Calculus with Applications (10th Edition)

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

Chapter 11 - Probability and Calculus - 11.3 Special Probability Density Functions - 11.3 Exercises - Page 597: 2


a. $\mu=\frac{23}{8}$ b. $\sigma\approx0.072$ c. 1.288

Work Step by Step

We are given $f(x)=4; [2.75;3]$ The mean of the distribution: $\mu=\frac{1}{2}(2.75+3)=\frac{23}{8}$ The standard deviation of X is $\sigma=\frac{1}{\sqrt 12}(3-2.75)=\frac{\sqrt 3}{24} \approx0.072$ One standard deviation above the mean indicates the length of $3+0.072=3.072$ $P(T\geq3.072)=\int^{3.072}_{2.75}4dt=4t|^{3.072}_{2.75}\approx1.288$
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