Explanation Given: A random variable X is said to be uniformly distributed over the interval [ a , b ] if it has probability density function f ( x ) = 1 b − a and x varies from a to b ( a ≤ x ≤ b ) . Concept used: For a probability density function f defined on the interval [ a , b ] associated with a continuous random variable X , the expected value E ( X ) of X is written as follows: E ( X ) = ∫ a b x f ( x ) d x ...... (1) The variance of X is written as follows: Var ( X ) = ∫ a b ( x − μ ) 2 f ( x ) d x ...... (2) Calculation: It is given that the probability density function f ( x ) = 1 b − a and x varies from a to b ( a ≤ x ≤ b ) . Substitute 1 b − a for f ( x ) in equation ( 1 ) to obtain the expected value E ( X ) as shown below: E ( X ) = ∫ a b x 1 b − a d x = 1 b − a [ x 2 2 ] a b = 1 b − a [ b 2 2 − a 2 2 ] = b 2 − a 2 2 ( b − a ) The above expression can be further simplified as follows: E ( X ) = b 2 − a 2 2 ( b − a ) = ( b − a ) ( b + a ) 2 ( b − a ) = ( b + a ) 2 Therefore, the expected value E ( X ) is ( b + a ) 2

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 10.2, Problem 23E

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The expected value E(X) and variance V(X) of the probability density function f(x)=1b−a as x varies from a to b.

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Chapter 10.2, Problem 22E

Chapter 10.2, Problem 24E

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Chapter 10 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

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The expected value E ( X ) and variance V ( X ) of the probability density function f ( x ) = 1 b − a as x varies from a to b .

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The expected value E(X) and variance V(X) of the probability density function f(x)=1b−a as x varies from a to b.

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