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Econ 250 Winter 2009
Assignment 1
Due at Midterm February 11, 2009
There are 9 questions with each one worth 10 marks.
1. The time (in seconds) that a random sample of employees took to
complete a task is
23
12
37
27
35
40
20
16
14
27
29
40
37
13
49
20
28
26
40
13
45
25
13
66
a. Find the mean time, and the standard deviation.
b. Plot the data in a histogram with 3, 6, and 9 categories. Is the
distribution skewed?
c. Find the five-number summary and draw the box-plot.
Solution:
a. Mean is 29.0 seconds and the standard deviation is 13.6 seconds.
b. See Figures 1, 2, 3. The distribution is right-skewed.
c. Max = 66, Min = 12, Q1 = 16, Q2 = 27, Q3 = 37. The box-plot
is shown below:
60
50
40
30
20
1
2. Verify the following using Venn diagrams:
a. Ā ∪ B̄ = A ∩ B
b. A ∪ B = Ā ∩ B̄
c. Ā ∪ B̄ = A ∩ B
Solution:
3. Two fair 6-sided dice are rolled.
a. What is the sample space?
b. What is the event “the sum of the faces is ≤5”?
c. What is the event “the sum of the two faces is an even number”?
d. What is the event “the difference of the two faces is equal to 3”?
Solution:
a. The sample space consists of all pairs of outcomes:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), ..., (6, 4), (6, 5), (6, 6)}
There are 36 pairs in total.
b. The event “less than or equal to 5” is the union of the events
“sum=1”, “sum=2”, “sum=3”, “sum=4”, “sum=5.” The event
is as follows:
E1 = {(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 2), (1, 4), (4, 1), (2, 3), 3, 2)}.
c. The sum range from 2 to 12. So we are looking for the event that
the sum of the two numbers is either 2, 4, 6, 8, 10 or 12. The event
is as follows: E2 = {(1, 1), (1, 3), (3, 1), (2, 2), (1, 5), (5, 1), (2, 4), (4, 2),
(3, 3), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4), (4, 6), (6, 4), (5, 5), (6, 6)}.
d. The event is as follows: E3 = {(1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3)}.
4. In a bolt factory, 30, 50, and 20% of production is manufactured by
machines I, II and III, respectively. If 4, 5 and 3% of the output of
these respective machines is defective, what is the probability that a
randomly selected bolt that is found to be defective is manufactured
by machine III?
2
Solution: This is simple extension to the example done in class and
of course an application of Bayes’s rule. We are interested in the probability that a random defective bolt is manufactured by machine III.
Denote this by P (MIII |D). Then, by Bayes’s rule we have:
P (MIII |D) =
P (MIII ) × P (D|MIII )
P (D)
We have the following information: P (D|MIII ) is just the fraction of
defective bolt produced by machine III, so P (D|MIII ) = 0.03. Moreover, P (MIII )isjust0.2. To find P (D) we note that producing bolts at
the different machines are mutually exclusive (can’t produce the bolt
simultaneously by two machines) and collectively exhaustive (there are
only three machines). Therefore,
P (D) = P (D|MI )P (MI ) + P (D|MII )P (MII ) + P (D|MIII )P (MIII )
= 0.04 × 0.3 + 0.05 × 0.5 + 0.03 × 0.2
= 0.043
Hence, P (MIII |D) =
0.03×0.2
0.043
= 0.14.
5. Cyrus and 27 other students are taking a course in probability this
semester. If their professor chooses eight students at random and with
replacement to ask them eight different questions, what is the probability that one of them is Cyrus?
Solution: Let P (C) denote the probability that Cyrus is one of them.
Then using the complement rule we know that
P (C) = 1 − P (Cyrus is not chosen)
Since the professor selects students with replacement, successive selections are independent. That is, the probability of not choosing Cyrus
. Hence, the probability of not
each time is the same and equal to 27
28
27 8
) . So,
selecting Cyrus 8 times in a row is just ( 28
P (C) = 1 − (
27 8
) = 1 − 0.748 = 0.252.
28
6. The chairperson of the industry-academic partnership of Kingston invites all 12 members of the board and their spouses to his house for a
3
Christmas party. If a board member may attend without his spouse,
but not vice-versa, how many different groups can the chairperson get?
