Explanation Formula used: Let f ( x ) be an exponential probability density function for a continuous random variable X then, f ( x ) = { ( 1 λ ) e − x λ if x ≥ 0 0 otherwise . Expected value or mean ( µ ) of the exponential density function: μ = λ . Standard deviation of the exponential density function: σ = λ . Calculation: Given that X is a continuous random variable that is exponentially distributed with m = 6 ln 2 which is of the form m = λ ln 2 . Substitute the value λ = 6 in the formula for the exponential density function, f ( x ) = ( 1 λ ) e − x λ = ( 1 6 ) e − x 6 = 1 6 e − x 6 Thus, the exponential density function is f ( x ) = { 1 6 e − x 6 x ≥ 0 0 otherwise . By using the above formula, compute the mean value. μ = λ = 6 Thus, the value of mean μ = 6

MyLab Math with Pearson eText - Stand Alone Access Card - for Calculus for Business, Economics, Life Sciences & Social Sciences, Brief Version (14th Edition)

14th Edition

ISBN: 9780134856599

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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Chapter 11.4, Problem 43E

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

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Chapter 11.4, Problem 42E

Chapter 11.4, Problem 44E

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Chapter 11 Solutions

MyLab Math with Pearson eText - Stand Alone Access Card - for Calculus for Business, Economics, Life Sciences & Social Sciences, Brief Version (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 ), the standard deviation ( σ ) and the probability P ( μ − σ ≤ X ≤ μ + σ ) .

Solution Summary: The author calculates the mean, standard deviation, and probability of the exponential density function for a continuous random variable X.

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

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Chapter 11 Solutions