## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 11 - Probability and Calculus - 11.2 Expected Value and Variance of Continuous Random Variables - 11.2 Exercises - Page 585: 5

#### Answer

$\mu=\frac{17}{6}$ $Var(X)=\frac{103}{180}$ $\sigma=\frac{\sqrt 515}{30}$

#### Work Step by Step

We are given $f(x)=1-\frac{1}{\sqrt x}; [1;4]$ The expected value $\mu=\int^{4}_{1}xf(x)dx$ $=\int^{4}_{1}x(1-\frac{1}{\sqrt x})dx$ $=\int^{4}_{1}(x-\sqrt x)dx$ $=(\frac{x^{2}}{2}-\frac{x^{\frac{3}{2}}}{\frac{3}{2}})|^{4}_{1}$ $=\frac{17}{6}$ The variance is $Var(X)=\int^{4}_{1}(x-\frac{17}{6})^{2}f(x)dx$ $=\int^{4}_{1}(x^{2}-\frac{17}{3}x+\frac{289}{36})(1-\frac{1}{\sqrt x})dx$ $=\int^{4}_{1}(x^{2}-x\sqrt x-\frac{17}{3}x+\frac{17}{3}\sqrt x+\frac{289}{36}-\frac{289}{36\sqrt x})dx$ $=\frac{x^{3}}{3}-\frac{2}{5}x^{\frac{5}{2}}-\frac{17}{6}x^{2}+\frac{34}{9}x^{\frac{3}{2}}+\frac{289}{36}x-\frac{289}{18}x^{\frac{1}{2}}|^{4}_{1}$ $=\frac{103}{180}$ The standard deviation of X is $\sigma=\sqrt Var(X)$ $=\sqrt \frac{103}{180}=\frac{\sqrt 515}{30}$

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