SOLUTIONS TO CHAPTER 4 HOMEWORK
Exercise 2. Suppose that you are offered the following “deal”. You roll a die. If
you roll a 6, you win $10. If you roll a 4 or 5, you win $5. If you roll a 1, 2, or 3,
you pay $6.
a. What are you ultimately interested in here (the value of the roll or the
money you win)?
Solution: The money you win.
b. In words, define the Random Variable X.
Solution: X = the number of dollars won or lost.
c. List the values that X may take on. Solution: X ∈ {−6, 5, 10}
d. Construct a PDF. Solution:
P (X = −6) = 3/6 = 1/2
P (X = 5) = 2/6 = 1/3
P (X = 10) = 1/6
e. Over the long run of playing this game, what are your expected average
winnings per game?
Solution: You should expect to lose $1.33 per game. This is found by
calculating the expected value of X:
µ = −6(3/6) + 5(2/6) + 10(1/6) = 2/6 ≈ 0.33
f. Based on numerical values, should you take the deal? Explain your decision
in complete sentences.
Solution: This looks like a pretty good deal to me because you should
win about 33 cents per game. But, then again is that really worth your
time?
Exercise 4. A theater group holds a fund-raiser. It sells 100 raffle tickets for $5
apiece. Suppose you purchase 4 tickets. The prize is 2 passes to a Broadway show,
worth a total of $150.
a. What are you interested in here? Solution: The net profit.
b. In words, dene the Random Variable X. Solution: X = net profit
c. List the values that X may take on. Solution: S = {−20, 130}
d. Construct a PDF. Solution:
The probability of winning is simply the proportion of tickets you hold.
P (X = −20) = 0.96
P (X = 130) = 0.04
e. If this fund-raiser is repeated often and you always purchase 4 tickets, what
would be your expected average winnings per game?
Solution: We should expect to lose $ 14 per game because
µX = (0.04)(130) + (0.96)(−20) = −14
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Exercise 6. Suppose that the PDF for the number of years it takes to earn a
Bachelor of Science (B.S.) degree is given below.
a. In words, define the Random Variable X.
X P (X = x)
Solution: X = the number of years taken to
3
0.05
complete a B.S.
4
0.40
b. What does it mean that the values 0, 1, and 2 are
5
0.30
not included for X on the PDF?
6
0.15
Solution: The probability is 0 that someone com7
0.10
pletes the degree in less than 3 years.
c. On average, how many years do you expect it to take for an individual to
earn a B.S.?
Solution: This is the expected value of X:
µ = 3(.05) + 4(.4) + 5(.3) + 6(.15) + 7(.1) = 4.85
Exercise 8. According to a 2003 publication by Waits and Lewis (source:
http://nces.ed.gov/pubs2003/2003017.pdf ), by the end of 2002, 92% of U.S. public
two- year colleges offered distance learning courses. Suppose you randomly pick 13
U.S. public two-year colleges. We are interested in the number that offer distance
learning courses.
a. In words, dene the Random Variable X. Solution: X = the number of 2
year colleges in the sample that offer distance learning courses.
b. List the values hat X may take on. Solution: S = {0, 1, 2, . . . , 13}
c. Give the distribution of X. Solution: X ∼ B(13, .92)
d. On average, how many schools would you expect to offer such courses?
Solution: µX = np = 13(0.92) = 11.96
e. Find the probability that at most 6 offer such courses.
Solution: binomcdf(13, 0.92, 6) = 0.00002
f. Is it more likely that 0 or that 13 will offer such courses? Use numbers to
justify your answer numerically and answer in a complete sentence. Solution: It should be intuitively clear that 13 is more likely and this is
supported by the calculations below from a calculator.
binompdf(13, 0.92, 0) = 6 × 10−15
binompdf(13, 0.92, 13) = 0.3383
Exercise 12. Approximately 8% of students at a local high school participate in
after-school sports all four years of high school. A group of 60 seniors is randomly
chosen. Of interest is the number that participated in after-school sports all four
years of high school.
a. In words, define the Random Variable X. Solution: X is the number of
students in the sample of 60 students who participate in after-school sports
all four years of high school.
