Explanation Given information: X be a continuous random variable with the density function f ( x ) = 3 x − 4 , x ≥ 1 . Formula used: Let X be a continuous random variable whose possible values lie between A and B , and f ( x ) be the probability density function for X . Then, the expected value (average) of X is the number E ( X ) defined by E ( X ) = ∫ A B x f ( x ) d x . And the variance of X is defined by, Var ( X ) = ∫ A B [ x − E ( X ) ] 2 f ( x ) d x . Calculation: Let X be the continuous random variable with density function, f ( x ) = 3 x − 4 , x ≥ 1 By using the formula of expected value, E ( X ) = ∫ 1 ∞ x f ( x ) d x = ∫ 1 ∞ x 3 x − 4 d x . = ∫ 1 ∞ 3 x − 3 d x . The upper limit of integration is ∞ that is indefinite form of integration, so take the limit as b → ∞ of the integration. E ( X ) = lim b → ∞ ∫ 1 b 3 x − 3 d x Apply power rule of integration, = lim b → ∞ [ 3 x − 2 − 2 ] | 1 b = lim b → ∞ [ − 3 2 x − 2 ] 1 b Plug the given limits, = lim b → ∞ [ − 3 2 b − 2 + 3 2 1 − 2 ] = lim b → ∞ [ − 3 2 b − 2 + 3 2 ] Apply limit as b → ∞ = [ − 3 2 ∞ − 3 + 3 2 ] = 0 + 3 2 ⇒ E ( X ) = 3 2 . By using the formula of variance, Var ( X ) = ∫ 1 ∞ [ x − E ( X ) ] 2 f ( x ) d x = ∫ 1 ∞ [ x − 3 2 ] 2 ⋅ 3 x − 4 d x The upper limit of integration is ∞ that is indefinite form of integration, so take limit as b → ∞ of the integration

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Chapter 12.3, Problem 18E

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To calculate: E(X) and Var(X) of the density function f(x) = 3x−4, x≥1

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Chapter 12.3, Problem 17E

Chapter 12.3, Problem 19E

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E ( X ) and Var ( X ) of the density function f ( x ) = 3 x − 4 , x ≥ 1

Solution Summary: The author explains how to calculate E(X), and the probability density function for X.

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To calculate: E(X) and Var(X) of the density function f(x) = 3x−4, x≥1

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Chapter 12 Solutions