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\title{Stat 311-Lecture 10 Normal Distribution}
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%\author{\Large Probability}
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Oct 20, 2005

\begin{enumerate}

\item The {\em expected value} of a continuous random variable X
with pdf $f(x)$ is $$\mathbb{E} X = \int_{-\infty}^{\infty} xf(x)
\;dx.$$
$$\mathbb{E} h(X) = \int_{-\infty}^{\infty} h(x)f(x)\;dx.$$
%\end{verbatim}
%\begin{figure}[h]
%\centering
%\includegraphics[width=50mm]{324.08.01.ps}
%\caption{Density function of $Unif(2,3)$.}\label{fig:hist}
%\end{figure}

\item The {\em cumulative distribution function} $F$ of a continuous
random variable $X$ is defined as
$$F(x) = P(X \leq x) = \int_{-\infty}^x f(x) \;dx.$$
Given $F$, we can recover the density
function by $f(x) = \frac{dF(x)}{dx}$ by the fundamental theorem of calculus.\\

%\item For given $p$, the $100p$-th {\em percentile} of the distribution of $X$ is a number
%$\eta(p)$ such that
%$$p = \int_{-\infty}^{\eta(p)} f(x) \;dx.$$

\item {\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 Problem. For $X \sim Unif(0,1)$, what is the probability density of
$Y=2X$? Answer. $Y \sim Unif(0,2)$.


\item A continuous random variable $Z$ is called a standard normal if
$$f(z) = \frac{1}{\sqrt{2\pi}}e^{-z^2/2}.$$
We denote $Z \sim N(0,1)$. $\mathbb{E} Z = 0, \mathbb{V} Z = 1$. The
cdf of the normal standard normal is
$$\Phi(z) = \int_{-\infty}^z
\frac{1}{\sqrt{2}}e^{-x^2/2}\;dx.$$ Let $X=\sigma Z + \mu$. Then $X$
is a normal distribution with parameters $\mu$ and $\sigma^2$. Its
density function is given by
$$f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/2\sigma^2}.$$ We
denote this distribution $X \sim N(\mu,\sigma^2).$


\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?  {\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. $P( 30 < X < 60) = P( -1.67 < Z < 0.83) = 0.75$.

%\begin{figure} \centering
%\includegraphics[width=50mm]{324.08.02.ps}
%\caption{Density function of standard normal
%$N(0,1)$}\label{fig:hist}
%\end{figure}
%\item The $100(1-\alpha)$-th percentile of $N(0,1)$ will be denoted
%by
% $z_{\alpha}$.


\begin{figure}
\centering
\includegraphics[width=100mm]{normal.eps}
\caption{Table for the cdf of standard normal
$N(0,1)$}\label{fig:hist}
\end{figure}



\end{enumerate}



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