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: 4


$\mu=\frac{1}{3}$ $Var(X)=\frac{1}{18}$ $\sigma=\frac{\sqrt 2}{6}$

Work Step by Step

We are given $f(x)=2(1-x); [0;1]$ The expected value $\mu=\int^{1}_{0}xf(x)dx$ $=2\int^{1}_{0}x(1-x)dx$ $=2\int^{1}_{0}(x-x^{2})dx$ $=2(\frac{x^{2}}{2}-\frac{x^{3}}{3})|^{1}_{0}$ $=2(\frac{1}{6})=\frac{1}{3}$ The variance is $Var(X)=\int^{1}_{0}(x-\frac{1}{3})^{2}f(x)dx$ $=2\int^{1}_{0}(x^{2}-\frac{2}{3}x+\frac{1}{9})(1-x)dx$ $=2\int^{1}_{0}(x^{2}-x^{3}-\frac{2}{3}x+\frac{2}{3}x^{2}+\frac{1}{9}-\frac{1}{9}x)dx$ $=2\int^{1}_{0}(-x^{3}+\frac{5}{3}x^{2}-\frac{7}{9}x+\frac{1}{9})dx$ $=2(\frac{-x^{4}}{4}+\frac{5x^{3}}{9}-\frac{7}{18}x^{2}+\frac{1}{9}x)|^{1}_{0}$ $=\frac{1}{18}$ The standard deviation of X is $\sigma=\sqrt Var(X)$ $=\sqrt \frac{1}{18}=\frac{\sqrt 2}{6}$
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