Explanation Given: The function is f ( x , y ) = k x 2 y and the region D is { 0 ≤ x ≤ 1 ; 1 ≤ y ≤ 2 } . Concept used: The joint probability density function of the random variables X and Y in an interval D is expressed as follows: ∬ D f ( x , y ) d A = 1 ...... ( 1 ) Calculation: The given function is expressed as follows: f ( x , y ) = k x 2 y If the given function is a joint probability density function on the region { 0 ≤ x ≤ 1 ; 1 ≤ y ≤ 2 } then the function must satisfy the equation ( 1 ) . To obtain the value of k , substitute k x 2 y for f ( x , y ) and apply the limits of x and y in equation ( 1 ) as shown below: ∫ 1 2 ∫ 0 1 k x 2 y d x d y =

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 10.1, Problem 33E

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The value of k such that the function f(x,y)=kx2y is a joint probability density function on the region D={0≤x≤1;1≤y≤2}.

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Chapter 10.1, Problem 32E

Chapter 10.1, Problem 34E

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Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

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The value of k such that the function f ( x , y ) = k x 2 y is a joint probability density function on the region D = { 0 ≤ x ≤ 1 ; 1 ≤ y ≤ 2 } .

Solution Summary: The author explains that f(x,y)=kx2y is a joint probability density function on the region D.

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The value of k such that the function f(x,y)=kx2y is a joint probability density function on the region D={0≤x≤1;1≤y≤2}.

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Chapter 10 Solutions