An Introduction to Mathematical Statistics and Its Applications (6th Edition)

Published by Pearson
ISBN 10: 0-13411-421-3
ISBN 13: 978-0-13411-421-7

Chapter 3 Random Variables - 3.4 Continuous Random Variables - Questions - Page 136: 3

Answer

$\color{blue}{\dfrac{13}{64}}$

Work Step by Step

$\begin{align*} P\left(\left\vert Y - \frac{1}{2}\right\vert\lt \frac{1}{4}\right) &= P\left(-\frac{1}{4}\lt Y - \frac{1}{2}\lt \frac{1}{4}\right) \\ &= P\left(\frac{1}{2}-\frac{1}{4}\lt Y \lt \frac{1}{2} + \frac{1}{4}\right) \\ &= P\left( \frac{1}{4} \lt Y \lt \frac{3}{4} \right) \\ &= \int_{1/4}^{3/4} \frac{3}{2}y^2\ dy \\ &= \left[ \frac{3}{2}\frac{y^3}{3} \right]_{1/4}^{3/4} \\ &= \left[ \frac{1}{2}y^3 \right]_{1/4}^{3/4} \\ &= \frac{1}{2}((3/4)^3 - (1/4)^3) \\ &= \frac{1}{2}(27/64 - 1/64) \\ &= \frac{1}{2}\cdot \frac{26}{64} \\ \color{blue}{P\left(\left\vert Y - \frac{1}{2}\right\vert\lt \frac{1}{4}\right)} &\color{blue}{= \frac{13}{64}} \end{align*}$
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