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Elementary Statistics
Practice Test 2, Spring 2016
Chapters 3 and 4
Chapter 3
1. On a baseball team, there are infielders and outfielders. Some players are great hitters,
and some players are not great hitters. Chapter 3 Homework (some of) 25-31.*
• Let I = the event that a player in an infielder.
• Let O = the event that a player is an outfielder.
• Let H = the event that a player is a great hitter.
Write the symbols for the following probabilities.
(a) the probability that a player is not an outfielder
(b) the probability that a player is an infielder and is not a great hitter
(c) the probability that a player is an infielder, given that the player is a great hitter
(d) the probability that of all the outfielders, a player is not a great hitter
(e) the probability that a player is an infielder or is not a great hitter
Solution: (a) P(O0 ) (b) P(I AND H0 ) (c) P(I|H) (d) P(H0 |O) (e) P(I OR H0 )
2. U and V are mutually exclusive events. P(U) = 0.26; P(V) = 0.37.
Chapter 3 Homework
42.
(a) P(U AND V) =
(b) P(U|V) =
(c) P(U OR V) =
Solution: (a) 0 (by the definition of mutually exclusive events) (b) 0 (c) 0.63
3. Approximately 281,000,000 people over age five live in the United States. Of these
people, 55,000,000 speak a language other than English at home. Of those who speak
another language at home, 62.3% speak Spanish. Let: E = speaks English at home; S
= speaks Spanish. Finish each probability statement by matching the correct answer.
Chapter 3 Homework 93.
(a) P(E0 ) =
(b) P(E) =
(c) P(S | E0 ) =
Elementary Statistics
Practice Test 2, Spring 2016 - Page 2
Chapters 3 and 4
(d) P(S AND E0 ) =
Solution: (a) 0.196 (b) 0.804 (c) 0.623 (d) 0.122 (=P(E0 )P(S | E0 ))
4. A college finds that 10% of students have taken a distance learning class and that 40%
of students are part time students. Of the part time students, 20% have taken a distance
learning class. Let D = event that a student takes a distance learning class and E =
event that a student is a part time student. Chapter 3 Homework 100.
(a) Find P(D AND E).
(b) Find P(E|D).
(c) Find P(D OR E).
(d) Using an appropriate test, show whether D and E are independent events.
(e) Using an appropriate test, show whether D and E are mutually exclusive events.
(f) Make a Venn diagram of the situation.
Solution: The tricky part of this problem is recognizing that the provided
probability of 0.2 applies to P (D|E).
(a) P (D and E) = P (E and D) = P (E)P (D|E) = (0.4)(0.2) = 0.08
(b) P (E|D) =
P (D and E)
P (D)
=
0.08
0.1
= 0.8
(c) P (D or E) = P (D) + P (E) − P (D and E) = 0.1 + 0.4 − 0.08 = 0.42
(d) P (D|E) = P (D)? No, 0.2 6= 0.1. The events are dependent.
(e) P (D and E) = 0? No, 0.08 6= 0. The events are not mutually exclusive.
(f) Venn diagram:
5. In a previous year, the weights of the players for the San Fransisco 49ers and the Dallas
Cowboys were published in the San Jose Mercury News. The factual data were compiled
into the following table. Chapter 3 Homework 113.
Jersey # ≤ 210 211-250 251-290 > 290
1-33
21
5
0
0
34-66
6
18
7
4
66-99
6
12
22
5
Elementary Statistics
Practice Test 2, Spring 2016 - Page 3
Chapters 3 and 4
For the following, suppose that you randomly select one player from the 49ers or Cowboys.
(a) Find the probability that his shirt number is from 1 to 33.
(b) Find the probability that he weighs at most 210 pounds.
(c) Find the probability that his shirt number is from 1 to 33 AND he weighs at most
210 pounds.
(d) Find the probability that his shirt number is from 1 to 33 OR he weighs at most
210 pounds.
(e) Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs
at most 210 pounds.
Solution: (a)
26
106
(b)
33
106
21
106
(c)
(d)
26
106
+
33
106
−
21
106
=
38
106
(e)
21
33
6. Given events G and H: P(G) = 0.43; P(H) = 0.26; P(H AND G) = 0.14.
Chapter 3 Homework
119.
(a) Make a Venn diagram of the situation.
