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 4 Special Distributions - 4.3 The Normal Distribution - Questions - Page 243: 6

Answer

$\color{blue}{1.349}$

Work Step by Step

$\underline{z_{.25}}$ $\begin{align*} P(Z\ge z_{.25}) &= .25 \\ 1-P(Z\ge z_{.25}) &= 1-.25 \\ P(Z \lt z_{.25}) &= 0.75 \\ P(Z\le z_{.25}) &= 0.75,\quad Z\sim N(0,1), \text{continuous pdf} \\ F_Z(z_{.25}) &= .75 \\ F_Z(z_{.25}) &\approx F_Z(0.6745) \qquad \text{[ see Appendix Table A.1, pp. 675-6; interpolate ]} \\ z_{.25} &\approx 0.6745 \qquad \text{[ since}\ F_Z(z)\ \text{is one-to-one for}\ 0 < F_Z(z) < 1\ ] \end{align*}$ $\underline{z_{.75}}$ $\begin{align*} P(Z\ge z_{.75}) &= .75 \\ 1-P(Z\ge z_{.75}) &= 1-.75 \\ P(Z \lt z_{.75}) &= 0.25 \\ P(Z\le z_{.75}) &= 0.25,\quad Z\sim N(0,1), \text{continuous pdf} \\ F_Z(z_{.75}) &= .25 \\ F_Z(z_{.75}) &\approx F_Z(-0.6745) \qquad \text{[ see Appendix Table A.1, pp. 675-6; interpolate ]} \\ z_{.25} &\approx -0.6745\qquad \text{[ since}\ F_Z(z)\ \text{is one-to-one for}\ 0 < F_Z(z) < 1\ ] \end{align*}$ $\underline{Q}$ $\begin{align*} Q &= z_{.25} - z_{.75} \\ &\approx 0.6745 - (-0.6745) \\ Q &\approx \color{blue}{1.349} \end{align*}$
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