## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 11 - Probability and Calculus - Chapter Review - Review Exercises - Page 602: 25

#### Answer

a) $\mu=4$; b) $V=0.5$; c) $\sigma=0.7071$; d) $m \approx 4.1213$; e) $F(x)=\frac{1}{9}x^2-\frac{4}{9}x+\frac{4}{9}$

#### Work Step by Step

We are given $f(x)=\frac{2}{9}(x-2), [2,5]$ a) The expected value is $\mu=\frac{2}{9}\int^5_2x(x-2)dx$ $=\frac{2}{9}\int^5_2(x^2-2x)dx$ $=\frac{2}{9}(\frac{x^3}{3}-x^2)|^5_2$ $=4$ b) The variance is $V=\frac{2}{9}\int^5_2x^2(x-2)dx-\mu^2$ $=\frac{2}{9}\int^5_2 (x^3-2x^2)dx-(4)^2$ $=\frac{2}{9}(\frac{x^4}{4}-\frac{2x^3}{3})|^5_2 - 16$ $=0.5$ c) The standar deviation is $\sigma=\sqrt V \approx 0.7071$ d) The median is $\int^m_2 f(x)dx=\frac{1}{2}$ $\frac{2}{9}(\frac{x^2}{2}-2x)|^m_2=\frac{1}{2}$ $\frac{m^2}{2}-2m+2=\frac{9}{4}$ $m^2-4m-\frac{1}{2}=0$ $m=\frac{4+3\sqrt 2}{2}$ and $m=\frac{4-3\sqrt 2}{2}$ $\rightarrow m =\frac{4+3\sqrt 2}{2} \approx4.1213$ e) The cumulative distribution is $F(x)=P(X \leq x)= \int^x_2 \frac{2}{9}(t-2)dx$ $= \frac{2}{9}( \frac{t^2}{2}-2t)|^x_2$ $= \frac{2}{9}( \frac{x^2}{2}-2x+2)$ $=\frac{1}{9}x^2-\frac{4}{9}x+\frac{4}{9}$

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