Explanation Given Information: The provided probability density function f ( x ) = 3 2 x 2 over the interval [ − 1 , 1 ] . Formula used: The mean, μ , is the expected value: μ = E ( x ) = ∫ a b x . f ( x ) d x . The variance σ 2 of x is; σ 2 = E ( x 2 ) − μ 2 = ∫ a b x 2 . f ( x ) d x − [ ∫ a b x . f ( x ) d x ] 2 The standard deviation is the square root of the variance σ = var i a n c e The integral of x n is ∫ x n d x = x n + 1 n + 1 . Here n ≠ 0 . Calculation: Consider the provided probability density function f ( x ) = 3 2 x 2 over the interval [ − 1 , 1 ] . Now to obtain the mean applying the formula of the mean, The mean, μ , is the expected value: μ = E ( x ) = ∫ a b x . f ( x ) d x . Substitute 3 2 x 2 for f ( x ), − 1 for a and 1 for b , then, μ = E ( x ) = ∫ − 1 1 x . 3 2 x 2 d x Now, use the formula ∫ x n d x = x n + 1 n + 1 , to get; E ( x ) = ∫ − 1 1 x 3 . 3 2 d x = 3 2 . [ x 3 + 1 ( 3 + 1 ) ] − 1 1 = 3 2 . [ x 4 4 ] − 1 1 Simplify further, to get; E ( x ) = 3 2 . [ ( 1 ) 4 4 − ( − 1 ) 4 4 ] = 3 2 [ 1 4 − 1 4 ] = 3 2 [ 0 ] = 0 Thus, the value of E ( x ) is 0

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ISBN: 9780136572671

Author: BITTINGER

Publisher: PEARSON C

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Chapter 5.5, Problem 8E

To determine

To calculate: The value of E(x),E(x2) and find the mean, the variance and the standard deviation for the provided probability density function f(x)=32x2 over the interval [−1,1].

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Chapter 5.5, Problem 7E

Chapter 5.5, Problem 9E

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The value of E ( x ) , E ( x 2 ) and find the mean, the variance and the standard deviation for the provided probability density function f ( x ) = 3 2 x 2 over the interval [ − 1 , 1 ] .

Solution Summary: The author calculates the probability density function f(x) = 32x