Understandable Statistics: Concepts and Methods
Understandable Statistics: Concepts and Methods
12th Edition
ISBN: 9781337119917
Author: Charles Henry Brase, Corrinne Pellillo Brase
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4.2, Problem 29P

a.

To determine

Find the values of P(S), P(S/A), and P(S/Pa).

a.

Expert Solution
Check Mark

Answer to Problem 29P

The value of P(S) is 0.5914.

The value of P(S/A) is 0.465.

The value of P(S/Pa) is 0.717.

Explanation of Solution

It is given that the events A, Pa, S, and N are defined as follows:

A=Aggressive approach, Pa=Passive approach

S=Sale, N=no sale

There are 686 customers in the event sale and 1160 are participating customers.

The probability of sale is as follows:

P(S)=Favorable casesTotal number of cases=6861,160=0.5914

Thus, the probability of sale is 0.5914.

There are 270 customers in sale and aggressive approach.

The probability of sale given that aggressive approach is used as follows:

P(S/A)=P(S and A)P(A)=270580=0.465

Thus, the probability of sale given that aggressive approach is 0.465.

The probability of sale given that passive approach is used is calculated as follows:

P(S/Pa)=P(S and Pa)P(Pa)=416580=0.717

The probability of sale given that passive approach is used is 0.717.

b.

To determine

Explain whether the events S=Sale and Pa=Passive approach are independent.

b.

Expert Solution
Check Mark

Answer to Problem 29P

The events Sale and Passive approach are not independent.

Explanation of Solution

From Part (a), the value of P(S) is 0.5914 and P(S/A) is 0.465.

It is noticed that P(S/A)P(S). Thus, the events sale and passive approach are not independent.

c.

To determine

Find the values of P(A and S) and P(Pa and S).

c.

Expert Solution
Check Mark

Answer to Problem 29P

The value of P(A and S) is 0.2327.

The value of P(Pa and S) is 0.3586

Explanation of Solution

There are 270 customers in both events sale and aggressive approach. The total number of participating customers is 1,160.

The value of P(A and S) is as follows:

P(A and S)=2701,160=0.2327

Thus, the value of P(A and S) is 0.2327.

The number of customers in both events sale and passive approach is 416.

The value of P(Pa and S) is computed as follows:

P(Pa and S)=4161,160=0.3586

Thus, the value of P(Pa and S) is 0.3586.

d.

To determine

Find the values of P(N) and P(N/A).

d.

Expert Solution
Check Mark

Answer to Problem 29P

The value of P(N) is 0.408.

The value of P(N/A) is 0.534.

Explanation of Solution

There are 474 customers in no sale and 1160 are participating customers.

The probability of no sale is as follows:

P(N)=4741,160=0.408

Thus, the probability of no sale is 0.408.

There are 270 customers in sale and aggressive approach.

The probability of no sale given that aggressive approach is used as follows:

P(N/A)=P(N and A)P(A)=310580=0.534

Thus, the probability of no sale given that aggressive approach is 0.534.

e.

To determine

Explain whether the events N=no sale and A=Aggressive approach are independent.

e.

Expert Solution
Check Mark

Answer to Problem 29P

The events no sale and aggressive approach are not independent.

Explanation of Solution

From Part (d), the value of P(N) is 0.408 4 and P(N/A) is 0.534.

It is noticed that P(N/A)P(N).

Therefore, the events no sale and aggressive approach are not independent.

f.

To determine

Find the value of P(A or S).

f.

Expert Solution
Check Mark

Answer to Problem 29P

The value of P(A or S) is 0.8586.

Explanation of Solution

The probability of aggressive approach or sale is as follows:

