Explanation Given: The probability density function is f ( x ) = 1 6 x on the interval [ 1 , 16 ] . The median of X is defined to be the number m such that P ( X ≤ m ) = 1 2 . Concept used: The probability that an observed value of the random variable X lies in the interval [ a , b ] can be expressed as below. P ( a ≤ X ≤ b ) = ∫ a b f ( x ) d x ...... ( 1 ) Here, f ( x ) is the probability density function. Calculation: It is given that the probability density function is f ( x ) = 1 6 x on the interval [ 1 , 16 ] . The median of X is defined to be the number m such that P ( X ≤ m ) and that can be written as follows: P ( X ≤ m ) = ∫ 1 m f ( x ) d x Substitute 1 6 x for f ( x ) in the above expression to obtain the probability as follows: P ( X ≤ m ) = ∫ 1 m 1 6 x d x = 1 6 [ 2 x 1 2 ] 1 m = 2 6

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 32E

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The median of the random variable X with the probability density function f(x)=16x on the interval [1,16].

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Chapter 10.2, Problem 31E

Chapter 10.2, Problem 33E

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Chapter 10 Solutions

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

Ch. 10.1 - Define the following terms in your own words: a....Ch. 10.1 - Prob. 2CQ Ch. 10.1 - Prob. 3CQ Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E

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The median of the random variable X with the probability density function f ( x ) = 1 6 x on the interval [ 1 , 16 ] .

Solution Summary: The probability that an observed value of the random variable X lies in the interval left[a,bright] can be expressed as follows.

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The median of the random variable X with the probability density function f(x)=16x on the interval [1,16].

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Chapter 10 Solutions