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