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: 20


a. $m=\sqrt[3] 2 \approx 1.26$ b. $\approx 0.2$

Work Step by Step

We are given $f(x)=3x^{-4}; [1;\infty)$ The expected value $\mu= \frac{3}{2}$ a. The median is $\int^{m}_{1}f(x)dx=\frac{1}{2}$ $\int^{m}_{1}3x^{-4}dx=\frac{1}{2}$ $-x^{-3}|^{m}_{1}=\frac{1}{2}$ $1-m^{-3}=\frac{1}{2}$ $m=\sqrt[3] 2 \approx 1.26$ b. The probability between mean and median is $P(\sqrt[3] 2 \leq X \leq \frac{3}{2})=\int ^{\frac{3}{2}}_{\sqrt[3] 2} 3x^{-4}dx=-x^{-3}|^{\frac{3}{2}}_{\sqrt[3] 2} \approx 0.2$
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