(b) Find P(H OR G).
(c) Find the probability of the complement of event (H AND G).
(d) Find the probability of the complement of event (H OR G).
(e) Find the probability of event H, given event G.
Solution: (a) Venn diagram:
(b) .55 (c) .86 (d) .45 (e)
.14
.43
= 0.326
7. Suppose that you have eight cards. Five are green and three are yellow. The cards
are well shuffled. Suppose that you randomly draw two cards, one at a time, with
replacement. Let G1 = first card is green, Y1 = first card is yellow, G2 = second card
is green, and Y2 = second card is yellow. Chapter 3 Homework 121.
(a) Draw a tree diagram of the situation.
(b) Find the probability that both cards are green.
(c) Find the probability that at least one card is green.
(d) Find the probability that the second card is green if the first is green.
Elementary Statistics
Practice Test 2, Spring 2016 - Page 4
Chapters 3 and 4
(e) Are G2 and G1 independent events? Explain why or why not.
Solution: (a)
9
(b) 25
(c) = 1−P(both are yellow) = 1 − 64
= 55
(d) 58 (e) Yes. The probability
64
64
of selecting a green card on the second draw is not affected by drawing a green
on the first draw: P(G2) = P(G2|G1) = 58 .
8. Suppose that you have eight cards. Five are green and three are yellow. The cards
are well shuffled. Suppose that you randomly draw two cards, one at a time, without
replacement. Let G1 = first card is green, Y1 = first card is yellow, G2 = second card
is green, and Y2 = second card is yellow. Chapter 3 Homework 122.
(a) Draw a tree diagram of the situation.
(b) Find the probability that both cards are green.
(c) Find the probability that at least one card is green.
(d) Find the probability that the second card is green if the first is green.
(e) Find the probability that the second card is green.
(f) Are G2 and G1 independent events? Explain why or why not.
Solution: (a)
Elementary Statistics
Practice Test 2, Spring 2016 - Page 5
Chapters 3 and 4
6
(c) = 1−P(both are yellow) = 1 − 56
= 50
(d) 47 (e) There are two events
(b) 20
56
56
in the sample space in which the second card is green. Since these events are
mutually exclusive, add the probabilities: P(G2) = P(G1 AND G2) + P(Y1
20
AND G2) = 56
+ 15
= 35
(e) No. The probability of selecting a green card on
56
56
6=
the second draw IS affected by drawing a green on the first draw: P(G2) = 35
56
4
P(G2|G1) = 7 .
Chapter 4
9. Javier volunteers in community events each month. He does not do more than five events
in a month. He attends exactly five events 35% of the time, four events 25% of the time,
three events 20% of the time, two events 10% of the time, one event 5% of the time, and
no events 5% of the time. Chapter 4 Practice 13-17 (edited).
(a) Define the random variable X.
Solution: X = the number of times Javier volunteers in a month
(b) What values does x take on?
Solution: x = {0, 1, 2, 3, 4, 5}
(c) Construct a PDF table.
Elementary Statistics
Solution:
Practice Test 2, Spring 2016 - Page 6
x
0
1
2
3
4
5
Chapters 3 and 4
P(x)
0.05
0.05
0.1
0.2
0.25
0.35
(d) Is this a valid probability distribution? Explain.
Solution: Yes. Each probability is between 0 and 1 and the sum of the probabilities is 1.
(e) Find the probability that Javier volunteers for less than three events each month.
Solution: P (x < 3) = P (x ≤ 2) = P (0) + P (1) + P (2) = 0.2
(f) Find the probability that Javier volunteers for at least one event each month.
Solution: P (x ≥ 1) = P (1) + P (2) + P (3) + P (4) + P (5) = 1 − P (0) = 0.95
(g) How many times per month does Javier volunteer on average?
Solution: This is the mean of the probability distribution:
µ = 0(0.05) + 1(0.05) + 2(0.1) + 3(0.2) + 4(0.25) + 5(0.35) = 3.6 times per month
10. You are playing a game by drawing a card from a standard deck and replacing it. If the
card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face
cards in a deck of 52 cards. Chapter 4 Practice 35 (edited).
(a) Suppose X is the amount that you win or lose when you play the game one time.
What values does x take on?
Solution: x = {−2, 30}
(b) Make a PDF table for X.
