Explanation Given: The probability density function is, f ( x ) = { 6 x − 6 x 2 if 0 ≤ x ≤ 1 0 otherwise . Formula used: Let f ( x ) be the probability density function for a continuous random variable X . Expected value or mean ( µ ) is, E ( x ) = ∫ − ∞ ∞ x f ( x ) d x . The alternative formula for variance is given by, V ( x ) = ∫ − ∞ ∞ x 2 f ( x ) d x − μ 2 . Calculation: From the above formula, obtain the mean as follows. E ( x ) = ∫ − ∞ ∞ x f ( x ) d x = ∫ 0 1 x ( 6 x − 6 x 2 ) d x = ∫ 0 1 ( 6 x 2 − 6 x 3 ) d x = [ 6 x 3 3 − 6 x 4 4 ] 0 1 On further simplification, E ( x ) = [ 2 x 3 − 3 x 4 2 ] 0 1 = ( 2 ( 1 ) 3 − 3 ( 1 ) 4 2 ) − ( 2 ( 0 ) 3 − 3 ( 0 ) 4 2 ) = ( 2 − 3 2 ) − 0 = 1 2 Thus, the expected value is μ = 1 2

MyLab Math with Pearson eText - Stand Alone Access Card - for Calculus for Business, Economics, Life Sciences & Social Sciences, Brief Version (14th Edition)

14th Edition

ISBN: 9780134856599

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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Chapter 11.3, Problem 3MP

To determine

To find: The variance of the given density function.

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Chapter 11.3, Problem 2MP

Chapter 11.3, Problem 4MP

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Chapter 11 Solutions

MyLab Math with Pearson eText - Stand Alone Access Card - for Calculus for Business, Economics, Life Sciences & Social Sciences, Brief Version (14th Edition)

Ch. 11.1 - Evaluate the following, if it converges: 3dx(x1)2.Ch. 11.1 - Prob. 2MP Ch. 11.1 - Prob. 3MP Ch. 11.1 - Prob. 4MP Ch. 11.1 - Prob. 5MP Ch. 11.1 - Prob. 6MP Ch. 11.1 - Prob. 1ED Ch. 11.1 - Prob. 2ED Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E

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