Explanation Given information: The density function is p ( x ) = 1 π 1 − x 2 on ( − 1 , 1 ) . Formula used: μ = ∫ − ∞ ∞ x . p ( x ) d x σ 2 = ∫ − ∞ ∞ ( x − μ ) 2 p ( x ) d x Calculation: p ( x ) = 1 π 1 − x 2 on ( − 1 , 1 ) . (Given) μ = ∫ − ∞ ∞ x . p ( x ) d x = ∫ − 1 1 x . 1 π 1 − x 2 d x = 1 π ∫ − 1 1 x 1 − x 2 d x = − 1 2 π ∫ − 1 1 − 2 x 1 − x 2 d x Consider the integral ∫ − 2 x 1 − x 2 d x . Let u = 1 − x 2 , then d u = − 2 x d x . Using this in the above integral, we get ∫ − 2 x 1 − x 2 d x = ∫ d u u = ∫ u − 1 2 d u = [ u − 1 2 + 1 − 1 2 + 1 ] So, ∫ − 2 x 1 − x 2 d x = [ u 1 2 1 2 ] = [ 2 u 1 2 ] = [ 2 ( 1 − x 2 ) 1 2 ] = 2 1 − x 2 Then, μ = − 1 2 π ∫ − 1 1 − 2 x 1 − x 2 d x = − 1 2 π [ 2 1 − x 2 ] − 1 1 = − 1 π [ 1 − 1 2 − 1 − ( − 1 ) 2 ] = − 1 π [ 0 − 0 ] = 0 Therefore, μ = 0 . σ 2 = ∫ − ∞ ∞ ( x − μ ) 2 p ( x ) d x = ∫ − 1 1 ( x − μ ) 2 1 π 1 − x 2 d x = ∫ − 1 1 ( x − 0 ) 2 1 π 1 − x 2 d x = 1 π ∫ − 1 1 x 2 1 − x 2 d x Now, consider the integral ∫ x 2 1 − x 2 d x

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Chapter 9.1, Problem 28E

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To calculate: The mean μ and the standard deviation σ for the density function p(x)=1π1−x2 on (−1,1) .

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Chapter 9.1, Problem 27E

Chapter 9.1, Problem 29E

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The mean μ and the standard deviation σ for the density function p ( x ) = 1 π 1 − x 2 on ( − 1 , 1 ) .

Solution Summary: The author calculates the mean and standard deviation for the density function p(x)=1pi sqrt

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To calculate: The mean μ and the standard deviation σ for the density function p(x)=1π1−x2 on (−1,1) .

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Chapter 9 Solutions