Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319013387
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4.5, Problem 114E
To determine

To graph: The Venn diagram that shows the relation between the events A and B in Exercise 4.112.

Expert Solution
Check Mark

Explanation of Solution

Calculation: The eventA” is defined as the event that the randomly selected American household is prosperous and the event “B” is defined as the event that the randomly selected American household is educated. The probabilities are provided as P(A)=0.138, P(B)=0.261, and P(Aand B), which can be represented as P(AB), is 0.082.

Graph: The Venn diagram indicating the above probabilities and depicting the events {A},{B},{AB},{ACB},{ABC},and{ACBC} in the universal set U is as follows:

Introduction to the Practice of Statistics, Chapter 4.5, Problem 114E

(a)

To determine

The probability for the event {AandB}.

Expert Solution
Check Mark

Answer to Problem 114E

Solution: The required probability is 0.082.

Explanation of Solution

From the Venn diagram, the probability of event {AandB} is ascertained as follows.

P(Aand B)=P(AB)=0.082

Hence, the required probability is 0.082.

To determine

To explain: THE event {AandB} in words.

Expert Solution
Check Mark

Answer to Problem 114E

Solution: The event {AandB} means that the household is prosperous (A) as well as educated (B).

Explanation of Solution

The event {AandB} means the intersection of A and B, that is, the household is prosperous (A) as well as educated (B).

(b)

To determine

The probability for the event {ACandB}.

(b)

Expert Solution
Check Mark

Answer to Problem 114E

Solution: The required probability is 0.179.

Explanation of Solution

From the Venn diagram, the probability of event {ACandB} is ascertained as follows.

P(ACand B)=P(ACB)=P(B)P(AB)=0.2610.082=0.179

Hence, the required probability is 0.179.

To determine

To explain: The event {ACandB} in words.

Expert Solution
Check Mark

Answer to Problem 114E

Solution: The event {ACandB} means that the household is educated (B) but not prosperous (A).

Explanation of Solution

The event {ACandB} means the intersection of B and the complement of A, that is, the household is educated (B) but not prosperous (A).

(c)

To determine

The probability for the event {AandBC}.

(c)

Expert Solution
Check Mark

Answer to Problem 114E

Solution: The required probability is 0.056.

Explanation of Solution

From the Venn diagram, the probability of event {AandBC} is ascertained as follows.

P(Aand BC)=P(ABC)=P(A)P(AB)=0.1380.082=0.056

Hence, the required probability is 0.056.

To determine

To explain: The event {AandBC} in words.

Expert Solution
Check Mark

Answer to Problem 114E

Solution: The event {AandBC} means that the household is prosperous (A) but not educated (B).

Explanation of Solution

The event {AandBC} means the intersection of A and the complement of B, that is, the household is prosperous (A) but not educated (B).

(d)

To determine

The probability for the event {ACandBC}.

(d)

Expert Solution
Check Mark

Answer to Problem 114E

Solution: The required probability is 0.683.

Explanation of Solution

From the Venn diagram, the probability of event {ACandBC} is ascertained as follows.

P(ACand BC)=P(ACBC)=1P(Aor B)=1P(A B)=10.317

=0.683

Hence, the required probability is 0.683.

To determine

To explain: The event {ACandBC} in words.

Expert Solution
Check Mark

Answer to Problem 114E

Solution: The event {ACandBC} means that the household is neither prosperous (A) nor educated (B).

Explanation of Solution

The event {ACandBC} means the intersection of complement of A and the complement of B, that is, the household is neither prosperous (A) nor educated (B).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 4 Solutions