P(A or S)=P(A)+P(S)P(A and S)=5801,160+6861,1602701,160=9961,160=0.8586

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

Understandable Statistics: Concepts and Methods

Ch. 4.1 - Critical Thinking Consider a family with 3...Ch. 4.1 - Prob. 12PCh. 4.1 - Prob. 13PCh. 4.1 - Critical Thinking (a) Explain why 0.41 cannot be...Ch. 4.1 - Myers-Briggs: Personality Types Isabel Briggs...Ch. 4.1 - General: Roll a Die (a) If you roll a single die...Ch. 4.1 - Psychology: Creativity When do creative people get...Ch. 4.1 - Agriculture: Cotton A botanist has developed a new...Ch. 4.1 - Expand Your Knowledge: Odds in Favor Sometimes...Ch. 4.1 - Expand Your Knowledge: Odds Against Betting odds...Ch. 4.1 - Business: Customers John runs a computer software...Ch. 4.2 - Statistical Literacy If two events are mutually...Ch. 4.2 - Statistical Literacy If two events A and B are...Ch. 4.2 - Basic Computation: Addition Rule Given P(A) = 0.3...Ch. 4.2 - Basic Computation: Addition Rule Given P(A) = 0.7...Ch. 4.2 - Prob. 5PCh. 4.2 - Prob. 6PCh. 4.2 - Prob. 7PCh. 4.2 - Prob. 8PCh. 4.2 - Prob. 9PCh. 4.2 - Prob. 10PCh. 4.2 - Prob. 11PCh. 4.2 - Prob. 12PCh. 4.2 - Prob. 13PCh. 4.2 - Prob. 14PCh. 4.2 - Prob. 15PCh. 4.2 - Environmental: Land Formations Arches National...Ch. 4.2 - Prob. 17PCh. 4.2 - General: Roll Two Dice You roll two fair dice, a...Ch. 4.2 - Prob. 19PCh. 4.2 - Prob. 20PCh. 4.2 - Prob. 21PCh. 4.2 - General: Deck of Cards You draw two cards from a...Ch. 4.2 - General: Deck of Cards You draw two cards from a...Ch. 4.2 - Prob. 24PCh. 4.2 - Marketing: Toys USA Today gave the information...Ch. 4.2 - Prob. 26PCh. 4.2 - Prob. 27PCh. 4.2 - Prob. 28PCh. 4.2 - Prob. 29PCh. 4.2 - Survey: Medical Tests Diagnostic tests of medical...Ch. 4.2 - Survey: Lung/Heart In an article titled Diagnostic...Ch. 4.2 - Survey: Customer Loyalty Are customers more loyal...Ch. 4.2 - Franchise Stores: Profits Wing Foot is a shoe...Ch. 4.2 - Education: College of Nursing At Litchfield...Ch. 4.2 - Medical: Tuberculosis The state medical school has...Ch. 4.2 - Prob. 36PCh. 4.2 - Brain Teasers Assume A and B are events such that...Ch. 4.2 - Brain Teasers Assume A and B are events such that...Ch. 4.2 - Prob. 39PCh. 4.2 - Prob. 40PCh. 4.2 - Prob. 41PCh. 4.2 - Prob. 42PCh. 4.2 - Prob. 43PCh. 4.2 - Prob. 44PCh. 4.2 - Prob. 45PCh. 4.2 - Prob. 46PCh. 4.2 - Prob. 47PCh. 4.2 - Prob. 48PCh. 4.2 - Prob. 49PCh. 4.2 - Prob. 50PCh. 4.2 - Prob. 51PCh. 4.2 - Prob. 52PCh. 4.3 - Prob. 1PCh. 4.3 - Prob. 2PCh. 4.3 - Prob. 3PCh. 4.3 - Prob. 4PCh. 4.3 - Tree Diagram (a) Draw a tree diagram to display...Ch. 4.3 - Tree Diagram (a) Draw a tree diagram to display...Ch. 4.3 - Prob. 7PCh. 4.3 - Prob. 8PCh. 4.3 - Prob. 9PCh. 4.3 - Multiplication Rule for Counting A sales...Ch. 4.3 - Counting: Agriculture Barbara is a research...Ch. 4.3 - Counting: Outcomes You toss a pair of dice. (a)...Ch. 4.3 - Compute P5,2.Ch. 4.3 - Prob. 14PCh. 4.3 - Compute P7,7.Ch. 4.3 - Compute P9,9.Ch. 4.3 - Compute C5,2.Ch. 4.3 - Compute C8,3.Ch. 4.3 - Prob. 19PCh. 4.3 - Prob. 20PCh. 4.3 - Counting: Hiring There are three nursing positions...Ch. 4.3 - Counting: Lottery In the Cash Now lottery game...Ch. 4.3 - Counting: Sports The University of Montana ski...Ch. 4.3 - Counting: Sales During the Computer Daze special...Ch. 4.3 - Counting: Hiring There are 15 qualified applicants...Ch. 4.3 - Counting: Grading One professor grades homework by...Ch. 4.3 - Prob. 27PCh. 4.3 - Counting: Powerball The Viewpoint of this section,...Ch. 4 - Prob. 1CRPCh. 4 - Prob. 2CRPCh. 4 - Prob. 3CRPCh. 4 - Prob. 4CRPCh. 4 - Prob. 5CRPCh. 4 - Prob. 6CRPCh. 4 - Prob. 7CRPCh. 4 - Critical Thinking For a class activity, your group...Ch. 4 - Prob. 9CRPCh. 4 - Prob. 10CRPCh. 4 - Prob. 11CRPCh. 4 - Prob. 12CRPCh. 4 - Prob. 13CRPCh. 4 - Prob. 14CRPCh. 4 - Prob. 15CRPCh. 4 - Prob. 16CRPCh. 4 - Prob. 17CRPCh. 4 - Prob. 18CRPCh. 4 - Prob. 19CRPCh. 4 - Prob. 20CRPCh. 4 - Look at Figure 4-11, Whos Cracking the Books? (a)...Ch. 4 - Prob. 2LCCh. 4 - Prob. 1UT
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
College Algebra
Algebra
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:Cengage Learning
Text book image
Algebra and Trigonometry (MindTap Course List)
Algebra
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:Cengage Learning
Text book image
College Algebra
Algebra
ISBN:9781337282291
Author:Ron Larson
Publisher:Cengage Learning
Text book image
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Text book image
College Algebra (MindTap Course List)
Algebra
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:Cengage Learning
Statistics 4.1 Point Estimators; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=2MrI0J8XCEE;License: Standard YouTube License, CC-BY
Statistics 101: Point Estimators; Author: Brandon Foltz;https://www.youtube.com/watch?v=4v41z3HwLaM;License: Standard YouTube License, CC-BY
Central limit theorem; Author: 365 Data Science;https://www.youtube.com/watch?v=b5xQmk9veZ4;License: Standard YouTube License, CC-BY
Point Estimate Definition & Example; Author: Prof. Essa;https://www.youtube.com/watch?v=OTVwtvQmSn0;License: Standard Youtube License
Point Estimation; Author: Vamsidhar Ambatipudi;https://www.youtube.com/watch?v=flqhlM2bZWc;License: Standard Youtube License