Calculus with Applications (10th Edition)

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

Chapter 11 - Probability and Calculus - 11.2 Expected Value and Variance of Continuous Random Variables - 11.2 Exercises - Page 585: 19


a. $m=\sqrt[4] 2 \approx1.189$ b. $\approx 0.184$

Work Step by Step

We are given $f(x)=4x^{-5}; [1;\infty)$ The expected value $\mu= \frac{4}{3}$ a. The median is $\int^{m}_{1}f(x)dx=\frac{1}{2}$ $\int^{m}_{1}4x^{-5}dx=\frac{1}{2}$ $-x^{-4}|^{m}_{1}=\frac{1}{2}$ $-m^{-4}+1=\frac{1}{2}$ $m^{-4}=\frac{1}{2}$ $m^{4}=2$ $m=\sqrt[4] 2 \approx1.189$ b. The probability between mean and median is $P(\sqrt[4] 2 \leq X \leq \frac{4}{3})=\int ^{\frac{4}{3}}_{\sqrt[4] 2} 4x^{-5}dx=-x^{-4}|^{\frac{4}{3}}_{\sqrt[4] 2} \approx 0.184$
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