% This is a simple LaTex sample document that gives a submission format
%   for IEEE PAMI-TC conference submissions.  Use at your own risk.

% Make two column format for LaTex 2e.


\documentclass[12pt,twocolumn]{article} %
\usepackage[dvips]{graphicx,graphics,color}
\usepackage{times,amsmath,amsfonts}

\setlength{\textheight}{10in} \setlength{\textwidth}{6.875in}
\setlength{\columnsep}{0.3125in} \setlength{\topmargin}{-0.5in}
\setlength{\headheight}{0in} \setlength{\headsep}{0in}
\setlength{\parindent}{1pc}
\setlength{\oddsidemargin}{-.2in}  % Centers text.
\setlength{\evensidemargin}{-.2in}

% Add the period after section numbers.  Adjust spacing.
\newcommand{\Section}[1]{\vspace{-8pt}\section{\hskip -1em.~~#1}\vspace{-3pt}}
\newcommand{\SubSection}[1]{\vspace{-3pt}\subsection{\hskip -1em.~~#1}
        \vspace{-3pt}}
\newcommand{\bqn}{\begin{eqnarray}}
\newcommand{\eqn}{\end{eqnarray}}
\newcommand{\bq}{\begin{eqnarray*}}
\newcommand{\eq}{\end{eqnarray*}}
\begin{document}

% Don't want date printed
\date{February 15, 2005}

% Make title bold and 14 pt font (Latex default is non-bold, 16pt)
\title{Stat 324: Lecture 9\\
Normal distributions}

% For single author (just remove % characters)
\author{Moo K. Chung\\
mchung@stat.wisc.edu}

% For two authors (default example)
\maketitle \thispagestyle{empty}
\begin{enumerate}



\item The $100(1-\alpha)$-th percentile of $N(0,1)$ will be denoted by
 $z_{\alpha}$. It is given by $100(1-\alpha) = \Phi(z_{\alpha})={\tt pnorm} (z_{\alpha},0,1)$. We
can find percentiles $z_{\alpha$ using $$>z_{\alpha}={\tt qnorm
(1-\alpha,0,1)}.$$

\begin{verbatim}
> alpha<-c(0,0.01,0.025,0.05,0.1)
> alpha
[1] 0.000 0.010 0.025 0.050 0.100
> qnorm(alpha,0,1)
[1]  -Inf -2.33 -1.96 -1.64 -1.28
> qnorm(1-alpha,0,1)
[1]      Inf 2.33 1.96 1.64 1.28
\end{verbatim}


\item The relationship between cdf and density function is $F(x) =
\int_{-\infty}^x f(y)\;dy$. Given $F$, we can recover the density
function by $f(x) = \frac{dF(x)}{dx}$ by the fundamental theorem of calculus.\\
{\em Example.} $f(x) = 2x$ for $0<x<1$ and $f(x)=0$ otherwise. The
cdf is given by $F(x) = x^2$ for $0<x<1$. $F(x)=0$ for $x <0$ and
$F(x)=1$ for $x>1$. By differentiating $F$, we recover the density
function.


\item Given $X \sim N(\mu,\sigma^2)$, let $Z=\frac{X-\mu}{\sigma}$.
It can be shown that this is a standard normal distribution. First
find the cdf of $Z$.
$$F_Z(z) = P(\frac{X-\mu}{\sigma} \leq z)=F_X(\sigma z + \mu).$$
By differentiating both sides,
$$f_Z(z) = f_X(\sigma z + \mu)\sigma =
\frac{1}{\sqrt{2\pi}}e^{-x^2/2}.$$ This is exactly the density
function of $N(0,1)$.
$$\mathbb{E} Z= (\mathbb{E} X -\mu)/\sigma=0, \mathbb{V} Z = \mathbb{V} X/\sigma^2 = 1.$

This implies that a linear function of a normal distribution is
again a normal distribution.


\item {\em Problem.} The length of time required to complete a college test is found to be normally distributed with mean 50 minutes and standard deviation 12 minutes. \\
(a) When should the test be terminated if we wish to allow
sufficient time for 90$\%$ of the students to complete the test? (b)
What proportion of students will finish the test between 30 and 60
minutes? You can use the following {\tt R} output.


{\em Solution.} Let $X$ be the length of time to complete the test.
Then $Z=\frac{X-50}{12} \sim N(0,1)$, the standard normal
distribution. Need to find $c$ such that $P(X < c) = P( Z <
\frac{c-50}{12}) =0.9$. From the {\tt R} output, $P(Z < 1.28) = {\tt
pnorm (1.28,0,1)}= 0.9$. $\frac{c-50}{12} = 1.28$. So at least
$c=65.36$ minutes should be given. $P( 30 < X < 60) = P( -1.67 < Z <
0.83) = {\tt pnorm (0.83,0,1) - pnorm(-1.67,0,1)} = 0.75$.

\item $100(1-\alpha)$ percentile of $N(\mu,\sigma^2)$ is $\sigma
Z_{\alpha} + \mu$. This idea will be used in probability plots
(Chapter 4.6).
\end{enumerate}
 {\bf Assigned Problems} Exercises 4.46. Exercise 4.53.


\end{document}