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
Question
Book Icon
Chapter 4.3, Problem 60E

(a)

To determine

To test: Whether the area under the curve is 1 or not.

(a)

Expert Solution
Check Mark

Answer to Problem 60E

Solution: Yes, the area under the curve is 1.

Explanation of Solution

Calculation: The area of a triangle is,

Area=12×base×height

The height is 1 and the base is equal to 2. Thus, the area can be calculated as:

Area=12×base×height=12×2×1=1

Hence, the area is 1. The provided curve is shown below:

Introduction to the Practice of Statistics, Chapter 4.3, Problem 60E , additional homework tip  1

(b)

Section 1:

To determine

To find: The probability P(Y<1).

(b)

Section 1:

Expert Solution
Check Mark

Answer to Problem 60E

Solution: The probability is 0.5.

Explanation of Solution

Calculation: First, obtain the area in the interval of [0,1]. Thus, from the provided triangle, base is 1 and height is equal to 1 because the triangle is symmetrical about its vertical line. Thus, the area can be calculated by the formula:

Area=12×base×height

In the graph, the total area is 1 and the dotted area which is also called the density of the curve. Thus, the area of ΔACD or density of curve Y<1 can be calculated as:

Density of (Y<1)=12×base×height=12×1×1=0.5

Now, the probability of Y is less than 1 can be calculated as:

P(Y<1)=Density of Y<1Density of curve=0.51=0.5

Hence, the probability is 0.5.

Section 2:

To determine

To graph: The density curve and shade the area which represents the probability.

Section 2:

Expert Solution
Check Mark

Explanation of Solution

Graph: In the provided triangle, draw a vertical line at 1. Construct a triangle on the left end of the triangle and shade the left area. Hence, the density curve is shown below:

Introduction to the Practice of Statistics, Chapter 4.3, Problem 60E , additional homework tip  2

Section 3:

To determine

To find: The area.

Section 3:

Expert Solution
Check Mark

Answer to Problem 60E

Solution: The area is 0.5.

Explanation of Solution

Calculation: From the provided triangle, base is equal to 1 and height is equal to 1. Thus, the area can be calculated as:

Area=12×base×height=12×1×1=0.5

Hence, the area is 0.5.

(c)

Section 1:

To determine

To find: The probability P(Y>1.5).

(c)

Section 1:

Expert Solution
Check Mark

Answer to Problem 60E

Solution: The probability is 0.875.

Explanation of Solution

First, obtain the area in the interval of [0,1.5]. Thus, from the triangle

Introduction to the Practice of Statistics, Chapter 4.3, Problem 60E , additional homework tip  3

base is equal to 0.5 and height is equal to 0.5. Thus, the area can be calculated by the formula:

Area=12×base×height

The area of ΔBEF or the density of curve Y>1.5 can be calculated as:

Area=12×base×height=12×0.5×0.5=0.125

Now, the probability of Y is greater than 1.5 can be calculated as:

P(Y>1.5)=1P(Y1.5)=1Area of ΔBEFDensity of curve=10.1251=10.125

=0.875

Hence, the probability is 0.875.

Section 2:

To determine

To graph: The density curve and shade the area which represents the probability.

Section 2:

Expert Solution
Check Mark

Explanation of Solution

Graph: In the provided triangle, draw a vertical line at 1.5. Construct a triangle on the right end of the triangle and shade the left area. Hence, the density curve is shown below:

Introduction to the Practice of Statistics, Chapter 4.3, Problem 60E , additional homework tip  4

Section 3:

To determine

To find: The area.

Section 3:

Expert Solution
Check Mark

Answer to Problem 60E

Solution: The area is 0.125.

Explanation of Solution

Calculation: From the provided triangle, base is equal to 0.5 and height is equal to 0.5. Thus, the area can be calculated as:

Area=12×base×height=12×0.5×0.5=0.125

Hence, the area is 0.125.

(d)

To determine

To find: The probability P(Y>0.5).

(d)

Expert Solution
Check Mark

Answer to Problem 60E

Solution: The probability is 0.875.

Explanation of Solution

Calculation: First, obtain the area in the interval of [0,0.5]. Thus, from the provided triangle, base is equal to 0.5 and height is equal to 0.5. Thus, the area can be calculated by the formula:

Area=12×base×height

The area of ΔAGH can be calculated as:

Area=12×base×height=12×0.5×0.5=0.125

Now, the probability of Y is greater than 0.5 can be calculated as:

P(Y>0.5)=1P(Y0.5)=1Area of ΔAGHDensity of curve=10.1251=10.125

=0.875

Hence, the probability is 0.875.

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
Homework Let X1, X2, Xn be a random sample from f(x;0) where f(x; 0) = (-), 0 < x < ∞,0 € R Using Basu's theorem, show that Y = min{X} and Z =Σ(XY) are indep. -
Homework Let X1, X2, Xn be a random sample from f(x; 0) where f(x; 0) = e−(2-0), 0 < x < ∞,0 € R Using Basu's theorem, show that Y = min{X} and Z =Σ(XY) are indep.
An Arts group holds a raffle.  Each raffle ticket costs $2 and the raffle consists of 2500 tickets.  The prize is a vacation worth $3,000.    a. Determine your expected value if you buy one ticket.     b. Determine your expected value if you buy five tickets.     How much will the Arts group gain or lose if they sell all the tickets?

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
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