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 586: 24


a. 310 hours; b. $\sigma \approx 267$; c. $P(\mu + 1\sigma \leq t)\approx0.206$ d. $m\approx240$

Work Step by Step

We are given $f(x)=\frac{1}{58\sqrt t}$ for t in $[1,900]$ a. $\mu=\int^{900}_{1} t(\frac{1}{58\sqrt t})dt$ $=\int^{900}_{1}( \frac{1}{58}\sqrt t)dt$ $=\frac{1}{87}t^\frac{3}{2}|^{900}_{1}$ $=\frac{931}{3} \approx 310$ b. $Var(X) =\int^{900}_{1} t^{2}(\frac{1}{58\sqrt t})dt - (310)^{2}$ $=\int^{900}_{1}( \frac{1}{58}t\sqrt t)dt-96306$ $=\frac{1}{145}.t^\frac{5}{2}|^{900}_{1} -96306$ $\approx 71280$ $\rightarrow \sigma= \sqrt 71280 \approx 267$ c. $P(\mu + 1\sigma \leq t)=\int^{900}_{\mu + 1\sigma}(\frac{1}{58\sqrt t})dt$ $=\frac{1}{29}t^\frac{1}{2}|^{900}_{577} \approx0.206$ d. $\int^{m}_{1}(\frac{1}{58\sqrt t})dt=\frac{1}{2}$ $\frac{1}{29}(m^{\frac{1}{2}}-1)=\frac{1}{2}$ $m^{\frac{1}{2}}=\frac{31}{2}$ $m\approx240$
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