Explanation Approach: If μ and σ are real numbers, σ > 0 , and if X is a random variable with probability density function defined by f ( x ) = 1 σ 2 π e − ( x − μ ) 2 / ( 2 σ 2 ) , then X is a normal random variable. Inflexion points of a function f ( x ) are the points where f ' ' ( x ) = 0 . Calculation: Find the first derivative of f ( x ) = 1 σ 2 π e − ( x − μ ) 2 / ( 2 σ 2 ) . f ' ( x ) = − ( x − μ ) σ 3 2 π e − ( x − μ ) 2 / ( 2 σ 2 ) = − ( x − μ ) f ( x ) σ 2 Find the second derivative of f ( x ) = 1 σ 2 π e − ( x − μ ) 2 / ( 2 σ 2 ) .

2nd Edition

ISBN: 9780321964038

Author: GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher: Pearson Addison Wesley,

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Chapter 13.3, Problem 30E

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To prove:

The inflexion points of the normal random variable are x=μ−σ and x=μ+σ.

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Chapter 13.3, Problem 28E

Chapter 13.3, Problem 35E

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Chapter 13 Solutions

Calculus For The Life Sciences

Ch. 13.1 - Repeat Example 1a for the function f(x)=2x2 on...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E Ch. 13.1 - Prob. 5E Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E

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To prove: The inflexion points of the normal random variable are x = μ − σ and x = μ + σ .

Solution Summary: The author explains how the inflexion points of the normal random variable are x=mu -sigma and

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To prove:

The inflexion points of the normal random variable are x=μ−σ and x=μ+σ.

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Chapter 13 Solutions

To prove: The inflexion points of the normal random variable are x = μ − σ and x = μ + σ .

Solution Summary: The author explains how the inflexion points of the normal random variable are x=mu -sigma and

Concept explainers

Videos

To prove: The inflexion points of the normal random variable are x=μ−σ and x=μ+σ.

Want to see the full answer?

Chapter 13 Solutions

To prove:

The inflexion points of the normal random variable are x=μ−σ and x=μ+σ.