Explanation Given: f ( x , y ) = { 0 , e l s e w h e r e k e − ( x 2 + y 2 ) , x ≥ 0 , y ≥ 0 Formula Used: For a function f ( x , y ) to be a probability distribution function ∫ − ∞ ∞ ∫ − ∞ ∞ f ( x , y ) d y d x = 1 Calculation: We can write the given problem in double integral as: ∫ 0 ∞ ∫ 0 ∞ k e − ( x 2 + y 2 ) d y d x Now, convert the double integral into polar coordinates by substituting: x = r cos θ y = r sin θ d x d y = r d r d θ The corresponding limits in polar coordinates will be: 0 ≤ θ ≤ π 2 <

Bundle: Calculus: Early Transcendental Functions, Loose-leaf Version, 6th + WebAssign Printed Access Card for Larson/Edwards' Calculus: Early Transcendental Functions, 6th Edition, Multi-Term

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ISBN: 9781305714045

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Chapter 14.3, Problem 69E

To determine

To Calculate: The value of k such that f(x,y)={0,elsewhereke−(x2+y2),x≥0,y≥0 is a probability distribution function

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Chapter 14.3, Problem 68E

Chapter 14.3, Problem 70E

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Chapter 14 Solutions

Bundle: Calculus: Early Transcendental Functions, Loose-leaf Version, 6th + WebAssign Printed Access Card for Larson/Edwards' Calculus: Early Transcendental Functions, 6th Edition, Multi-Term

Ch. 14.1 - Evaluate the iterated integral: 0433cosrdrd Ch. 14.1 - Prob. 1E Ch. 14.1 - Evaluate the integral: xx2yxdy Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Evaluate the integral: 04x2x2ydy Ch. 14.1 - Evaluate the integral: x3x(x2+3y2)dy Ch. 14.1 - Evaluate the integral: eyyylnxxdx;y0 Ch. 14.1 - Evaluate the integral: 1y21y2(x2+y2)dx Ch. 14.1 - Evaluate the integral: 0x2yeyxdy

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The value of k such that f ( x , y ) = { 0 , e l s e w h e r e k e − ( x 2 + y 2 ) , x ≥ 0 , y ≥ 0 is a probability distribution function

Solution Summary: The author explains how to convert the double integral into polar coordinates by substituting lx=rmathrm

To Calculate: The value of k such that f(x,y)={0,elsewhereke−(x2+y2),x≥0,y≥0 is a probability distribution function

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Chapter 14 Solutions