\documentclass[11pt]{amsart} \usepackage[charter]{mathdesign} \begin{document} \newcommand{\smean}{\overline X} \newcommand{\Ceil}[1]{\lceil{#1}\rceil} \newcommand{\Parens}[1]{\left({#1}\right)} \theoremstyle{remark} \newtheorem*{problem*}{Problem} \newenvironment{sol}{\begin{proof}[Solution]}{\end{proof}} \renewcommand{\labelenumi}{(\alph{enumi})} \title{HW Solutions} \author{Melikamp} \date{\today} \maketitle \section{Problems due by July 13} All drawings are omitted. All equalities in computations should be assumed to be approximate, even though $=$ is used. \begin{problem*}[2.6.1] \end{problem*}\begin{sol} $n=6$, $(85,92,103,108,128,156)$. \begin{enumerate} \item \[\smean = \frac{\sum_{i=1}^n X_i}n = 112.\] \item \[s = \sqrt{\frac{\sum_{i=1}^n(X_i-\smean)^2}{n-1}} = 26.15.\] \item \[Q_2 = \frac{103+108}{2} = 105.5.\] \item Using $\Ceil{\frac n4}$ to find the index of the first quartile, $Q_1 = 92$, $Q_3 = 128$. \end{enumerate} \end{sol} \begin{problem*}[2.6.5] \end{problem*}\begin{sol} $n=8$, $(46,77,83,84,85,90,91,94)$. \begin{enumerate} \item $\smean = 81.25$. \item $Q_2 = 84.5$. \item $s^2 = 231.36$. \item $s = 15.21$. \item $Q_1 = 77$, $Q_3 = 91$. \end{enumerate} \end{sol} \begin{problem*}[2.6.9] \end{problem*}\begin{sol} Here $n$ is the sum of frequencies, $n = 52$. \begin{enumerate} \item $Q_2 = 128$. \item The index is $\Ceil{52/4} = 13$, $Q_1 = 120$, $Q_3 = 136$. \end{enumerate} \end{sol} \begin{problem*}[2.6.18] \end{problem*}\begin{sol} $n = 25$. \begin{center} \begin{tabular}{rrrrr} 65 & 72 & 76 & 77 & 78\\ \hline 79 & \textbf{81} & 82 & 83 & 83\\ \hline 84 & 84 & \textbf{85} & 85 & 86\\ \hline 88 & 89 & 89 & \textbf{92} & 93\\ \hline 94 & 94 & 97 & 97 & 99 \end{tabular} \end{center} \begin{enumerate}\setcounter{enumi}2 \item The median is the $13$-th data point, $Q_2 = 85$. \item $\Ceil{25/4} = 7$, $Q_1 = 81$, $Q_3 = 92$. \item The sample range is $\max - \min = 99 - 65 = 34$. \end{enumerate} \end{sol} \begin{problem*}[3.9.4] \end{problem*}\begin{sol} Completed table: \begin{center} \begin{tabular}{rrrr|r} Gender & Private & Medicaid & None &\\ \hline Female & 250 & 452 & 208 & 910\\ Male & 128 & 680 & 157 & 965\\ \hline & 378 &1132 & 365 & 1875 \end{tabular} \end{center} \begin{enumerate} \item $P($no insurance$) = 365/1875 = 0.195$. \item $P($female $\cap$ no insurance$) = 208/1875 = 0.111$. \item $P($female $|$ no insurance$) = 208/365 = 0.57$. \item $P($female$) = 910/1875 = 0.485 \neq P($female $|$ no insurance$)$, so the events are dependent. \end{enumerate} \end{sol} \begin{problem*}[3.9.5] \end{problem*}\begin{sol} $P(T|D) = 0.8$, $P(D) = 0.3$, $P(T') = 0.76$. One can assemble the following table by having putting $1$ for total (since $P(\Omega) = 1$). \begin{center} \begin{tabular}{rrr|r} & $D$ & $D'$ &\\ \hline $T$ &&& 0.24\\ $T'$ &&& 0.76\\ \hline & 0.3 & 0.7 & 1 \end{tabular} \end{center} Also, $P(T|D) = 0.8 = \frac{P(T\cap D)}{P(D)}$, so \[P(T\cap D) = 0.8\cdot P(D) = 0.8\cdot 0.3 = 0.24,\] and the rest of the table can now be completed: \begin{center} \begin{tabular}{rrr|r} & $D$ & $D'$ &\\ \hline $T$ & 0.24 & 0 & 0.24\\ $T'$ & 0.06 & 0.7 & 0.76\\ \hline & 0.3 & 0.7 & 1 \end{tabular} \end{center} \begin{enumerate} \item $P(D\cap T) = 0.24$. \item $P(T) = 0.24$. \item $P(T'\cap D)\neq 0$, so these events are not mutually exclusive. \item $P(T)P(D) = 0.24\cdot0.3 = 0.072$, while $P(T\cap D) = 0.24$, so these events are dependent. \end{enumerate} \end{sol} \begin{problem*}[3.9.13] \end{problem*}\begin{sol} The answers follow directly from the completed table, not shown. \begin{enumerate} \item \begin{eqnarray*} P(\mbox{eligible}) &=& P(\mbox{eligible-enrolled}\cup\mbox{eligible-refused})\\ &=& (150+55)/300 = 0.683. \end{eqnarray*} \item $P($ineligible$) = 95/300 = 0.317.$ \item $P($enrolled $|$ male$) = 80/135 = 0.593$. \item $P($female $|$ ineligible$) = 60/95 = 0.631$. \item $P($eligible and male$) = (80+20)/300 = \frac13$, while \begin{eqnarray*} P(\mbox{eligible})P(\mbox{male}) &=& 0.683\cdot(135/300)\\ &=& 0.683\cdot0.45 = 0.307 \neq \frac13, \end{eqnarray*} so these events are dependent. \end{enumerate} \end{sol} \begin{problem*}[3.9.14] \end{problem*}\begin{sol} This is a binomial experiment with $n=12$ and $p=0.8$. \begin{enumerate} \item $P(X=10) = 0.2835$. \item If $p=0.82$, then \begin{eqnarray*} P(X=10) &=& {_{12}C}_{10}\cdot0.82^{10}\cdot0.18^2\\ &=& \frac{12!}{2!10!}\cdot0.82^{10}\cdot0.18^2\\ &=& \frac{11\cdot12}{2}\cdot0.82^{10}\cdot0.18^2\\ &=& 0.294. \end{eqnarray*} \item If $p=0.8$ again, then the expected value of $X$, $EX = pn = 0.8\cdot12 = 9.6$. \end{enumerate} \end{sol} \begin{problem*}[3.9.18] \end{problem*}\begin{sol} This is a binomial experiment with $p=0.4$. \begin{enumerate} \item If $n=10$, then \begin{eqnarray*} P(X\geq 4) &=& 1 - P(X\leq3)\\ &=& 1 - (P(X=0) + P(X=1) + P(X=2) + P(X=3))\\ &=& 1 - (0.0060 + 0.0403 + 0.1209 + 0.2150)\\ &=& 0.6178. \end{eqnarray*} (use the table). Here and below we use the probability axiom for complementary events in order to reduce the number of terms in our calculations. \item With $n=10$, \begin{eqnarray*} P(X\leq8) &=& 1 - P(X>8)\\ &=& 1 - (P(X=9) + P(X=10))\\ &=& 0.9983. \end{eqnarray*} \item $EX = 5250\cdot0.4 = 2100$. \end{enumerate} \end{sol} \begin{problem*}[3.9.25] \end{problem*}\begin{sol} $X\sim N(24,2.5^2)$. \begin{enumerate} \item \begin{eqnarray*} P(X>22) &=& 1 - P(X\leq22)\\ &=& 1 - P(Z\leq\frac{22-24}{2.5})\\ &=& 1 - P(Z\leq-0.8)\\ &=& 1 - 0.2119\\ &=& 0.7881. \end{eqnarray*} \item \begin{eqnarray*} P(15