Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 11 - Probability and Calculus - 11.1 Continuous Probability Models - 11.1 Exercises - Page 575: 22

Answer

$\frac{1}{63}x^{3}-\frac{27}{98}$

Work Step by Step

We are given $f(x)=\frac{3}{98}x^{2}$ for $3\leq x \leq 5$ The cumulative distribution function is given by $F(x)=P(X\leq x)=\int^{x}_{3}(\frac{3}{98}t^{2})dt$ $=\frac{1}{98}t^{3}|^{x}_{3}$ $=\frac{1}{63}x^{3}-\frac{27}{98}$
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