\magnification = \magstep1
\centerline{Mathematics 134 -- Intensive Calculus for Science 2}
\centerline{Discussion 4 -- Distributions and Densities}
\centerline{Thursday March 15, 2001}
\bigskip
\noindent
{\it Background}
\bigskip
\noindent
Last time, we introduced:
\medskip
\item{1)} the {\it density function} $p(x)$ for a quantity $x$
in a population.  This is the function that satisfies the equation:
(fraction of the population with 
$a \le x \le b$)  $= \int_a^b p(x)\ dx$.  
\item{2)}  The {\it cumulative distribution function} of $x$:
$$P(t) = \int_{-\infty}^t p(x)\ dx$$
which gives the fraction of the population with $x \le t$.
\item{3)}  The {\it median value of $x$}:  The value $t=T$ such that
half the population has $x \le T$ and the other half has $x \ge T$:
$P(T) = \int_{-\infty}^T p(x)\ dx = .5$
\item{4)}  The {\it mean value of $x$}:  The value
$$\overline{x} = \int_{-\infty}^\infty xp(x)\ dx$$
(analogous to the center of mass, since $\int_{-\infty}^\infty p(x)\ dx = 1$
here!)
\bigskip
\noindent
{\it Discussion Questions}
\bigskip
\item{A)}  Give qualitative sketches of graphs of a density function
and a cumulative distribution function that could represent
the distribution of the quantity $x = ${\it annual income} in a population 
where
\medskip
\itemitem{1)}  There is a large middle class 
\itemitem{2)}  There is a large group of poor people, and small 
middle and upper classes
\bigskip
\item{B)}  Recall, last time we introduced the density function
for phone call lengths:  $p(x) = .4e^{-.4x}$ ($x$ = length in minutes).
\bigskip
\itemitem{1)} What are the median and mean phone call lengths here?
\itemitem{2)} Suppose you selected a random phone call out of the
whole ``population''.  What is the {\it probability} that 
the length of that call is between $5$ and $10$ minutes?  Explain.
\bigskip
\item{C)}  Suppose that $x$ is the time (in hours) it takes students
in a large calculus lecture class at a state university to finish
an exam.  Assume that all students finish after 2 hours and that
the density function for $x$ is given by 
$$p(x) = \cases{x^3/4& if $0 < x < 2$\cr
                  0& otherwise\cr}$$
\medskip
\itemitem{1)}  What fraction of the students take between $1.5$ and 
$2$ hours to finish the exam?
\itemitem{2)}  What is the {\it median time} to finish the exam?
\itemitem{3)}  What is the {\it mean time} to finish the exam?
\bigskip
\item{D)}  The mean and median values of a quantity $x$ do 
not have to be equal (although they certainly could be in 
some cases).  Explain why the mean and median can be different
in intuitive terms.  (Think about means and medians for test scores 
in a class, for example.  Here the density function $p(x)$ would give 
the fraction of the class earning a score of $x$.  You can picture 
this with a histogram like our height histogram.)
\bigskip
\item{E)}  A basic principle of statistics is that (for 
large enough populations) many quantities
have density functions of the form
$$p(x) = {1\over \sigma\sqrt{2\pi}} e^{-(x - \mu)^2/(2\sigma^2)}.$$
If so, we say the quantity $x$ is {\it normally distributed}.
The constants $\sigma, \mu$ have the following names here --
$\mu$ is the {\it mean value} of $x$ (it is also the median in this case!).
$\sigma$ is another quantity called the {\it standard deviation}.
\bigskip
\itemitem{1)}  Plot $p(x)$ above with $\mu = 1, \sigma = 1$, 
then with $\mu = 1, \sigma = 2$, 
and finally with $\mu = 1, \sigma =  3$.  
What happens to the graph as $\sigma$ increases?
Describe in words what the standard deviation of a normal 
distribution tells you.
\itemitem{2)}  Let $\mu = 1, \sigma = 1$ as in your first plot above.
Sketch the cumulative distribution function for this density function.
\itemitem{3)}  The distribution of IQ scores for people is modeled
by a normal distribution with $\mu = 100$, and $\sigma = 15$.
Write the integral giving the fraction of the population
with IQ scores between 115 and 120.  Approximate this fraction
using a LHS with $n = 10$.
\bigskip
\noindent
{\it Assignment}
\bigskip
Group write-ups due in class on Tuesday, March 20.
\bye