Explanation Given: The function is: f ( x , y ) = { 1 3 , 0 ≤ x ≤ 1 , 1 ≤ y ≤ 4 0 , elsewhere P ( 0 ≤ x ≤ 1 , 1 ≤ y ≤ 3 ) Calculation: A function f ( x , y ) is said to be a joint density function if the following properties are satisfied: f ( x , y ) ≥ 0 for all ( x , y ) ∫ − ∞ ∞ ∫ − ∞ ∞ f ( x , y ) d A = 1 P [ ( x , y ) ∈ R ] = ∬ R f ( x , y ) d A Now, Since the given function is f ( x , y ) = { 1 3 , 0 ≤ x ≤ 1 , 1 ≤ y ≤ 4 0 , elsewhere , f ( x , y ) ≥ 0 for all ( x , y )

11th Edition

ISBN: 9781337275378

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Question

Chapter 14.2, Problem 61E

To determine

To calculate: The probability for the function

f(x,y)={13,0≤x≤1,1≤y≤40, elsewhereP(0≤x≤1,1≤y≤3),

where the function is a joint density function.

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Chapter 14.2, Problem 60E

Chapter 14.2, Problem 62E

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Chapter 14 Solutions

Multivariable Calculus

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The probability for the function f ( x , y ) = { 1 3 , 0 ≤ x ≤ 1 , 1 ≤ y ≤ 4 0 , elsewhere P ( 0 ≤ x ≤ 1 , 1 ≤ y ≤ 3 ) , where the function is a joint density function.

Solution Summary: The author explains that the function f(x,y) is a joint density function if the following properties are satisfied.

To calculate: The probability for the function

f(x,y)={13,0≤x≤1,1≤y≤40, elsewhereP(0≤x≤1,1≤y≤3),

where the function is a joint density function.

Want to see the full answer?

Chapter 14 Solutions

The probability for the function f ( x , y ) = { 1 3 , 0 ≤ x ≤ 1 , 1 ≤ y ≤ 4 0 , elsewhere P ( 0 ≤ x ≤ 1 , 1 ≤ y ≤ 3 ) , where the function is a joint density function.

Solution Summary: The author explains that the function f(x,y) is a joint density function if the following properties are satisfied.

To calculate: The probability for the function f(x,y)={13,0≤x≤1,1≤y≤40, elsewhereP(0≤x≤1,1≤y≤3), where the function is a joint density function.

Want to see the full answer?

Chapter 14 Solutions

To calculate: The probability for the function

f(x,y)={13,0≤x≤1,1≤y≤40, elsewhereP(0≤x≤1,1≤y≤3),

where the function is a joint density function.