2379C-midterm1-23B Solution
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MAT 2379C
Midterm Examination
Version B
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
B
2
E
3
E
4
A
5
C
6
A
1. Consider the following R output:
> pbinom(20,50,0.3)
[1] 0.9522362
> pbinom(19,50,0.3)
[1] 0.9151974
> pbinom(18,50,0.3)
[1] 0.8594401
> pbinom(15,50,0.3)
[1] 0.5691784
> pbinom(14,50,0.3)
[1] 0.4468316
> dbinom(15,50,0.3)
[1] 0.1223469
Let
X
be a binomial random variable with
n
= 50
and
p
= 0
.
3
.
Using the R output above,
calculate
P
(15
< X
≤
19)
.
A) 0.4684
B) 0.3460
C) 0.2903
D) 0.9445
E) 0.9187
Solution:
We have
P
(15
< X
≤
19) =
P
(
X
= 16) +
P
(
X
= 17) +
P
(
X
= 18) +
P
(
X
= 19) =
P
(
X
≤
19)
-
P
(
X
≤
15) = 0
.
9151974
-
0
.
5691784 = 0
.
3460190
The answer is B.
2. A simple urine test was developed for a particular disease.
A study involved 250 patients with
the disease and 150 patients without the disease. Among the patients with the disease 245 had a
positive result, while there were only 10 positive results among the subjects without the disease.
Compute the sensitivity of the test.
A) 0.68
B) 0.46
C) 0.29
D) 0.37
E) 0.98
2
True +
True
-
Total
Test +
245
10
255
Test
-
5
140
145
Total
250
150
400
Solution:
The test results are summarized in the table below:
sensitivity
=
P
(Test +
|
True+) =
P
(
{
Test+
} ∩ {
True+
}
)
P
(True+)
=
245
/
400
250
/
400
= 0
.
98
The answer is E.
3. Let
X
be a discrete random variable with values
0
,
1
,
2
,
3
,
4
. The cumulative distribution function
of
X
is given by the table below:
x
0
1
2
3
4
P
(
X
≤
x
)
0.5
0.6
0.7
0.9
1
Find the expected value of
X
and the variance of
X
.
A)
E
(
X
) = 1
.
3
,
Var(
X
) = 0
.
61
B)
E
(
X
) = 0
.
7
,
Var(
X
) = 2
.
21
C)
E
(
X
) = 1
.
3
,
Var(
X
) = 0
.
61
D)
E
(
X
) = 0
.
5
,
Var(
X
) = 1
.
1
E)
E
(
X
) = 1
.
3
,
Var(
X
) = 2
.
21
Solution:
First, we find the table of the probability mass function of
X
:
x
0
1
2
3
4
P
(
X
=
x
)
0.5
0.1
0.1
0.2
0.1
E
(
X
) = 0(0
.
5) + 1(0
.
1) + 2(0
.
1) + 3(0
.
2) + 4(0
.
1) = 1
.
3
Var(
X
) = 0
2
(0
.
5) + 1
2
(0
.
1) + 2
2
(0
.
1) + 3
2
(0
.
2) + 4
2
(0
.
1)
-
(1
.
3)
2
= 3
.
9
-
1
.
69 = 2
.
21
The answer is E.
4. The October 2013 issue of the National Geographic Magazine features the survival story of the
Kihanzi spray toad from Tanzania, which was saved from extinction due to conservation efforts.
In 2009, the species was declared extinct in the wild.
Luckily, 500 toads had been captured in
2004, half of them being sent to the Bronx Zoo, and the other half to the Toledo Zoo. Some of
these toads became infected with a deadly fungus which was devastating amphibian populations
worldwide.
Assume that the infection rate is 85.7% among the spray toads at the Bronx Zoo,
and 73.3% at the Toledo Zoo. We select randomly a toad among the 500. What is the proba-
bility that it is infected? Is the fungus infection of the spray toads independent of the zoo location?
A) 0.795, no
B) 0.795, yes
C) 0.5, no
D) 0.5, yes
E) not enough information given
3
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Solution:
Let
I
be the event that a randomly chosen toad is infected. Let
B
be the event that the
toad comes from the Bronx Zoo and t
T
the event that the toad comes from the Toledo Zoo. We
know that
P
(
B
) =
P
(
T
) = 0
.
5
,
P
(
I
|
B
) = 0
.
857
and
P
(
I
|
T
) = 0
.
733
. By the total probability
rule,
P
(
I
) =
P
(
I
|
B
)
P
(
B
) +
P
(
I
|
T
)
P
(
T
) = (0
.
857)(0
.
5) + (0
.
733)(0
.
5) = 0
.
795
Since
P
(
I
)
6
=
P
(
I
|
B
)
,
I
is not independent of
B
. The answer is A.
5. 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 25% 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) 0.6
D) 0.7
E) 0.5
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
.
25
. Hence
P
(
A
∪
B
) = 1
-
P
(
A
0
∩
B
0
) = 1
-
0
.
25 = 0
.
75
. 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
.
75 = 0
.
15
The desired probability is
P
(
A
∩
B
0
) +
P
(
A
0
∩
B
) =
P
(
A
∪
B
)
-
P
(
A
∩
B
) = 0
.
75
-
0
.
15 = 0
.
6
The answer is C.
6. In humans, the eye color is determined by a gene whose allele for brown eyes is dominant over the
allele for blue eyes. A man and a woman have brown eyes, and are heterozygous for this gene.
