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: 1


$\mu=5$ $Var(X)=\frac{4}{3}$ $\sigma=\frac{2\sqrt 3}{3}$

Work Step by Step

We are given $f(x)=\frac{1}{4}; [3;7]$ The expected value $\mu=\int^{7}_{3}xf(x)dx$ $=\int^{7}_{3}x(\frac{1}{4})dx$ $=\frac{1}{4}(\frac{x^{2}}{2})|^{7}_{3}$ $=\frac{1}{4}(\frac{49}{2}-\frac{9}{2})=5$ The variance is $Var(X)=\int^{7}_{3}(x-5)^{2}f(x)dx$ $=\int^{7}_{3}(x^{2}-10x+25)\frac{1}{4}dx$ $=\frac{1}{4}\int^{7}_{3}(x^{2}-10x+25)dx$ $=\frac{1}{4}(\frac{x^{3}}{3}-5x^{2}+25x)|^{7}_{3}$ $=\frac{4}{3}$ The standard deviation of X is $\sigma=\sqrt Var(X)$ $=\sqrt \frac{4}{3}=\frac{2\sqrt 3}{3}$
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