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\centerline{Mathematics 136 -- Calculus 2}
\centerline{Discussion Day -- Distribution Functions}
\centerline{March 16, 2020}
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\noindent
{\it Background}
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We now look at applications of integration describing
the {\it distribution of various quantities
in populations}.  This kind of approach gives the foundations
for {\it mathematical probability and statistics}.  Statistical reasoning
is used to analyze data from experiments in almost all of the 
physical and social sciences, and is also used extensively in 
business and political decision-making. Over the next several days
we will look at a few of the basic ideas here and see how calculus
enters into these questions.  

Our starting point will be today's discussion, where we will look
at the distribution of heights in a hypothetical college class.  
Here is the heights data set (all heights in inches):
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\noindent
Women ($N = 16$):  71, 69, 68, 67, 67, 66, 66, 66, 66, 64, 63, 63, 63, 63, 63, 62
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\noindent
Men ($N = 10$):  72, 72, 72, 71, 70, 70, 69, 68, 68, 67
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\noindent
One way to represent the distribution of values in numerical data
like this is a graph called a {\it relative frequency histogram}.
To make a relative frequency histogram, we:
\begin{itemize}
\item First decide on a finite subdivision of the range
of possible values of the quantity we are measuring.  For instance
for the {\it women's heights} in the class, we can use intervals
of length $\Delta h =  1$ inch on the range $61.5 \le h \le 71.5$, {\it centered at
the integer values}. 
\item On each interval, we plot a rectangle whose 
height is {\it the fraction of the total population} whose measurement
(here the height) falls into that range.  For instance,  5 of
the women have height 63 inches, so the rectangle in the relative
frequency histogram for the women's heights centered at $h = 63$ 
has length $5/16 = .3125$.  
\item We do this for all the intervals and assemble the
rectangles (see plot on next page).
\item Note that this makes the sum of the areas of all of the rectangles
\emph{equal to one}.
\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{WHeightHist.png}
\caption{Relative frequency histogram for heights of the women.}
\end{center}
\end{figure}
\end{itemize}
\noindent
{\it Discussion Questions}
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\begin{enumerate}
\item[A)]  Using the data given above, construct relative frequency histograms
for the men's heights and then for the whole class (women and men).
First decide on appropriate ranges of $h$-values for each.  Your
histograms should include everyone in the appropriate categories, of 
course!
\item[] {\it Solution:}  See the graphs on the next page.
\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{MHeightHist.png}
\caption{Relative frequency histogram for heights of the men.}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{MWHeightHist.png}
\caption{Relative frequency histogram for heights of the whole class.}
\end{center}
\end{figure}


\item[B)]  Using these histograms, determine the 
fraction of the whole class that had heights between 
64 and 69 inches.  Determine the fraction of the women that had heights 
63 inches or less.  Determine the fraction of the men
that had heights 70 inches or more.
\item[] {\it Solution:}  $13/26 = .5$ of the class had heights between 64 and 
69 inches (inclusive).  This is the sum of the heights of the bars for 
heights $64,65,66,67,68,69$. $6/16 = .375$ of the women had heights 
$63$ inches or less.  $6/10 = .6$ of the men had heights 70 inches or more.
\item[C)]  Thinking about what you did in question B, describe in general
how the fraction of a population with heights $h$ in a range $a \le h \le b$, 
or $h \le b$, or $h \ge a$ relates to {\it areas} of rectangles
in that histogram. 
\item[] {\it Solution:}  You would add the heights of the rectangles
for $h$ in any one of these ranges to get the fraction of the population.
\item[D)]  It might not be the first way you would describe what is
happening with the histogram, but we can think of the 
histogram as a {\it graph} of a {\it ``height density function''}
$p(h)$.  
(Think of the ``tops''
of the rectangles -- horizontal
line segments -- as portions of the graph.)  
\begin{enumerate}
\item[1)] Give ``split-domain'' formula(s) for the height
density function for the men in the class.
\item[] {\it Solution:} The height density function for the men is 
$$p(h) = \begin{cases}
.1 & 66.5 < h < 67.5\\
.2 & 67.5 < h < 68.5\\
.1 & 68.5 < h < 69.5\\
.2 & 69.5 < h < 70.5\\
.1 & 70.5 < h < 71.5\\
.3 & 71.5 < h < 72.5
\end{cases}$$
and $0$ outside the interval $[66.5,72.5]$.
The endpoints of the intervals could also be included as the upper
limits of these bins.
\item[2)] Taking your answer from question C
one step farther, describe how you could compute the fraction of a
population with heights $h$ in a range $a \le h \le b$, 
or $h \le b$, or $h \ge a$ using {\it integrals}.
\item[] {\it Solution:}
These would be computed by 
$$\int_a^b p(h)\ dh, \quad \int_{-\infty}^b p(h)\ dh, \quad \int_a^\infty p(h)\ dh$$
Note that for the second integral, the lower limit could also be
given as $66.5$, since $p(h) = 0$ for all $h < 66.5$.  Similarly, the upper
limit in the third integral could be written as $72.5$.   
\end{enumerate}
\item[] If we had a large sample of people from 
a population (much larger than the 26 people in this
class), and we measured the heights more precisely (say to the 1/10)
of an inch, then the height relatively frequency histogram would 
look ``smoother'', with lots more thinner rectangles.  Mathematically
speaking, we can even imagine taking a limit 
with rectangles with $\Delta h \to 0$, and when we did this, we would
obtain the graph of 
what is called the population {\it height density function} $p(h)$.
\item[E)]  How would you express the fraction of the population 
with heights $a \le h \le b$ as an integral of $p(h)$?
Why must it be true that $p(h) \ge 0$ for all $h$, and 
$p(h) = 0$ for all $h < 0$.
What would be true about the value of 
$$\int_{-\infty}^\infty p(h)\ dh = \int_0^\infty p(h)\ dh?$$
\item[] {\it Solution:}  The fraction of the population with $a \le h \le b$
would be computed by 
$$\int_a^b p(h)\ dh.$$  
This last integral should equal $1$, since the 
whole population would have heights in the range $[0,\infty)$.  
\end{enumerate}
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\noindent
{\it Assignment}
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Writeups due Wednesday, March 18.
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