2379C-midterm1-23A Solution
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Jan 9, 2024
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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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