b. List the values hat X may take on. Solution: {0, 1, 2, 3, . . . , 60}
c. Give the distribution of X. Solution: X ∼ B(60, .08)
d. How many seniors are expected to have participated in after-school sports
all four years of high school? Solution: np = (60)(.08) = 4.8
e. Based on numerical values, would you be surprised if none of the seniors
participated in after-school sports all four years of high school? Justify your
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answer numerically. Solution: The probability that none participate is
(1 − p)n = (.92)60 = .0067, so we should be surprised if no one participates.
f. Based upon numerical values, is it more likely that 4 or that 5 of the seniors
participated in after-school sports all four years of high school? Justify your
answer numerically. Solution: The relevant probabilities are
60
60
P (X = 4) =
(.08)4 (.92)58 ≈ .1873 P (X = 5) =
(.08)5 (.92)55 ≈ .1824
4
5
so it is more likely that 4 will participate. . . but not much more likely.
Exercise 14. There are two games played for Chinese New Year and Vietnamese
New Year. They are almost identical. In the Chinese version, fair dice with numbers
1, 2, 3, 4, 5, and 6 are used, along with a board with those numbers. In the
Vietnamese version, fair dice with pictures of a gourd, fish, rooster, crab, crayfish,
and deer are used. The board has those six objects on it, also. We will play with
bets being $1. The player places a bet on a number or object. The house rolls three
dice. If none of the dice show the number or object that was bet, the house keeps
the $1 bet. If one of the dice shows the number or object bet (and the other two
do not show it), the player gets back his $1 bet, plus $1 profit. If two of the dice
show the number or object bet (and the third die does not show it), the player gets
back his $1 bet, plus $2 profit. If all three dice show the number or object bet, the
player gets back his $1 bet, plus $3 profit.
Let X = number of matches and Y = profit per game.
a. In words, define the Random Variable X. Solution: X = the number of
matches between the rolled dice and your chosen number (picture).
b. List the values that X may take on. Solution: x ∈ {0, 1, 2, 3}
c. GIve the distribution of X. Solution: X ∼ B(3, 1/6)
d. List the values that Y may take on. Then, construct one PDF table that
includes both X and Y and their probabilities.
X Y P (X = x) = P (Y = y) binompdf(3, 1/6, x)
0 −1
(5/6)3
.579
1
1
3(1/6)(5/6)2
.347
2
2
3(1/6)2 (5/6)
.069
3
3
(1/6)3
.005
e. Calculate the average expected matches over the long run of playing this
game for the player.
Solution: This is the expected value of X: µ = np = 3(1/6) = 1/2.
f. Calculate the average expected earnings over the long run of playing this
game for the player.
Solution: This is the monetary value of the expected value of Y . The
answer turns out to be a loss of about 8 cents per game because
µ = −1(.579) + 1(.347) + 2(.069) + 3(.005) = −.079
g. Determine who has the advantage, the player or the house.
Solution: Not surprisingly, the house has the advantage and should make
money in the long run.
Exercise 22. The chance of a IRS audit for a tax return with over $25,000 in
income is about 2% per year. We are interested in the expected number of audits a
person with that income has in a 20 year period. Assume each year is independent.
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a. In words, dene the Random Variable X. Solution: X = the number of
audits a person with income over $25,000 per annum has in a 20 year period.
b. List the values that X may take on. Solution: S = {0, 1, 2, . . . , 19, 20}
c. Give the distribution of X.
Solution: X ∼ B(20, 0.02) or we could use the mean value calculated
below and approximate with the Poisson distribution X ∼ P (.4).
d. How many audits are expected in a 20 year period?
Solution: µX = np = 20(0.02) = .4
e. Find the probability that a person is not audited at all.
Solution: (.98)20 or binompdf(20, 0.02, 0) = 0.668 or poissonpdf(.4, 0) =
.670.
f. Find the probability that a person is audited more than twice.
Solution: 1− binomcdf(20, 0.02, 2) = 0.007 or 1−poissoncdf(.4, 2) = .008.
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