Calculus with Applications (10th Edition)

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

Chapter 11 - Probability and Calculus - Chapter Review - Review Exercises - Page 602: 26


$\mu=\frac{13}{2}$ $Var(X)=\frac{25}{12}$ $\sigma=\frac{5\sqrt 3}{6}$ $m=\frac{13}{2}$ The cumulative distribution is $F(x)=\frac{1}{5}x-\frac{4}{5}$

Work Step by Step

We are given $f(x)=\frac{1}{5}; [4;9]$ The expected value $\mu=\int^{9}_{4}xf(x)dx$ $=\int^{9}_{4}x(\frac{1}{5})dx$ $=\frac{1}{5}(\frac{x^{2}}{2})|^{9}_{4}$ $=\frac{1}{5}(\frac{81}{2}-8)=\frac{13}{2}$ The variance is $Var(X)=\int^{9}_{4}(x-\frac{13}{2})^{2}f(x)dx$ $=\int^{9}_{4}(x^{2}-13x+\frac{169}{4})\frac{1}{5}dx$ $=\frac{1}{5}\int^{9}_{4}(x^{2}-13x+\frac{169}{4})dx$ $=\frac{1}{5}(\frac{x^{3}}{3}-\frac{13}{2}x^{2}+\frac{169}{4}x)|^{9}_{4}$ $=\frac{25}{12}$ The standard deviation of X is $\sigma=\sqrt Var(X)$ $=\sqrt \frac{25}{12}=\frac{5\sqrt 3}{6}$ The median is $\int^{m}_{4}f(x)dx=\frac{1}{2}$ $\frac{1}{5}m-\frac{4}{5}=\frac{1}{2}$ $m=\frac{13}{2}$ The cumulative distribution is $F(x)=P(X\leq x)=\int ^{x} _{4}\frac{1}{5}dt=\frac{1}{5}t| ^{x} _{4}=\frac{1}{5}x-\frac{4}{5}$
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