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 . The Mean, μ = E ( X ) = ∫ − ∞ ∞ x f ( x ) d x . The variance, V ( X ) = ∫ − ∞ ∞ ( x − μ ) 2 f ( x ) d x . The standard deviation, 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 mean of the random variable X 2 is greater than X 1 . By using the above 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 . Find 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 ( ∵ w = − x 2 ) Integrate μ 1 by using the above formula. 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

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Chapter 11, Problem 35RE

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To check: The prediction in Problem 33 by using the numerical integration to approximate the means of X1 and X2.

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Chapter 11, Problem 34RE

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Chapter 11 Solutions

FIN 108 ISU LOOSE >IP<

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 33 by using the numerical integration to approximate the means of X 1 and X 2 .

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Chapter 11 Solutions