Explanation Given information: The integral ∫ − ∞ ∞ x 2 f ( x ) d x = 1 , where f ( x ) is the standard normal density function. The f ( x ) approaches zero rapidly as x gets large in magnitude; hence, the value of the improper integral is nearly the same as the definite integral of x 2 f ( x ) from x = − 8 to x = 8 . Formula used: ∫ − ∞ ∞ x 2 f ( x ) d x = 1 , where f ( x ) is the standard normal density function. Proof: A random variable X with standard normal density function f ( x ) is f ( x ) = 1 2 π e − ( 1 / 2 ) x 2 . The integral ∫ − ∞ ∞ x 2 f ( x ) d x = 1 is written as ∫ − ∞ ∞ x 2 1 2 π e − ( 1 / 2 ) x 2 d x = 1 . Since f ( x ) approaches zero rapidly as x gets large in magnitude, the value of the improper integral is nearly the same as the definite integral of x 2 f ( x ) from x = − 8 to x = 8 . Hence, the limits are from x = − 8 to x = 8 , and substitute them in the above integral. ⇒ ∫ − 8 8 x 2 1 2 π e − ( 1 / 2 ) x 2 d x = 1 To find the value of the above integral, use Ti-84 graphing calculator steps as shown below: Step 1: Press the [MATH] to access the math menu

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Author: Goldstein

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Chapter 12.4, Problem 37E

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To prove: The standard deviation of the standard normal random variable is 1 by using the integral ∫−∞∞x2f(x) dx=1, where f(x) is the standard normal density function.

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Chapter 12.4, Problem 36E

Chapter 12.5, Problem 1CYU

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To prove: The standard deviation of the standard normal random variable is 1 by using the integral ∫ − ∞ ∞ x 2 f ( x ) d x = 1 , where f ( x ) is the standard normal density function.

Solution Summary: The author proves that the standard deviation of the normal random variable is 1 by using the integral displaystyle

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To prove: The standard deviation of the standard normal random variable is 1 by using the integral ∫−∞∞x2f(x) dx=1, where f(x) is the standard normal density function.

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Chapter 12 Solutions