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: 18


a. $m=\frac{2-\sqrt 2}{2}$ b. $=\sqrt 2 +\frac{3-2\sqrt 2}{2} -\frac{13}{9}$

Work Step by Step

We are given $f(x)=2(1-x); [0;1]$ The expected value $\mu= \frac{1}{3}$ a. The median is $\int^{m}_{0}f(x)dx=\frac{1}{2}$ $\int^{m}_{0}2(1-x)dx=\frac{1}{2}$ $2x-x^{2}|^{m}_{0}=\frac{1}{2}$ $2m-m^{2}=\frac{1}{2}$ $m=\frac{2-\sqrt 2}{2}$ b. The probability between mean and median is $P(\frac{2-\sqrt 2}{2} \leq X \leq \frac{1}{3})=\int ^{\frac{1}{3}}_{\frac{2-\sqrt 2}{2}} 2(1-x)dx=2x-x^{2}|^{\frac{1}{3}}_{\frac{2-\sqrt 2}{2}}=\frac{5}{9}-2+\sqrt 2+\frac{3-2\sqrt 2}{2}=\sqrt 2 +\frac{3-2\sqrt 2}{2} -\frac{13}{9}$
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