\documentstyle{book} \begin{document} \section{Alternative definition of the mean}

It is useful to present an alternative formulation of the definition of the mean (which is also valid for the variance and standard deviation; i.e., $\sigma^2$ and $\sigma$, not $s^2$ and $s$). The usual definition can be rewritten: $$ \bar{x} = \frac{1}{n}\sum x_{i} = \sum \frac{1}{n} x_{i} $$ in the latter form the mean of the data set $\{ 2, 7, 4, 9, 4, 3, 6, 3, 7 \}$ is $$ \frac{1}{9}2 + \frac{1}{9} 7 + \frac{1}{9} 4 + \frac{1}{9} 9 + \frac{1}{9} 4 + \frac{1}{9} 3 + \frac{1}{9} 6 + \frac{1}{9} 3 + \frac{1}{9} 7 $$ which we can rewrite as $$ \frac{1}{9} 2 + \frac{2}{9} 3 + \frac{2}{9} 4 + \frac{1}{9} 6 + \frac{2}{9} 7 + \frac{1}{9} 9. $$ Instead of multiplying each datum by $\frac{1}{n}$, we multipy each value taken by the data set by the fraction of the time it occurs. In summation notation this formula is $$ \bar{x} = \sum p_{i}x_{i} $$ where the $x_{i}$ are the different values taken by the data set rather than different points in the data set, $p_{i} \geq 0$, and $\sum p_{i} = 1$. In this notation the variance is $\sigma^2 = \sum p_{i}(x_{i} - \bar{x})^{2}$.

\section{Histograms and summary statistics}

The construction of histograms entails grouping data together into classes for
better visual presentation. This grouping loses some of the original
information, specifically the values of data are replaced by ranges within
which the values lie. It is not possible to find the mean or median of the
data which provided a histogram, but a ``best'' estimate for the mean or
median can be calculated, and bounds on where the mean or median can be are
obtainable. The ``best'' estimates are obtained by assuming that the data is
uniformly spread within each class.

\noindent {\sl Example}: Consider a histogram which has 10 data in the class with class mark 100, 12 data in the class with class mark 125, 20 data in the class with class mark 150, 8 data in the class with class mark 175, and 5 data in the class with class mark 200. What can you say about the mean and median of the data?

The ``best'' estimate for the mean is obtained by assuming the data is uniformly spread within each interval; for purposes of calculating the mean, this is equivalent to assuming that all the data lie on the class marks. In this example the ``best'' estimate for the mean is $$ \mu = \frac{10 \times 100 + 12 \times 125 + 20 \times 150 + 8 \times 175 + 5 \times 200}{55} = 143\frac{7}{11}. $$

In order to get bounds on the mean, it is necessary to know the class boundaries, which are halfway between the class marks. Adding or subtracting $\frac{25}{2}$ from the class marks provides the class boundaries 87.5, 112.5, 137.5, etc. The least possible mean would occur if all of the data in each class were at the lower class boundary. In this example the least possible mean is $$ \mu = \frac{10 \times 87.5 + 12 \times 112.5 + 20 \times 137.5 + 8 \times 162.5 + 5 \times 187.5}{55} = 131\frac{3}{22}. $$ Similarly, the greatest possible value for the mean is $156\frac{3}{22}$. \vspace*{217pt}

The median is the middle value; uniformly spread data will provide that the area of the histogram on each side of the median will be equal. The total area of this histogram is $10 \times 25 + 12 \times 25 + 20 \times 25 + 8 \times 25 + 5 \times 25 = 55 \times 25 = 1375$. Of this sum, 250 comes from the first class, 300 comes from the second class, hence $\frac{1375}{2} - 550 = 137.5$ is needed from the third class to account for half the area. The area 137.5 is obtained from the third class by going $\frac{137.5}{20} = 6.875$ into it. The ``best'' estimate for the median is 144.375, which provides equal area in the histogram on either side of it. Since there are 55 data, the median is the value of the $28^{\rm th}$ in rank order. This datum will lie in the third class, which contains the $23^{\rm rd}$ through $42^{\rm nd}$ data. It is possible that all (most) of the data in the third class would be at the bottom or top of that class, hence the actual value of the median can be anywhere in the range 137.5 to 162.5. \vspace*{145pt}

\section{Other statistics}

There are many other statistics which are used. Two which denote relative
position which are often encountered are the rank and the z-score. The {\bf
rank} is just what you are familiar with as class rank: it gives the
position among the other data, but no raw score. The {\bf z-score} measures
how far a datum is from the mean in terms of standard deviation units.
Specifically, $z_{i} = \frac{x_{i} - \bar{x}}{\sigma} $. The z-score is a
measure of relative position.

\noindent {\sl Example}: If a set of data has mean $\bar{x} = 7$ and standard deviation $\sigma = 4$, what is the z-score corresponding to $x = 5$? The z-score is $\frac{5 - 7}{4} = -.5$.

\section{Exercises} \begin{enumerate}

\item The weights of students in a class are 130, 154, 210, 190, 200, 106, 180, 160, 125, 185, 210, 128, 150, 120, 150, 120, 140, 195, 235, 180, 148, 115, 155, 190, 126, 125, 125, 170, 140, 110, 148, 185, 120, 230, 190, 170, 135, 175, 168, 140, 202, 190, 120, 125, 167, 134, 200, 117, 160, 170, 180, 170, 190, and 120 pounds. Display this information in a histogram. Why did you choose your class marks? Why did you label it the way you did?

\item The heights of students in a class are 67, 68, 70, 72, 75.5, 63, 69, 71, 63, 70, 73, 69, 70.5, 64, 73, 63, 68, 77, 74, 71, 66.5, 67, 74, 76, 67, 63, 67, 68, 69, 66, 63, 74, 67, 69, 74, 74, 64, 72, 72, 69, 70, 73, 62, 64, 69, 68, 73.5, 64, 72, 73, 72, 71, 75, 66, 64.5, and 68 inches. Display this information in a histogram. Why did you choose your class marks? Why did you label it the way you did?

\item Give the mean, median, standard deviation, first quartile, third quartile, and inter-quartile range for the weights in problem 1.

\item Give the mean, median, standard deviation, first quartile, third quartile, and inter-quartile range for the heights in problem 2.

\item If 10\% of the marbles in a jar weigh .4 ounces each, 30\% weigh .5 ounces each, 40\% weigh .6 ounces each, and 20\% weigh .7 ounces each; what are the mean and median weight? What are the standard deviation and inter-quartile range?

\item If a class has several people between 60 and 78 inches tall, and one midget who is 36 inches tall, which average (mean, median, or midrange) will have the lowest value? Which average will have the highest value?

\item Why would you expect the mean income of students in your class to be greater than the median? Why would you expect the mean age of students in your class to be higher than the median?

\item Consider a histogram with 2 data in the class 7.5 -- 12.5, 5 data in the interval 12.5 -- 17.5, and 4 data in the class 17.5 -- 22.5. What is the best estimate for the mean, and what do you know for certain about the mean? What is the best estimate for the median, and what do you know for certain about the median?

\item For the data in problem 1, what are the z-scores corresponding to the weights 130, 154, and 210 pounds?

\item For the data in problem 2, what are the z-scores corresponding to the heights 60, 72, and 76 inches?

\end{enumerate} \end{document}