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 242: 2

Answer

(a) $\color{blue}{0.4808}$ (b) $\color{blue}{0.1951}$ (c) $\color{blue}{0.8554}$ (d) $\color{blue}{0.0099}$ (e) $\color{blue}{0.0000}$

Work Step by Step

(a) $\begin{align*} P(0\le Z\le 2.07) &= P(Z\le 2.07) - P(Z\lt 0) \\ &= P(Z\le 2.07) - P(Z\le 0) ,\quad Z\sim N(0,1), \text{continuous pdf} \\ &= F_Z(2.07) - F_Z(0) \\ &= 0.9808 - 0.5000 \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\ P(0\le Z\le 2.07) &= \color{blue}{0.4808} \end{align*}$ (b) $\begin{align*} P(-0.64 \le Z \lt -0.11) &= P(Z\lt -0.11) - P(Z\lt -0.64) \\ &= P(Z\le -0.11) - P(Z\le -0.64),\quad Z\sim N(0,1), \text{continuous pdf} \\ &= F_Z(-0.11) - F_Z(-0.64) \\ &= 0.4562 - 0.2611 \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\ P(-0.64\le Z\lt -0.11) &= \color{blue}{0.1951} \end{align*}$ (c) $\begin{align*} P(Z \gt -1.06) &= 1 - P(Z\le -1.06),\quad Z\sim N(0,1), \text{continuous pdf} \\ &= 1 - F_Z(-1.06) \\ &= 1 - 0.1446 \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\ P(Z\gt -1.06) &= \color{blue}{0.8554} \end{align*}$ (d) $\begin{align*} P(Z \lt -2.33) &= P(Z\le -2.33),\quad Z\sim N(0,1), \text{continuous pdf} \\ &= F_Z(-2.33) \\ &= 0.0099 \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\ P(Z\lt -2.33) &= \color{blue}{0.0099} \end{align*}$ (e) $\begin{align*} P(Z \ge 4.61) &= P(Z\gt 4.61),\quad Z\sim N(0,1) , \text{continuous pdf} \\ &= 1 - F_Z(4.61) \\ &= 1 - 1.0000 \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\ P(Z\ge 4.61) &= \color{blue}{0.0000} \end{align*}$
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