 0
.
45
.
(a)
P
(
X
= 0) =
10
0
(0
.
45)
0
(0
.
55)
10
= 0
.
0025
(b)
P
(
X
≥
2) = 1
-
P
(
X
= 0)
-
P
(
X
= 1) = 1
-
0
.
0025
-
0
.
0207 = 0
.
9768
(c)
P
(2
≤
X
≤
4) =
P
(
X
= 2) +
P
(
X
= 3) +
P
(
X
= 4)
=
10
2
(0
.
45)
2
(0
.
55)
8
+
10
3
(0
.
45)
3
(0
.
55)
7
+
10
4
(0
.
45)
4
(0
.
55)
6
= 0
.
4811
2. In a study, 268 women with breast cancer were followed up for at least two years after the initial
diagnosis.
In this study, 87 women suffered from inoperable locoregional recurrent disease, 140
from distant metastases and 41 from both.
(a) (5 marks) What is the probability that a randomly chosen woman in this study suffered from
inoperable locoregional recurrent disease, but did not have distant metastases?
5
(b) (5 marks) What is the probability that a randomly chosen woman in this study suffered from
distant metastases, but did not have inoperable locoregional recurrent disease?
Solution:
Let A be the event that a randomly chosen woman in this study suffers from inoperable
locoregional recurrent disease, and B the event that she has distant metastases. We know that
P
(
A
) =
87
268
= 0
.
325
, P
(
B
) =
140
268
= 0
.
522
,
and
P
(
A
∩
B
) =
41
268
= 0
.
153
.
(a) The probability that a randomly chosen woman in this study suffers from inoperable locoregional
recurrent disease, but does not have distant metastases is:
P
(
A
∩
B
0
) =
P
(
A
)
-
P
(
A
∩
B
) =
87
-
41
268
=
46
268
= 0
.
172
.
(b) The probability that a randomly chosen woman in this study suffers from distant metastases
but does not have inoperable locoregional recurrent disease is:
P
(
A
0
∩
B
) =
P
(
B
)
-
P
(
A
∩
B
) =
140
-
41
268
=
99
268
= 0
.
369
.
3. In a city,
22
% of adult citizens are young (age 18-39),
32
% of adult citizens are middle-aged (age
40-64), and
46
% of adult citizens are seniors (age 65+).
Diabetes has a prevalence of
4
% for
young people,
7
% for middle-aged people, and
12
% for seniors. (The prevalence is the percentage
of the people in the corresponding age group who have the disease.) We select at random an adult
citizen from this city.
(a) What is the probability that this person has diabetes?
(b) If we find that this person has diabetes, what is the probability that this person is a senior?
Solution:
Let
D
be the event that a randomly selected citizen who has diabetes. Let
Y
,
M
,
S
denote the event that the citizen is young, mid-aged, or senior, respectively. Note that
Y
,
M
,
S
form a partition of the sample space.
(a) By the total probability rule,
P
(
D
)
=
P
(
Y
)
P
(
D
|
Y
) +
P
(
M
)
P
(
D
|
M
) +
P
(
S
)
P
(
D
|
S
)
=
(0
.
22)
×
(0
.
04) + (0
.
32)
×
(0
.
07) + (0
.
46)
×
(0
.
12)
=
0
.
0864
(b) By Bayes rule,
P
(
S
|
D
) =
P
(
S
)
P
(
D
|
S
)
P
(
D
)
=
(0
.
46)(0
.
12)
0
.
0864
= 0
.
6389
4. We are interested in the frizzle character of fowls. A frizzled fowl has genotype FF, a normal fowl
has genotype ff and a fowl with genotype Ff is slightly frizzled. This is an example of codominant
alleles.
Both parents are slightly frizzled.
Use the tree diagram to illustrate all the possible
genotypes and phenotypes of the offspring and the associated probabilities.
6
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a) (5 marks) What is the probability that their child is slightly frizzled?
b) (5 marks) If this couple has 4 children, what is the probability that exactly 3 of their children
are normal?
Solution:
a) Both man and woman have the genotype
Ff
. We draw the tree diagram:
F
F
‘
‘
‘
‘‘
f
‘
‘
‘
‘
‘
f
F
‘
‘
‘
‘‘
f
Female
Gamete
Male
Gamete
Offspring
Genotype
FF
Ff
fF
ff
Offspring
Phenotype
frizzled
slightly frizzled
slightly frizzled
normal
Probability
1
/
4
1
/
4
1
/
4
1
/
4
The child is slightly frizzled if the genotype is
Ff
or
fF
. The probability that the child is slightly
frizzled is
1
/
4 + 1
/
4 = 1
/
2
.
b) Let
X
be the number of children who are normal, among the 4.
Then
X
has a binomial
distribution with
n
= 4
trials and probability of success
p
= 0
.
25
. The probability is
P
(
X
= 3) =
4
3
(0
.
25)
3
(0
.
75)
1
= 0
.
046875
.
7
Table 18.1 Binomial Coefficients
(
n
k
)
k
n
0
1
2
3
4
5
1
1
1
2
1
2
1
3
1
3
3
1
4
1
4
6
4
1
5
1
5
10
10
5
1
6
1
6
15
20
15
6
7
1
7
21
35
35
21
8
1
8
28
56
70
56
9
1
9
36
84
126
126
10
1
10
45
120
210
252
11
1
11
55
165
330
462
12
1
12
66
220
495
792
13
1
13
78
286
715
1,287
14
1
14
91
364
1,001
2,002
15
1
15
105
455
1,365
3,003
16
1
16
120
560
1,820
4,368
17
1
17
136
680
2,380
6,188
18
1
18
153
816
3,060
8,568
19
1
19
171
969
3,876
11,628
20
1
20
190
1,140
4,845
15,504
k
n
6
7
8
9
10
6
1
7
7
1
8
28
8
1
9
84
36
9
1
10
210
120
45
10
1
11
462
330
165
55
11
12
924
792
495
220
66
13
1,716
1,716
1,287
715
286
14
3,003
3,432
3,003
2,002
1,001
15
5,005
6,435
6,435
5,005
3,003
16
8,008
11,440
12,870
11,440
8,008
17
12,376
19,448
24,310
24,310
19,448
18
18,564
31,824
43,758
48,620
43,758
19
27,132
50,388
75,582
92,378
92,378
20
38,760
77,520
125,970
167,960
184,756
8