EBK INTRODUCTION TO THE PRACTICE OF STA
EBK INTRODUCTION TO THE PRACTICE OF STA
9th Edition
ISBN: 8220103674638
Author: Moore
Publisher: YUZU
Question
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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:

EBK INTRODUCTION TO THE PRACTICE OF STA, 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:

EBK INTRODUCTION TO THE PRACTICE OF STA, 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

EBK INTRODUCTION TO THE PRACTICE OF STA, 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:

EBK INTRODUCTION TO THE PRACTICE OF STA, 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.

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Chapter 4 Solutions

EBK INTRODUCTION TO THE PRACTICE OF STA

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
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