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 576: 40


a) $P(1 \leq t \leq 5)= \approx 0.537$ b) $P(t\gt10) \approx 0.231$

Work Step by Step

We are given $f(t)=\frac{1}{(\ln20)t}$ for $t$ in $[1,20]$ a) $P(1 \leq t \leq 5)=\int^{5}_1\frac{1}{(\ln20)t}dt=\frac{\ln t}{\ln 20}|^5_1=\frac{\ln 5}{\ln 20} \approx 0.537$ b) $P(t\gt10)=\frac{1}{\ln 20}\int^{20}_{10}\frac{1}{t}dt=\frac{\ln t}{\ln 20} |^{20}_{10} \approx 0.231$
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