Explanation Given information: The density function is p ( x ) = 1 3 e − x 3 on [ 0 , ∞ ) . Formula used: An exponential density function is given by p ( x ) = 1 r e − x r where r is the mean. Standard deviation, σ 2 = ∫ − ∞ ∞ ( x − μ ) 2 p ( x ) d x Calculation: p ( x ) = 1 3 e − x 3 on [ 0 , ∞ ) (Given) This equation is of the form of exponential density p ( x ) = 1 r e − x r , where r = 3 is the mean. Hence, mean μ = 3 ⇒ μ = 3 σ 2 = ∫ − ∞ ∞ ( x − μ ) 2 p ( x ) d x = ∫ 0 ∞ ( x − 3 ) 2 1 3 e − x 3 d x = 1 3 ∫ 0 ∞ ( x 2 + 9 − 6 x ) e − x 3 d x = 1 3 ∫ 0 ∞ ( x 2 e − x 3 + 9 e − x 3 − 6 x e − x 3 ) d x = 1 3 { ∫ 0 ∞ x 2 e − x 3 d x + ∫ 0 ∞ 9 e − x 3 d x − ∫ 0 ∞ 6 x e − x 3 d x } = 1 3 { ∫ 0 ∞ x 2 e − x 3 d x + 9 ∫ 0 ∞ e − x 3 d x − 6 ∫ 0 ∞ x e − x 3 d x } ..... ( 1 ) Take the second integral, ∫ 0 ∞ e − x 3 d x ∫ 0 ∞ e − x 3 d x = [ − 3 e − x 3 ] 0 ∞ = − 3 ( 0 − 1 ) = 3 ⇒ ∫ 0 ∞ e − x 3 d x = 3 ..... ( 2 ) Take the third integral, ∫ 0 ∞ x e − x 3 d x ∫ 0 ∞ x e − x 3 d x = [ − 3 x e − x 3 ] 0 ∞ + 3 ∫ 0 ∞ e − x 3 d x ( By using integration by parts ) = − 3 ( 0 − 0 ) + 3

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 9.1, Problem 29E

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The mean μ and the standard deviation σ for the density function p(x)=13e−x3 on [0,∞) .

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Chapter 9.1, Problem 28E

Chapter 9.1, Problem 30E

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Chapter 9 Solutions

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The mean μ and the standard deviation σ for the density function p ( x ) = 1 3 e − x 3 on [ 0 , ∞ ) .

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The mean μ and the standard deviation σ for the density function p(x)=13e−x3 on [0,∞) .

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Chapter 9 Solutions