Introduction to the Practice of Statistics

Ch. 4.2 - Prob. 11UYKCh. 4.2 - Prob. 12UYKCh. 4.2 - Prob. 13UYKCh. 4.2 - Prob. 14UYKCh. 4.2 - Prob. 15UYKCh. 4.2 - Prob. 16UYKCh. 4.2 - Prob. 17ECh. 4.2 - Prob. 18ECh. 4.2 - Prob. 19ECh. 4.2 - Prob. 20ECh. 4.2 - Prob. 21ECh. 4.2 - Prob. 22ECh. 4.2 - Prob. 23ECh. 4.2 - Prob. 24ECh. 4.2 - Prob. 25ECh. 4.2 - Prob. 26ECh. 4.2 - Prob. 27ECh. 4.2 - Prob. 28ECh. 4.2 - Prob. 29ECh. 4.2 - Prob. 30ECh. 4.2 - Prob. 31ECh. 4.2 - Prob. 32ECh. 4.2 - Prob. 33ECh. 4.2 - Prob. 34ECh. 4.2 - Prob. 35ECh. 4.2 - Prob. 36ECh. 4.2 - Prob. 37ECh. 4.2 - Prob. 38ECh. 4.2 - Prob. 39ECh. 4.2 - Prob. 40ECh. 4.2 - Prob. 41ECh. 4.3 - Prob. 42UYKCh. 4.3 - Prob. 43UYKCh. 4.3 - Prob. 44UYKCh. 4.3 - Prob. 45ECh. 4.3 - Prob. 46ECh. 4.3 - Prob. 47ECh. 4.3 - Prob. 48ECh. 4.3 - Prob. 49ECh. 4.3 - Prob. 50ECh. 4.3 - Prob. 51ECh. 4.3 - Prob. 52ECh. 4.3 - Prob. 53ECh. 4.3 - Prob. 54ECh. 4.3 - Prob. 55ECh. 4.3 - Prob. 56ECh. 4.3 - Prob. 57ECh. 4.3 - Prob. 58ECh. 4.3 - Prob. 59ECh. 4.3 - Prob. 60ECh. 4.3 - Prob. 61ECh. 4.3 - Prob. 62ECh. 4.4 - Prob. 63UYKCh. 4.4 - Prob. 64UYKCh. 4.4 - Prob. 65UYKCh. 4.4 - Prob. 66UYKCh. 4.4 - Prob. 67UYKCh. 4.4 - Prob. 68ECh. 4.4 - Prob. 69ECh. 4.4 - Prob. 70ECh. 4.4 - Prob. 71ECh. 4.4 - Prob. 72ECh. 4.4 - Prob. 73ECh. 4.4 - Prob. 74ECh. 4.4 - Prob. 75ECh. 4.4 - Prob. 76ECh. 4.4 - Prob. 77ECh. 4.4 - Prob. 78ECh. 4.4 - Prob. 79ECh. 4.4 - Prob. 80ECh. 4.4 - Prob. 81ECh. 4.4 - Prob. 82ECh. 4.4 - Prob. 83ECh. 4.4 - Prob. 84ECh. 4.4 - Prob. 85ECh. 4.4 - Prob. 86ECh. 4.4 - Prob. 87ECh. 4.4 - Prob. 88ECh. 4.5 - Prob. 89UYKCh. 4.5 - Prob. 90UYKCh. 4.5 - Prob. 91UYKCh. 4.5 - Prob. 92UYKCh. 4.5 - Prob. 93UYKCh. 4.5 - Prob. 94UYKCh. 4.5 - Prob. 95UYKCh. 4.5 - Prob. 96ECh. 4.5 - Prob. 97ECh. 4.5 - Prob. 98ECh. 4.5 - Prob. 99ECh. 4.5 - Prob. 100ECh. 4.5 - Prob. 101ECh. 4.5 - Prob. 102ECh. 4.5 - Prob. 103ECh. 4.5 - Prob. 104ECh. 4.5 - Prob. 105ECh. 4.5 - Prob. 106ECh. 4.5 - Prob. 107ECh. 4.5 - Prob. 108ECh. 4.5 - Prob. 109ECh. 4.5 - Prob. 110ECh. 4.5 - Prob. 111ECh. 4.5 - Prob. 112ECh. 4.5 - Prob. 113ECh. 4.5 - Prob. 114ECh. 4.5 - Prob. 115ECh. 4.5 - Prob. 116ECh. 4.5 - Prob. 117ECh. 4.5 - Prob. 118ECh. 4.5 - Prob. 119ECh. 4.5 - Prob. 120ECh. 4.5 - Prob. 121ECh. 4.5 - Prob. 122ECh. 4.5 - Prob. 123ECh. 4 - Prob. 124ECh. 4 - Prob. 125ECh. 4 - Prob. 126ECh. 4 - Prob. 127ECh. 4 - Prob. 128ECh. 4 - Prob. 129ECh. 4 - Prob. 130ECh. 4 - Prob. 131ECh. 4 - Prob. 132ECh. 4 - Prob. 133ECh. 4 - Prob. 134ECh. 4 - Prob. 135ECh. 4 - Prob. 136ECh. 4 - Prob. 137ECh. 4 - Prob. 138ECh. 4 - Prob. 139E
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
The Fundamental Counting Principle; Author: AlRichards314;https://www.youtube.com/watch?v=549eLWIu0Xk;License: Standard YouTube License, CC-BY
The Counting Principle; Author: Mathispower4u;https://www.youtube.com/watch?v=qJ7AYDmHVRE;License: Standard YouTube License, CC-BY