Explanation Let Y is an exponential random variable and the probability density function is as follows: f ( y ) = { 1 β e − y β , y > 0 0 , elsewhere The mean and standard deviation of Y is as follows: μ = β σ 2 = β 2 The P ( | Y − μ | ≤ 2 σ ) value can be obtained as follows: P ( | Y − μ | ≤ 2 σ ) = P ( | Y − β | ≤ 2 β ) = P (

Mathematical Statistics with Applications

7th Edition

ISBN: 9780495110811

Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer

Publisher: Cengage Learning

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Chapter 4.10, Problem 150E

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Find P(|Y−μ|≤2σ) for the exponential random variable.

Compare with the corresponding probabilistic statement given by Tchebysheff’s theorem and the empirical rule.

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Chapter 4.10, Problem 149E

Chapter 4.10, Problem 151E

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Chapter 4 Solutions

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Find P ( | Y − μ | ≤ 2 σ ) for the exponential random variable. Compare with the corresponding probabilistic statement given by Tchebysheff’s theorem and the empirical rule .

Solution Summary: The author compares Tchebysheff's theorem and the empirical rule for the exponential random variable Y.

Find P(|Y−μ|≤2σ) for the exponential random variable.

Compare with the corresponding probabilistic statement given by Tchebysheff’s theorem and the empirical rule.

Want to see the full answer?

Chapter 4 Solutions

Find P ( | Y − μ | ≤ 2 σ ) for the exponential random variable. Compare with the corresponding probabilistic statement given by Tchebysheff’s theorem and the empirical rule .

Solution Summary: The author compares Tchebysheff's theorem and the empirical rule for the exponential random variable Y.

Find P(|Y−μ|≤2σ) for the exponential random variable. Compare with the corresponding probabilistic statement given by Tchebysheff’s theorem and the empirical rule.

Want to see the full answer?

Chapter 4 Solutions

Find P(|Y−μ|≤2σ) for the exponential random variable.

Compare with the corresponding probabilistic statement given by Tchebysheff’s theorem and the empirical rule.