Solution:
x
-2
30
P(x)
40/52
12/52
(c) What is the expected gain (or loss) per game in the long run?
Elementary Statistics
Practice Test 2, Spring 2016 - Page 7
Chapters 3 and 4
Solution: This is the expected value of X.
+ (30) 12
= 5.3846
(−2) 40
52
52
The expected gain per game is $5.38.
(d) Is it a good idea to play this game?
Solution: Yes!
11. The Higher Education Research Institute at UCLA collected data from 203,967 incoming
first-time, full-time freshmen from 270 four-year colleges and universities in the U.S.
71.3% of those students replied that, yes, they believe that same-sex couples should
have the right to legal marital status. Suppose that you randomly pick 25 first-time,
full-time freshmen from the survey. You are interested in the number that believes that
same sex-couples should have the right to legal marital status. Chapter 4 Practice 37-44 (edited).
(a) In words, define the random variable X.
Solution: X = the number of freshmen who answer “yes” out of 25
(b) Describe a “success”.
Solution: A “success” is a surveyed freshman answering “yes”, because this is
the outcome that the random variable X is counting.
(c) What is the distribution of X?
Solution: X ∼ B(25, 0.713)
(d) How would you describe the shape of a histogram of a “perfect” data set of values
from X?
Solution: The binomial distribution is symmetrical and bell shaped, like the
normal distribution, with the greatest concentration of data values near the
mean.
(e) What values does the random variable X take on?
Solution: x = {0, 1, 2, . . . , 24, 25}
(f) On average, about how many of the freshmen would you expect to answer “yes”?
(Round to the nearest whole number of students.)
Elementary Statistics
Practice Test 2, Spring 2016 - Page 8
Chapters 3 and 4
Solution: µ = (25)(0.713) ≈ 18
(g) What is the standard deviation of X? (Round to one decimal place.)
Solution: q = 1 − 0.713 = 0.287 σ =
p
25(0.713)(0.287) ≈ 2.3
(h) Would it be unusual for 12 of the freshmen to answer yes? Explain.
Solution: Yes. Twelve is less than the minimum number of “yes” responses,
which is about 18 − 2(2.3) ≈ 13.
(i) What is the probability that exactly 19 of the freshmen reply “yes”?
Solution: P (x = 19) = binompdf(25, 0.713, 19) ≈ 0.160
(Use three decimal places for probabilities.)
(j) What is the probability that at most 15 of the freshmen reply “yes”?
Solution: P (x ≤ 15) = binomcdf(25, 0.713, 15) ≈ 0.152
(k) What is the probability that at least 20 of the freshmen reply “yes”?
Solution: P (x ≥ 20) = 1 − P (x ≤ 19) = 1 − binomcdf(25, 0.713, 19) ≈ 0.234
12. According to The World Bank, only 9% of the population of Uganda had access to
electricity as of 2009. Chapter 4 Homwork 34 (edited).
(a) Find the probability that in a 2009 random sample of 150 people from Uganda, at
least 26 have access to electricity.
Solution: First we must recognize that this is a binomial scenario with X =
the number of Ugandans out of 150 who have acces to electricity, and that
X ∼ B(150, 0.09).
P (x ≥ 26) = 1 − binomcdf(150, 0.09, 25) ≈ 9.05 × 10−4 ≈ 0.001
Caution: It is important to pay attention to the “E-4” in the calculator display.
I often see students ignore that part of the display and answer with “9.05”, which
makes no sense as a probability because probabilities must be between 0 and 1.
(b) Suppose that in 2015, a random sample of 150 people in Uganda contains 26 who
have access to electricity. Is there statistical evidence that access to electricity has
improved since 2009? Explain.
Elementary Statistics
Practice Test 2, Spring 2016 - Page 9
Chapters 3 and 4
Solution: Yes. As we saw in part (a), it would be very rare (only 1 in 1000
chance) to find so many people in the sample who have access to electricity,
assuming that the true percentage of the population with electricity is 9%.
Since the sample doesn’t jive with the assumption, the assumption is most
likely wrong. This means that there is statisitical evidence that the percentage
of the population who have access to electricity in 2015 is higher than 9%.
*Problems are from Barbara Illowsky & Susan Dean. “Introductory Statistics.” OpenStax College, 2013. iBooks. https://itun.es/us/kFeL1.l