Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
Book Icon
Chapter 4.1, Problem 15E
To determine

a)

To find:

The experimental probability.

Expert Solution
Check Mark

Answer to Problem 15E

Solution:

The probability that the next respondent selected by the telemarketer’s computer will be over 45 years of age is 0.2914.

Explanation of Solution

Given:

The table is given as,

Number of Respondents by Age
18-25 26-35 36-45 Over 45
29 40 55 51

Formula used:

Let E denote the event that the selected respondent is over 45 years of age. Then, P(E) denotes the probability that E occurs. The probability is given by,

P(E)=fn

Where, f denotes the frequency of the event E and n denotes the number of times the experiment is repeated.

Calculation:

In this case, f denotes the number of respondents of age over 45 years and n denotes the total number of respondents selected. From the table,

f=51n=29+40+55+51=175

Thus, the probability that the next respondent selected will be over 45 years of age is given by,

P(E)=fn=51175=0.2914

To determine

b)

To find:

The experimental probability.

Expert Solution
Check Mark

Answer to Problem 15E

Solution:

The probability that the next respondent selected by the telemarketer’s computer will be aged between 26 and 35 years is 0.2286.

Explanation of Solution

Given:

The table is given as,

Number of Respondents by Age
18-25 26-35 36-45 Over 45
29 40 55 51

Formula used:

Let E denote the event that the selected respondent is aged between 26 and 35 years. Then, P(E) denotes the probability that E occurs. The probability is given by,

P(E)=fn

Where, f denotes the frequency of the event E and n denotes the number of times the experiment is repeated.

Calculation:

In this case, f denotes the number of respondents of age between 26 - 35 and n denotes the total number of respondents selected. From the table,

f=40n=29+40+55+51=175

Thus, the probability that the next respondent selected will be aged between 26-35 years is given by,

P(E)=fn=40175=0.2286

To determine

c)

To find:

The experimental probability.

Expert Solution
Check Mark

Answer to Problem 15E

Solution:

The probability that the next respondent selected by the telemarketer’s computer will be at least 36 years in age is 0.6057.

Explanation of Solution

Given:

The table is given as,

Number of Respondents by Age
18-25 26-35 36-45 Over 45
29 40 55 51

Formula used:

Let E denote the event that the selected respondent is at least 36 years of age. Then, P(E) denotes the probability that E occurs. The probability is given by,

P(E)=fn

Where, f denotes the frequency of the event E and n denotes the number of times the experiment is repeated.

Calculation:

In this case, f denotes the number of respondents of age at least 36 years and n denotes the total number of respondents selected. From the table, a respondent at least 36 years of age will either belong to the interval ’36-45’ or ‘Over 45’. So,

f=55+51=106n=29+40+55+51=175

Thus, the probability that the next respondent selected will be at least 36 years of age is given by,

P(E)=fn=106175=0.6057

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!
Students have asked these similar questions
Using the accompanying Home Market Value data and associated regression​ line, Market ValueMarket Valueequals=​$28,416+​$37.066×Square ​Feet, compute the errors associated with each observation using the formula e Subscript ieiequals=Upper Y Subscript iYiminus−ModifyingAbove Upper Y with caret Subscript iYi and construct a frequency distribution and histogram.   LOADING... Click the icon to view the Home Market Value data.       Question content area bottom Part 1 Construct a frequency distribution of the​ errors, e Subscript iei.   ​(Type whole​ numbers.) Error Frequency minus−15 comma 00015,000less than< e Subscript iei less than or equals≤minus−10 comma 00010,000 0   minus−10 comma 00010,000less than< e Subscript iei less than or equals≤minus−50005000 5   minus−50005000less than< e Subscript iei less than or equals≤0 21   0less than< e Subscript iei less than or equals≤50005000 9…
The managing director of a consulting group has the accompanying monthly data on total overhead costs and professional labor hours to bill to clients. Complete parts a through c   Overhead Costs    Billable Hours345000    3000385000    4000410000    5000462000    6000530000    7000545000    8000
Using the accompanying Home Market Value data and associated regression​ line, Market ValueMarket Valueequals=​$28,416plus+​$37.066×Square ​Feet, compute the errors associated with each observation using the formula e Subscript ieiequals=Upper Y Subscript iYiminus−ModifyingAbove Upper Y with caret Subscript iYi and construct a frequency distribution and histogram. Square Feet    Market Value1813    911001916    1043001842    934001814    909001836    1020002030    1085001731    877001852    960001793    893001665    884001852    1009001619    967001690    876002370    1139002373    1131001666    875002122    1161001619    946001729    863001667    871001522    833001484    798001589    814001600    871001484    825001483    787001522    877001703    942001485    820001468    881001519    882001518    885001483    765001522    844001668    909001587    810001782    912001483    812001519    1007001522    872001684    966001581    86200

