Explanation Given: The two probability density functions are, f 1 ( x ) = { 0.25 x e − x 2 if x ≥ 0 0 otherwise and f 2 ( x ) = { 0.0625 x 2 e − x 2 if x ≥ 0 0 otherwise . Formula used: Let f ( x ) be the probability density function for a continuous random variable X . Expected value or mean ( µ ): E ( x ) = ∫ − ∞ ∞ x f ( x ) d x . Variance: V ( x ) = ∫ − ∞ ∞ x 2 f ( x ) − μ 2 . Standard deviation ( σ ): σ = v ( x ) . Integration by parts : ∫ a b u ( d v d x ) d x = [ u v ] a b − ∫ a b v ( d u d x ) d x . Calculation: From Problem 33, the prediction is that the variance of the random variable X 2 is greater than X 1 . From the above mentioned formula, obtain the mean μ 1 as follows. μ 1 = ∫ − ∞ ∞ x f 1 ( x ) d x = ∫ 0 ∞ x ( 0.25 x e − x 2 ) d x Assume that the interval varies from 0 to 30. μ 1 = 0.25 ∫ 0 30 x 2 e − x 2 d x Let u = x 2 and v ′ = e − x 2 d x . That is, d u = 2 x . Obtain v = ∫ e − x 2 d x by using the substitution method. Assume that the variable w = − x 2 . Differentiate on both sides. d d x ( w ) = d d x ( − x 2 ) 2 d w = − d x d x = − 2 d w Substitute w = − x 2 , d x = − 2 d w in the integral v = ∫ e − x 2 d x , v = ∫ e w ( − 2 d w ) = − 2 e w = − 2 e − x 2 Integrate μ 1 by using the formula mentioned above. Here, u = x 2 and v = − 2 e − x 2 . μ 1 = 0.25 [ ( − 2 e − x 2 x 2 ) | 0 30 − ( ∫ 0 30 − 4 e − x 2 x d x ) | ] = 0.25 { ( − 2 e − x 2 x 2 ) | 0 30 + 4 [ ( − 2 e − x 2 x − 4 e − x 2 ) | 0 30 ] } = 0.25 { ( − 2 e − 30 2 ( 30 ) 2 ) − 0 + 4 [ ( − 2 e − 30 2 ( 30 ) − 4 e − 30 2 ) − ( 0 − 4 ) ] } = 0.25 { ( − 1 , 800 e − 15 ) − 0 + [ ( − 240 e − 15 − 16 e − 15 ) + 16 ] } On further simplification, μ 1 = 0.25 [ − 2 , 056 e − 15 + 16 ] ≈ 3.999 = 4 Thus, the mean of μ 1 is 4 . By using the above formula, obtain the variance V 1 ( x ) as follows. V 1 ( x ) = ∫ − ∞ ∞ x 2 f 1 ( x ) − μ 1 2 = ∫ 0 ∞ x 2 ( 0 ) d x + ∫ 0 ∞ x 2 ( 0.25 x e − x 2 ) d x − ( 4 ) 2 = ∫ 0 ∞ x 2 ( 0.25 x e − x 2 ) d x − 16 Assume that the interval varies from 0 to 30. V 1 ( x ) = ∫ 0 30 ( 0.25 ) x 3 e − x 2 d x − 16 Let u = x 3 and v ′ = e − x 2 d x . That is, d u = 3 x . Integrate V 1 ( x ) by using the above formula. Here, u = x 3 and v = − 2 e − x 2 . V 1 ( x ) = 0.25 [ ( − 2 e − x 2 x 3 ) | 0 30 − ( ∫ 0 30 − 2 e − x 2 ( 3 x 2 ) d x ) | ] − 16 = 0.25 [ ( − 2 e − x 2 x 3 ) | 0 30 + 6 ( ( − 2 e − x 2 x 2 ) | 0 30 − ∫ 0 30 − 2 e − x 2 ( 2 x ) d x ) ] − 16 = 0.25 [ ( − 2 e − x 2 x 3 ) | 0 30 + 6 ( − 2 e − x 2 x 2 ) | 0 30 + 24 ∫ 0 30 e − x 2 ( x ) d x ] − 16 = 0.25 [ ( − 2 e − x 2 x 3 ) | 0 30 − 12 ( e − x 2 x 2 ) | 0 30 + 24 [ ( − 2 e − x 2 x ) − 4 e − x 2 ] | 0 30 ] − 16 On further simplification, V 1 ( x ) = 0.25 [ ( − 2 e − 30 2 ( 30 ) 3 − 0 ) − ( 12 e − 30 2 ( 30 ) 2 − 0 ) − ( 48 e − 30 2 ( 30 ) − 0 ) − 24 ( 4 e − 30 2 − 4 ) ] − 16 = 0

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

14th Edition

ISBN: 9780134668574

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

Publisher: PEARSON

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Chapter 11, Problem 36RE

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To check: The prediction in Problem 34 by using the numerical integration to approximate the variance of X1 and X2.

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Chapter 11, Problem 35RE

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Chapter 11 Solutions

Calculus for Business, Economics, Life Sciences, and Social Sciences (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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To check: The prediction in Problem 34 by using the numerical integration to approximate the variance of X 1 and X 2 .

Solution Summary: The author explains how to check the prediction in Problem 34 by using the numerical integration to approximate the variance of X_1.

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To check: The prediction in Problem 34 by using the numerical integration to approximate the variance of X1 and X2.

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Chapter 11 Solutions