Explanation Formula used: Let f ( x ) be a uniform probability density function for a continuous random variable X then, f ( x ) = { 1 b − a if a ≤ x ≤ b 0 otherwise . Expected value or mean ( µ ) of the uniform density function: μ = 1 2 ( a + b ) . Standard deviation ( σ ) of the uniform density function: σ = 1 12 ( b − a ) . Calculation: Given that X is a continuous random variable that is uniformly distributed on the interval [ − 2 , 2 ] . Substitute the values a = − 2 and b = 2 in the formula for the uniform density function. f ( x ) = 1 b − a = 1 2 − ( − 2 ) = 1 2 + 2 = 1 4 Thus, the uniform density function is f ( x ) = { 1 4 if − 2 ≤ x ≤ 2 0 otherwise . Substitute the values a = − 2 and b = 2 in μ = 1 2 ( a + b ) , μ = 1 2 ( a + b ) = 1 2 ( − 2 + 2 ) = 1 2 ( 0 ) = 0 Thus, the value of the mean is μ = 0 . Substitute the values a = − 2 and b = 2 in σ = 1 12 ( b − a ) to compute the standard deviation

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 11.4, Problem 42E

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To find: The mean (µ), standard deviation (σ) and the probability P(μ−σ≤X≤μ+σ).

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Chapter 11.4, Problem 41E

Chapter 11.4, Problem 43E

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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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The mean ( µ ), standard deviation ( σ ) and the probability P ( μ − σ ≤ X ≤ μ + σ ) .

Solution Summary: The author explains how to find the mean, standard deviation, and probability of a continuous random variable X.

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To find: The mean (µ), standard deviation (σ) and the probability P(μ−σ≤X≤μ+σ).

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Chapter 11 Solutions