Answer
Yes
Work Step by Step
$\int^ {\infty}_{-\infty}6(x-x^2) =\int^ {1}_{0}1.5(x-x^2)=6\times\frac{1}{6}=1$
And the function is positive everywhere.
Therefore it could be a probability density function.
Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition
by LaTorre, Donald R.; Kenelly, John W.; Reed, Iris B.; Carpenter, Laurel R.; Harris, Cynthia R.
Published by
Brooks Cole
ISBN 10:
1-43904-957-2
ISBN 13:
978-1-43904-957-0
Chapter 6 - Analyzing Accumulated Change: Integrals in Action - 6.5 Activities - Page 474: 22
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
