Explanation Given: The probability density function is f ( x ) = 1 4 ( 3 x 2 + 1 ) over [ − 1 , 1 ] . Concept used: The probability density function f be defined on the interval [ a , b ] associated with a continuous random variable X . Then the expected value μ of X is written as follows. μ = ∫ a b x f ( x ) d x ...... (1) The variance of X is written as follows. Var ( X ) = ∫ a b ( x − μ ) 2 f ( x ) d x ...... (2) The standard deviation of X is written as follows. σ = Var ( X ) ...... (3) Calculation: The probability density function is given as f ( x ) = 1 4 ( 3 x 2 + 1 ) over [ − 1 , 1 ] . Substitute − 1 for a , 1 for b and 1 4 ( 3 x 2 + 1 ) for f ( x ) in equation ( 1 ) to obtain the mean value as shown below. μ = ∫ − 1 1 x ( 1 4 ( 3 x 2 + 1 ) ) d x = 1 4 ∫ − 1 1 x ( 3 x 2 + 1 ) d x = 1 4 ∫ − 1 1 ( 3 x 3 + x ) d x = 1 4 [ 3 x 4 4 + x 2 2 ] − 1 1 The above expression can be further simplified as follows. μ = 1 4 [ 3 ( 1 ) 4 4 + ( 1 ) 2 2 − ( 3 ( − 1 ) 4 4 + ( − 1 ) 2 2 ) ] = 1 4 [ 3 4 + 1 2 − ( 3 4 + 1 2 ) ] = 0 Therefore, the mean of x is 0

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, Problem 15RE

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The mean, variance and standard deviation of the random variable x associated with the probability density function f(x)=14(3x2+1) over [−1,1].

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Chapter 10, Problem 14RE

Chapter 10, Problem 16RE

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

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The mean, variance and standard deviation of the random variable x associated with the probability density function f ( x ) = 1 4 ( 3 x 2 + 1 ) over [ − 1 , 1 ] .

Solution Summary: The author explains the probability density function f be defined on the interval left[a,bright] associated with a continuous random variable

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The mean, variance and standard deviation of the random variable x associated with the probability density function f(x)=14(3x2+1) over [−1,1].

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Chapter 10 Solutions