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: 14


a) 0.75; b) 0.2375; c) 0.487; d) 0.464; e) 0.63

Work Step by Step

We are given $f(x)=\frac{3}{16}(4-x^2), [0,2]$ a) The expected value is $\mu=\int^2_0x(\frac{3}{16}(4-x^2))dx$ $=\frac{3}{16}\int^2_0(4x-x^3)dx$ $=\frac{3}{16}(2x^2-\frac{x^4}{4})|^2_0$ $=\frac{3}{4}=0.75$ b) The variance is $V=\frac{3}{16}\int^2_0x^2(4-x^2)dx-\mu^2$ $=\frac{3}{16}\int^2_0 (4x^2-x^4)dx-(0.75)^2$ $=\frac{3}{16}(\frac{4x^3}{3}-\frac{x^5}{5})|^2_0-(0.75)^2$ $=0.2375$ c) The standard deviation is $\sigma=\sqrt V \approx 0.487$ d) $P(x \geq \mu)=\int^2_{0.75}\frac{3}{16}(4-x^2)dx$ $=\frac{3}{16}(4x-\frac{x^3}{3})|^2_{0.75}$ $=0.464$ e) $P(\mu - \sigma \leq x \leq \mu + \sigma)=P(0.263 \leq x \leq1.237)$ $=\int^{1.237}_{0.263}\frac{3}{16}(4-x^2)dx$ $=\frac{3}{16}(4x-\frac{x^3}{3})|^{1.273}_{0.263}\approx 0.63$
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