PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 5.3, Problem 87E

(a)

To determine

Probability for the person contributed to the charity.

(a)

Expert Solution
Check Mark

Answer to Problem 87E

Probability that the person contributed to the charity is 0.2240.

Explanation of Solution

Given information:

Figure showing probabilities for a charity by potential donors:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 87E , additional homework tip  1

According to multiplication rule for independent events,

  P(AandBandC)=P(ABC)=P(A)×P(B)×P(C)

According to the addition rule for mutually exclusive events:

  P(ABC)=P(AorBorC)=P(A)+P(B)+P(C)

Let

R: Recent donor

P: Past donor

N: New prospect

PL: Pledge to contribute

C: Contributor

Now,

For recent donors:

Probability for recent donor,

  P(R)=0.5

Probability for recent donor pledge to contribute,

  P(PL)=0.4

Probability for recent donor makes contribution,

  P(C)=0.8

For past donors:

Probability for past donor,

  P(P)=0.3

Probability for past donor pledge to contribute,

  P(PL)=0.3

Probability for past donor makes contribution,

  P(C)=0.6

For new prospects:

Probability for new prospect,

  P(N)=0.2

Probability for new prospect pledge to contribute,

  P(PL)=0.1

Probability for new prospect makes contribution,

  P(C)=0.5

Now,

Since each person cannot become all three types of contributor,

Apply multiplication rule for each of the above three cases separately:

For recent donors:

  P(RPLC)=P(R)×P(PL)×P(C)=0.5×0.4×0.8=0.1600

For past donors:

  P(PPLC)=P(P)×P(PL)×P(C)=0.3×0.3×0.6=0.0540

For new prospect:

  P(NPLC)=P(N)×P(PL)×P(C)=0.2×0.1×0.5=0.0100

To get the probability for contributor,

Apply the addition rule for mutually exclusive events:

  P(C)=P(RPLC)+P(PPLC)+P(NPLC)=0.1600+0.0540+0.0100=0.2240

Thus,

The probability for the randomly selected person contributes to the charity is 0.2240.

(b)

To determine

Probability for the person contributed is a recent donor.

(b)

Expert Solution
Check Mark

Answer to Problem 87E

Probability that the person contributed is a recent donor is approx. 0.7143.

Explanation of Solution

Given information:

Figure showing probabilities for a charity by potential donors:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 87E , additional homework tip  2

Conditional probability definition:

  P(A|B)=P(AB)P(B)

From Part (a),

We have

Probability for the person contributes to charity,

  P(C)=0.2240

Probability for recent donor and pledged to contribute and contributor,

  P(RPLC)=0.1600

It is understood that

Recent donor who contributed was pledged to contribute.

Thus,

  P(RPLC)=P(RC)=0.1600

Apply the conditional probability:

  P(R|C)=P(RC)P(C)=0.16000.22400.7143

Thus,

The conditional probability for the person contributed is a recent donor is approx. 0.7143.

Chapter 5 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.1 - Prob. 22ECh. 5.1 - Prob. 23ECh. 5.1 - Prob. 24ECh. 5.1 - Prob. 25ECh. 5.1 - Prob. 26ECh. 5.1 - Prob. 27ECh. 5.1 - Prob. 28ECh. 5.1 - Prob. 29ECh. 5.1 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - Prob. 32ECh. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.2 - Prob. 37ECh. 5.2 - Prob. 38ECh. 5.2 - Prob. 39ECh. 5.2 - Prob. 40ECh. 5.2 - Prob. 41ECh. 5.2 - Prob. 42ECh. 5.2 - Prob. 43ECh. 5.2 - Prob. 44ECh. 5.2 - Prob. 45ECh. 5.2 - Prob. 46ECh. 5.2 - Prob. 47ECh. 5.2 - Prob. 48ECh. 5.2 - Prob. 49ECh. 5.2 - Prob. 50ECh. 5.2 - Prob. 51ECh. 5.2 - Prob. 52ECh. 5.2 - Prob. 53ECh. 5.2 - Prob. 54ECh. 5.2 - Prob. 55ECh. 5.2 - Prob. 56ECh. 5.2 - Prob. 57ECh. 5.2 - Prob. 58ECh. 5.2 - Prob. 59ECh. 5.2 - Prob. 60ECh. 5.3 - Prob. 61ECh. 5.3 - Prob. 62ECh. 5.3 - Prob. 63ECh. 5.3 - Prob. 64ECh. 5.3 - Prob. 65ECh. 5.3 - Prob. 66ECh. 5.3 - Prob. 67ECh. 5.3 - Prob. 68ECh. 5.3 - Prob. 69ECh. 5.3 - Prob. 70ECh. 5.3 - Prob. 71ECh. 5.3 - Prob. 72ECh. 5.3 - Prob. 73ECh. 5.3 - Prob. 74ECh. 5.3 - Prob. 75ECh. 5.3 - Prob. 76ECh. 5.3 - Prob. 77ECh. 5.3 - Prob. 78ECh. 5.3 - Prob. 79ECh. 5.3 - Prob. 80ECh. 5.3 - Prob. 81ECh. 5.3 - Prob. 82ECh. 5.3 - Prob. 83ECh. 5.3 - Prob. 84ECh. 5.3 - Prob. 85ECh. 5.3 - Prob. 86ECh. 5.3 - Prob. 87ECh. 5.3 - Prob. 88ECh. 5.3 - Prob. 89ECh. 5.3 - Prob. 90ECh. 5.3 - Prob. 91ECh. 5.3 - Prob. 92ECh. 5.3 - Prob. 93ECh. 5.3 - Prob. 94ECh. 5.3 - Prob. 95ECh. 5.3 - Prob. 96ECh. 5.3 - Prob. 97ECh. 5.3 - Prob. 98ECh. 5.3 - Prob. 99ECh. 5.3 - Prob. 100ECh. 5.3 - Prob. 101ECh. 5.3 - Prob. 102ECh. 5.3 - Prob. 103ECh. 5.3 - Prob. 104ECh. 5.3 - Prob. 105ECh. 5.3 - Prob. 106ECh. 5.3 - Prob. 107ECh. 5.3 - Prob. 108ECh. 5 - Prob. R5.1RECh. 5 - Prob. R5.2RECh. 5 - Prob. R5.3RECh. 5 - Prob. R5.4RECh. 5 - Prob. R5.5RECh. 5 - Prob. R5.6RECh. 5 - Prob. R5.7RECh. 5 - Prob. R5.8RECh. 5 - Prob. T5.1SPTCh. 5 - Prob. T5.2SPTCh. 5 - Prob. T5.3SPTCh. 5 - Prob. T5.4SPTCh. 5 - Prob. T5.5SPTCh. 5 - Prob. T5.6SPTCh. 5 - Prob. T5.7SPTCh. 5 - Prob. T5.8SPTCh. 5 - Prob. T5.9SPTCh. 5 - Prob. T5.10SPTCh. 5 - Prob. T5.11SPTCh. 5 - Prob. T5.12SPTCh. 5 - Prob. T5.13SPTCh. 5 - Prob. T5.14SPT
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
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