Explanation Given: The function f ( x ) = { 20 x 4 9 if 0 ≤ x ≤ 1 20 9 x 5 if x > 1 is a probability density function. Definition used: If X is a continuous random variable with probability function f on [ a , b ] , then the expected value of X is E ( X ) = μ = ∫ a b x f ( x ) d x The variance of X is V a r ( X ) = ∫ a b x 2 f ( x ) d x − ( μ ) 2 . And the standard deviation value of X is σ ≈ V a r ( X ) . Calculation: Consider the given function f ( x ) = { 20 x 4 9 if 0 ≤ x ≤ 1 20 9 x 5 if x > 1 Use the above mentioned definition and find expected value as follows, E ( X ) = μ = ∫ 0 ∞ x f ( x ) d x = ∫ 0 1 x f ( x ) d x + ∫ 1 ∞ x f ( x ) d x = ∫ 0 1 x ( 20 x 4 9 ) d x + ∫ 1 ∞ x ( 20 9 x 5 ) d x On further simplification the above expression becomes, E ( X ) = 20 9 ∫ 0 1 x 5 d x + 20 9 ∫ 1 ∞ ( 1 x 4 ) d x = 20 9 [ x 6 6 ] 0 1 + 20 9 [ − 1 3 x 3 ] 1 ∞ = 20 54 [ 1 6 − 0 5 ] − 20 27 [ 1 ∞ − 1 1 3 ] = 20 54 + 20 27 ( ∵ 1 ∞ = 0 ) On further simplification the above expression becomes, E ( X ) = 60 54 = 10 9 Thus, the expected value is E ( X ) = 10 9

Calculus with Applications (11th Edition)

11th Edition

ISBN: 9780321979421

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 11.2, Problem 22E

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To find: The expected value, the variance and the standard deviation.

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Chapter 11.2, Problem 21E

Chapter 11.2, Problem 23E

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Chapter 11 Solutions

Calculus with Applications (11th Edition)

Ch. 11.1 - YOUR TURN 1 Repeat Example 1(a) for the function...Ch. 11.1 - Prob. 2YT Ch. 11.1 - Prob. 3YT Ch. 11.1 - Prob. 4YT Ch. 11.1 - Prob. 1WE Ch. 11.1 - Prob. 2WE Ch. 11.1 - Prob. 3WE Ch. 11.1 - Decide whether the functions defined as follows...Ch. 11.1 - Decide whether the functions defined as follows...Ch. 11.1 - Decide whether the functions defined as follows...

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