Explanation Formula used: Let f ( x ) be a 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 ) : μ = λ Standard deviation: σ = λ Calculation: Let X be continuous random variable that is exponentially distributed with m = 4 ln 2 which is of the form m = λ ln 2 . Substitute the value λ in the formula for the exponential density function as, f ( x ) = ( 1 λ ) e − x λ = ( 1 4 ) e − x 4 = 1 4 e − x 4 Thus, the uniform density function is f ( x ) = { 1 4 e − x 4 x ≥ 0 0 otherwise . Use the above mentioned formula and obtain the mean as follows. μ = λ = 4 Thus, the mean μ = 4 . Similarly, obtain the standard deviation as, σ = λ = 4 Thus, the standard deviation σ = 4

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

15th Edition

ISBN: 9781323047620

Author: Barnett, Ziegler, Byleen

Publisher: Pearson Education

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Chapter 11.4, Problem 44E

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

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Chapter 11.4, Problem 43E

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Chapter 11 Solutions

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

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 then the probability P ( μ − σ ≤ X ≤ μ + σ ) .

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