Explanation Given information: f ( x ) = k x 2 ( 1 − x ) , 0 ≤ x ≤ 1 . Formula used: 1) Power rule of integration . ∫ x n d x = x n + 1 n + 1 , n ≠ − 1 2) Sum rule of integration. ∫ f ( x ) ± g ( x ) d x = ∫ f ( x ) d x ± ∫ g ( x ) d x Calculation: A probability density function is a function f defined by an interval ( a , b ) and having the following properties. i) f ( x ) ≥ 0 for every x . ii) ∫ a b f ( x ) d x = 1 Probability density function f ( x ) = k x 2 ( 1 − x ) , 0 ≤ x ≤ 1 Therefore ∫ a b f ( x ) d x = 1 ⇒ ∫ 0 1 k x 2 ( 1 − x ) d x = 1 Take the constant k outside the integral, ⇒ k ∫ 0 1 x 2 ( 1 − x ) d x = 1 Multiplying inside the bracket by

12th Edition

ISBN: 9780137590810

Author: Larry J. Goldstein, David C. Lay, David I. Schneider, Nakhle H. Asmar, William Edward Tavernetti

Publisher: PEARSON

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Chapter 12.2, Problem 11E

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To calculate: The value of k , such that the function f(x)=kx2(1−x), 0≤x≤1 is a probability density function on the specified interval.

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Chapter 12.2, Problem 10E

Chapter 12.2, Problem 12E

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The value of k , such that the function f ( x ) = k x 2 ( 1 − x ) , 0 ≤ x ≤ 1 is a probability density function on the specified interval.

Solution Summary: The author explains how a probability density function is defined by an interval (a,b) and having the following properties.

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To calculate: The value of k , such that the function f(x)=kx2(1−x), 0≤x≤1 is a probability density function on the specified interval.

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Chapter 12 Solutions