Solution: Denote the individual board members M1 , M2 , . . . , M12 and
their wives by W1 , . . . W12 , respectively. Then, the ith board member
can show up by himself (Mi ) or with his wife ((M1 , Wi ) or (Wi , Mi )).
That is, there are three ways each board member and his wife can show
up. There are 12 board members and they show up independently
of each other. Therefore, the total number of different groups the
chairperson can get is 312 = 531441.
7. In a lottery the tickets are numbered 1 through N . A person purchases
n (1 ≤ n ≤ N ) tickets at random. What is the probability that the
ticket numbers are consecutive?
Solution: Let x be the number of ways n consecutive numbers can
be drawn from the first N integers. The number
of ways of drawing
any n numbers from the first N integers is Nn . Therefore the relevant
probability is Nx . Now, how many consecutive sequences of n numbers
(n)
are there in first N integers? Well, we can enumerate them as follows:
1st sequence: 1, 2, 3, . . . , n
2nd sequence: 2, 3, 4, . . . , n + 1
3rd sequence: 3, 4, 5, . . . , n + 2
......
Last sequence: N − n + 1, N − n + 2, N − n + 3, . . . , N
As the above argument suggests, there are N − n + 1 sequences of n
consecutive integers. Therefore, the probability that the ticket numbers
are consecutive is N −n+1
.
(Nn )
8. Suppose that jury members decide independently, and that each with
probability p makes the correct decision. If the decision of the majority
is final, which is preferable: a three-person jury or a single juror?
Solution: Let X denote the number of persons who decide correctly
among a three-person jury. Then X is a binomial random variable
4
with n = 3 and success probability p. Hence, the probability that a
three-person jury decides correctly is
P (X ≥ 2) = P (X = 2) + P (X = 3)
3 2
3 3
=
p (1 − p) +
p (1 − p)0
2
3
2
3
= 3p (1 − p) + p
= 3p2 − 2p3
Since the probability is p that a single juror decides correctly, a threeperson jury is preferable to a single juror if and only if
3p2 − 2p3 > p
This is equivalent to 3p − 2p2 > 1, so −2p2 + 3p − 1 > 0.
But −2p2 + 3p − 1 = 2(1 − p)(p − 1/2) > 0 only if p > 1/2 as 1 − p
is always greater than 0. Hence a three-person jury is preferable if
p > 1/2. If p < 1/2, the decision of a single juror is preferable. For
p = 1/2 there is no difference.
9. If X is a random number selected from the first 10 positive integers,
what is E[X(11 − X)]? What is V ar[X(11 − X)]?
Solution: One might be tempted to approach the problem algebraically.
That is, by expanding the expectation as follows:
E(X(11 − X)) = E(11X) − E(X 2 )
= 11 × E(X) − E(X 2 )
and then computing E(X) and E(X 2 ) for the random variable in question. However, while this works, it is tedious since you need to carry
out the calculations for E(X 2 ) and in the second part of the question,
for V (X 2 ). A somewhat less tedious approach is to simply calculate
the realized values of the random variable X(11 − X):
5
X X(11-X)
1
10
2
18
3
24
4
28
5
30
6
30
7
28
8
24
9
18
10
10
Hence, computing the expectation we have:
E(X(11 − X) =
1
10
× 10 +
1
10
× 18 + · · · +
1
10
× 10 = 22.
Similarly,
1
1
1
V ar(X(11 − X) = 10−1
× (10 − 22)2 + 10−1
× (18 − 22)2 + · · · + 10−1
×
2
(10 − 22) = 58.67.
Bonus Question - 5 Marks
10. What is the probability that a student chosen at random in our class
will do this assignment? Recall, the assignment is NOT mandatory.
Solution: The relevant probability is subjective as repeated trials cannot be conducted with the current class. Any answer between 0 and 1
is valid with the subjective justification.
6
Frequency
15
7
2
12
30
48
66
Figure 1: 3 Categories
7
Frequency
8
7
4
3
1
12
21
30
39
48
57
66
Figure 2: 6 Categories
8
Frequency
6
5
3
1
12
18
24
30
36
42
48
54
60
66
Figure 3: 9 Categories
9