\documentclass[30pt,landscape]{foils}

\usepackage{amsmath}

\usepackage{pause}               \usepackage{background}
\usepackage{graphicx}           \usepackage{geometry}
\usepackage{hyperref}

\usepackage{pp4slide}
\geometry{headsep=3ex,hscale=0.9}

\newtheorem{example}{Example} 
\newtheorem{question}[example]{Question}
\newtheorem{definition}[example]{Definition}
\newtheorem{theorem}[example]{Theorem}
\newtheorem{fact}[example]{Fact}
\newtheorem{notation}[example]{Notation}

\newcommand{\pz}{\pause}
\newcommand{\pb}{\pagebreak}

\definecolor{bgblue}{rgb}{0.03,0.39,0.23}
\vpagecolor{bgblue}

\hypersetup{pdftitle={lecture},
  pdfsubject={A mathematics lecture},
  pdfauthor={Dev Sinha, University of Oregon},
  pdfkeywords={pdftex, acrobat},
  pdfpagemode={FullScreen},
  linkcolor={red}
}

\begin{document}

{\Large Standard deviation}

We have seen that mean and median both measure the ``middle'' of
a set of data.  But the mean is easier to compute, while the median
is reliably in the middle. \pz

The situation is similar when measuring the ``spread'' of data. 
Last time we defined the inter-quartile ratio $Q_3 - Q_1$, which
gives us how much of a spread is needed to account for half 
of the data. \pz  The way to measure spread in a computable way,
akin to the average, is with the {\em standard deviation}. \pz

\begin{definition}
Let 
\[
x_{1},..,x_{n}
\]
be a list of data.  \pz Let $\overline{x}$ be the mean.  \pz The
standard deviation is given by 
\[
\sigma  = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}
(x_{i}-\overline{x})^{2}} 
\] \pb
\end{definition}

Why is this a reasonable measure of spread? \pz
If it is small then one expects the quartiles to be close
together, \pz and if it is large then the quartiles should be spread apart. \pz

\begin{example}[Excel example]
Excel can compute standard deviation with the STDEV command. \pz
We can see how the deviation changes for data sets with larger
and smaller ``spread.'' \pb
\end{example}

{\large Which description of data is better: five-number summary or 
mean and standard deviation?} \pz

The mean and standard deviation are always easier to compute; 
the five-number summary is always more accurate. \pz
Use $\overline{x}$ and $\sigma$ when you have a symmetric distribution of
data.  Use the five-number summary otherwise. \pz

If the distribution is approximately symmetric, the median and the
mean will be close, and the quartiles will be about equally placed
around the mean.  In that case, the mean, and the standard deviation
provide a similar level of information.  \pb



{\Large First manipulations with normal distributions} \pz

Last time we were introduced to the all-important ``Bell curve,''
otherwise known as a normal distribution.  \pz 
There are only  three numbers needed to describe a normal
distribution: \pz

The total number of data points.  \pz This is usually dealt with by ``setting
it to one and multiplying at the end.'' \pz

$\mu$, its mean, which is the center of the distribution, around which it
is symmetric \pz

and $\sigma$, its standard deviation 
(hinted at towards the end of the first lecture). \pz

We will learn better what $\sigma$ is, but now let's see how it can be used. \pb

\begin{theorem}
In a normal distribution, with mean $\mu$ and standard deviation $\sigma$,
\begin{enumerate}
\item 68\% of the observations fall within $\sigma$ of $\mu$  (within
one standard deviation of the mean). \pz
\item 95\% of the observations fall within $2\sigma$ of $\mu$ (within
two standard deviations of the mean). \pz
\item 99.7\% of the observations fall within $3\sigma$ of $\mu$
(within 3 standard deviations of the mean). \pz
\end{enumerate}
\end{theorem}
\includegraphics{Figures/gauds.jpg} \pz


\begin{example}
The height of adult males in the U.S. is normally
distributed with (measured in inches) $\mu = 69.3$ and 
$\sigma = 2.8$.  Based on this, \pz
\begin{enumerate}
\item In our class, how many men should be between 5'6'' and 6'?   (And how 
many men are between those heights?) \pz
\item What percentage of men are over six feet tall?\pz
\item If you were designing a piece of sports equipment with a minimum height needed
(golf clubs, hockey sticks), where should you set that height so that over 95\%
of men could use your equipment? \pb
\end{enumerate}
\end{example}

\begin{example}
Birth weight of babies born in the US is normally
distributed with 
\[
\overline{x} = 7.31 \mbox{ pound, and } s = 1.26 \mbox{ pounds.}  \pz
\]
\pz  Prof. Sinha's daughter Kiri was born at 6.12 pounds (6 pounds, 2 ounces).  
Roughly, what 
percentage of babies are born smaller than she?\pb
\end{example}

{\Large $z$-scores and comparing data on different distributions} \pz

Let's start with an example question:
Who's taller for their gender?  A 75 inch tall man, or a 72 inch tall
woman? \pz  (Who ranks more highly in terms of percentiles?) \pz

Male height distribution is $N (69.3,2.8)$.  So our man is height 5.7
inches taller than the mean $\mu$.  \pz But in order to figure
out where he is in terms of percentiles, we would need to know
how many $\sigma$'s (standard deviations) he was away from average. \pz
This is a matter of arithmetic:  $5.7 = 2.04\sigma$.  So our man's height
is $\mu +2.04\sigma$ because $75 = 69.3 + 2.04 \times 2.8$. \pz

Female height distribution is $N (64,2.7)$.  So our woman is 8 inches
taller than average.  We similarly compute that $8=2.96\sigma$, and
thus our woman's height is $\mu + 2.96 \sigma$. \pz

We don't need to compute the actual percentages to deduce that 
our woman is taller for a woman (more standard deviations above the
mean) than our man is for a man. \pause   

Notice that the numbers we are
looking (the multiples of the standard deviation by which our
observation is above or below the mean) are given by subtracting
the mean and then dividing by the standard deviation. \pz
The resulting numbers are called \emph{$z$-scores}. \pz

\end{document}




\end{document}
 