They have four children. Find the probability that
at most
one of their children has blue eyes.
A) 0.7383
B) 0.3164
C) 0.4218
D) 0.6835
E) 0.2617
Solution:
From the class notes, we know that the probability that a child has blue eyes is 1/4=0.25.
Let
X
be the number of children with blue eyes.
X
has a binomial distribution with
n
= 4
and
p
= 0
.
25
. The desired probability is
P
(
X
≤
1)
=
P
(
X
= 0) +
P
(
X
= 1)
=
4
0
(0
.
25)
0
(1
-
0
.
25)
4
-
0
+
4
1
(0
.
25)
1
(1
-
0
.
25)
4
-
1
=
(0
.
75)
4
+ 4(0
.
25)(0
.
75)
3
= 0
.
7383
The answer is A.
4
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. 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.
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 2 of their children
are frizzled?
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
= 2) =
4
2
(0
.
25)
2
(0
.
75)
2
= 0
.
2109375
.
2. The Rideau Canal connects the Ottawa River with Lake Ontario. In the winter, a section of 7.8
km of the canal which passes through the city of Ottawa is open for public skating, being the
world’s largest naturally frozen skating rink. Typically, it takes 10 to 14 days of continuous cold
temperature (
-
15
degrees Celsius or colder) to form safe ice, which would allow the canal to be
open for public skating. Meteorological data collected during the past 40 years shows that, during
the 120 days of winter season, on average, the canal was open for public skating for 50 days, and
the weather was sunny for 54 days. On average, a winter season would have 33 sunny days when
the canal is open for public skating.
(a) What is the probability that on a randomly chosen day during the winter season, the canal is
open for public skating, but the weather is not sunny?
(b) What is the probability that on a randomly chosen day during the winter season, the weather
is sunny, but the canal is not open for public skating?
5
Solution
Let A be the event that the canal is open, and B be the event that the weather is sunny.
We know that
P
(
A
) =
50
120
, P
(
B
) =
54
120
and
P
(
A
∩
B
) =
33
120
.
(a)
P
(
A
∩
B
0
) =
P
(
A
)
-
P
(
A
∩
B
) = 50
/
120
-
33
/
120 = 17
/
120 = 0
.
142
.
(b)
P
(
A
0
∩
B
) =
P
(
B
)
-
P
(
A
∩
B
) = 54
/
120
-
33
/
120 = 21
/
120 = 0
.
175
.
3. A screening test which measures the level of a prostate specific antigen (PSA) is a commonly used
tool for the detection of prostate cancer. Men with PSA levels greater than 10 (ng/ml) have a
chance of 67% of prostate cancer, whereas men whose PSA levels are between 4 and 10 have
a 25% chance of having prostate cancer.
For those whose PSA levels are below 4, the risk of
developing prostate cancer is only 5%. Suppose that 15% of men have PSA levels greater than
10, 10% of men have PSA levels between 4 and 10, and 75% of men have PSA levels lower than
4.
(a) What is the probability that a randomly chosen man will develop prostate cancer?
(b) What is the probability that a randomly chosen man has a PSA level greater than 10, given
that he has prostate cancer?
Solution
Let F be the event that a randomly chosen man has prostate cancer. Let A be the event
that his PSA level is greater than
10
, B be the event that his PSA level is between
4
and
10
, and C
be the event that his PSA level is smaller than 4. We know that
P
(
F
|
A
) = 0
.
67
,
P
(
F
|
B
) = 0
.
25
and
P
(
F
|
C
) = 0
.
05
. By the total probability rule,
(a) By the total probability rule,
P
(
F
)
=
P
(
F
|
A
)
P
(
A
) +
P
(
F
|
B
)
P
(
B
) +
P
(
F
|
C
)
P
(
C
)
=
(0
.
67)
×
(0
.
15) + (0
.
25)
×
(0
.
10) + (0
.
05)
×
(0
.
75)
=
0
.
163
(b) By Bayes rule,
P
(
A
|
F
) =
P
(
F
|
A
)
P
(
A
)
P
(
F
)
=
0
.
1005
0
.
163
= 0
.
62
.
4. In all major Canadian cities, hospitals are experiencing an increase in the number of patients
admitted 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 8 ICU patients,
(a) (2 marks) no patient have Covid-19?
(b) (3 marks) at least 2 patients have Covid-19?
(c) (5 marks) between 3 and 5, inclusively, have Covid-19?
Solution:
Let
X
be the number of patients with Covid-19 in this room. Then
X
has a binomial
distribution with
n
= 8
trials and probability of success
p
= 0
.
35
. The desired probability is
(a)
P
(
X
= 0) =
8
0
(0
.
35)
0
(0
.
65)
8
= 0
.
0319
6
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(b)
P
(
X
≥
2)
=
1
-
P
(
X
≤
1) = 1
-
P
(
X
= 0)
-
P
(
X
= 1)
=
1
-
8
0
(0
.
35)
0
(0
.
65)
8
-
8
1
(0
.
35)
1
(0
.
65)
7
=
1
-
(0
.
65)
8
-
8(0
.
35)(0
.
65)
7
= 0
.
8309
(c)
P
(3
≤
X
≤
5) =
P
(
X
= 3) +
P
(
X
= 4) +
P
(
X
= 5)
=
8
3
(0
.
35)
3
(0
.
65)
5
+
8
4
(0
.
35)
4
(0
.
65)
4
+
8
5
(0
.
35)
5
(0
.
65)
3
= 0
.
5469
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