Chapter 4 Solutions

Beginning Statistics, 2nd Edition

Ch. 4.1 - Prob. 11ECh. 4.1 - Prob. 12ECh. 4.1 - Prob. 13ECh. 4.1 - Prob. 14ECh. 4.1 - Prob. 15ECh. 4.1 - Prob. 16ECh. 4.1 - Prob. 17ECh. 4.1 - Prob. 18ECh. 4.1 - Prob. 19ECh. 4.1 - Prob. 20ECh. 4.1 - Prob. 21ECh. 4.1 - Prob. 22ECh. 4.1 - Prob. 23ECh. 4.1 - Prob. 24ECh. 4.1 - Prob. 25ECh. 4.1 - Prob. 26ECh. 4.2 - Prob. 1ECh. 4.2 - Prob. 2ECh. 4.2 - Prob. 3ECh. 4.2 - Prob. 4ECh. 4.2 - Prob. 5ECh. 4.2 - Prob. 6ECh. 4.2 - Prob. 7ECh. 4.2 - Prob. 8ECh. 4.2 - Prob. 9ECh. 4.2 - Prob. 10ECh. 4.2 - Prob. 11ECh. 4.2 - Prob. 12ECh. 4.2 - Prob. 13ECh. 4.2 - Prob. 14ECh. 4.2 - Prob. 15ECh. 4.2 - Prob. 16ECh. 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. 4.3 - Prob. 1ECh. 4.3 - Prob. 2ECh. 4.3 - Prob. 3ECh. 4.3 - Prob. 4ECh. 4.3 - Prob. 5ECh. 4.3 - Prob. 6ECh. 4.3 - Prob. 7ECh. 4.3 - Prob. 8ECh. 4.3 - Prob. 9ECh. 4.3 - Prob. 10ECh. 4.3 - Prob. 11ECh. 4.3 - Prob. 12ECh. 4.3 - Prob. 13ECh. 4.3 - Prob. 14ECh. 4.3 - Prob. 15ECh. 4.3 - Prob. 16ECh. 4.3 - Prob. 17ECh. 4.3 - Prob. 18ECh. 4.3 - Prob. 19ECh. 4.3 - Prob. 20ECh. 4.3 - Prob. 21ECh. 4.3 - Prob. 22ECh. 4.3 - Prob. 23ECh. 4.3 - Prob. 24ECh. 4.3 - Prob. 25ECh. 4.3 - Prob. 26ECh. 4.3 - Prob. 27ECh. 4.3 - Prob. 28ECh. 4.3 - Prob. 29ECh. 4.3 - Prob. 30ECh. 4.3 - Prob. 31ECh. 4.4 - Prob. 1ECh. 4.4 - Prob. 2ECh. 4.4 - Prob. 3ECh. 4.4 - Prob. 4ECh. 4.4 - Prob. 5ECh. 4.4 - Prob. 6ECh. 4.4 - Prob. 7ECh. 4.4 - Prob. 8ECh. 4.4 - Prob. 9ECh. 4.4 - Prob. 10ECh. 4.4 - Prob. 11ECh. 4.4 - Prob. 12ECh. 4.4 - Prob. 13ECh. 4.4 - Prob. 14ECh. 4.4 - Prob. 15ECh. 4.4 - Prob. 16ECh. 4.4 - Prob. 17ECh. 4.4 - Prob. 18ECh. 4.4 - Prob. 19ECh. 4.4 - Prob. 20ECh. 4.4 - Prob. 21ECh. 4.4 - Prob. 22ECh. 4.4 - Prob. 23ECh. 4.4 - Prob. 24ECh. 4.4 - Prob. 25ECh. 4.4 - Prob. 26ECh. 4.4 - Prob. 27ECh. 4.4 - Prob. 28ECh. 4.4 - Prob. 29ECh. 4.4 - Prob. 30ECh. 4.4 - Prob. 31ECh. 4.4 - Prob. 32ECh. 4.4 - Prob. 33ECh. 4.4 - Prob. 34ECh. 4.4 - Prob. 35ECh. 4.4 - Prob. 36ECh. 4.4 - Prob. 37ECh. 4.4 - Prob. 38ECh. 4.4 - Prob. 39ECh. 4.4 - Prob. 40ECh. 4.4 - Prob. 41ECh. 4.4 - Prob. 42ECh. 4.4 - Prob. 43ECh. 4.4 - Prob. 44ECh. 4.4 - Prob. 45ECh. 4.4 - Prob. 46ECh. 4.4 - Prob. 47ECh. 4.4 - Prob. 48ECh. 4.4 - Prob. 49ECh. 4.4 - Prob. 50ECh. 4.4 - Prob. 51ECh. 4.4 - Prob. 52ECh. 4.5 - Prob. 1ECh. 4.5 - Prob. 2ECh. 4.5 - Prob. 3ECh. 4.5 - Prob. 4ECh. 4.5 - Prob. 5ECh. 4.5 - Prob. 6ECh. 4.5 - Prob. 7ECh. 4.5 - Prob. 8ECh. 4.5 - Prob. 9ECh. 4.5 - Prob. 10ECh. 4.5 - Prob. 11ECh. 4.5 - Prob. 12ECh. 4.5 - Prob. 13ECh. 4.5 - Prob. 14ECh. 4.CR - Prob. 1CRCh. 4.CR - Prob. 2CRCh. 4.CR - Prob. 3CRCh. 4.CR - Prob. 4CRCh. 4.CR - Prob. 5CRCh. 4.CR - Prob. 6CRCh. 4.CR - Prob. 7CRCh. 4.CR - Prob. 8CRCh. 4.CR - Prob. 9CRCh. 4.CR - Prob. 10CRCh. 4.CR - Prob. 11CRCh. 4.CR - Prob. 12CRCh. 4.CR - Prob. 13CRCh. 4.CR - Prob. 14CRCh. 4.CR - Prob. 15CRCh. 4.CR - Prob. 16CRCh. 4.CR - Prob. 17CRCh. 4.CR - Prob. 18CRCh. 4.CR - Prob. 19CRCh. 4.CR - Prob. 20CRCh. 4.CR - Prob. 21CRCh. 4.CR - Prob. 22CRCh. 4.CR - Prob. 23CRCh. 4.CR - Prob. 24CRCh. 4.CR - Prob. 25CR
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
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