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\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<X<25) &=& P(X<25) - P(X<15)\\
    &=& P(Z<0.4) - P(Z<-3.6)\\
    &=& 0.6554 - 0 = 0.6554.
  \end{eqnarray*}
  \item Look up $0.05$ in the middle of the table (the 5-th
    percentile) in order to find that \[P(-1.64<Z<1.64) = 0.9.\]
    Then the 5-th percentile for $X$ is given by
    \[-1.64\cdot2.5+24 = 19.9,\]
    and the 95-th percentile is
    \[1.64\cdot2.5+24 = 28.1\]
  \end{enumerate}
\end{sol}

\begin{problem*}[3.9.26]
\end{problem*}\begin{sol}
  $X\sim N(500, 100^2)$.

  \begin{enumerate}
  \item \begin{eqnarray*}
    P(580<X<620) &=& P\Parens{\frac{580-500}{100} < Z < \frac{620-500}{100}}\\
    &=& P(0.8 < Z < 1.2)\\
    &=& 0.8849 - 0.7881 = 0.0968.
  \end{eqnarray*}
  \item $P(400<X<800) = 0.84$.
  \item The 90-th percentile for $Z$ is $1.28$ (look up 0.9 in the
    middle of the $Z$ table), so 90-th percentile for $X$ is
    \[1.28\cdot 100 + 500 = 628.\]
  \end{enumerate}
\end{sol}

\end{document}
