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The Central Limit Theorem is one of the most profound and useful results in all of statistics and probability, and yet it isn't all that hard to understand.
The idea is this: take some quantity that isn't necessarily normally distributed. For instance, consider annual salaries in the U.S.
\begin{enumerate}
\item Take a large sample of size $n$, say where $n=1000$.
\item Find the average salary of everyone in that sample, and record that average.
\item Take another sample and repeat many times.
\item If you take the records of all those \emph{averages} and use them to create a histogram, you'll see a symmetric, bell-shaped distribution.
\end{enumerate}
\begin{proc}{Central Limit Theorem}
The point of the Central Limit Theorem is this:
\begin{center}
If you take ``large'' ($n > 30$) samples from any sort of distribution,\\
the sample means will follow a normal distribution.
\end{center}
\begin{itemize}
\item The mean of this distribution will be the population mean.
\[\mu_{\overline{X}} = \mu_X\]
\item The standard deviation of this distribution will be the population standard deviation divided by the square root of the sample size.
\[\sigma_{\overline{X}} = \dfrac{\sigma_X}{\sqrt{n}}\]
\end{itemize}
\end{proc}
\vfill
\pagebreak
\subsection{Illustrating the Central Limit Theorem}
The following table records the number of days that a cookie recipe lasted at a diner.
\begin{center}
\begin{tabular}{c c | c c | c c | c c}
Recipe \# & $X$ & Recipe \# & $X$ & Recipe \# & $X$ & Recipe \# & $X$\\
\hline
& & & \\
1 & 1 & 16 & 2 & 31 & 3 & 46 & 2\\
2 & 5 & 17 & 2 & 32 & 4 & 47 & 2\\
3 & 2 & 18 & 4 & 33 & 5 & 48 & 11\\
4 & 5 & 19 & 6 & 34 & 6 & 49 & 5\\
5 & 6 & 20 & 1 & 35 & 6 & 50 & 5\\
6 & 1 & 21 & 6 & 36 & 1 & 51 & 4\\
7 & 2 & 22 & 5 & 37 & 1 & 52 & 6\\
8 & 6 & 23 & 2 & 38 & 2 & 53 & 5\\
9 & 5 & 24 & 5 & 39 & 1 & 54 & 1\\
10 & 2 & 25 & 1 & 40 & 6 & 55 & 1\\
11 & 5 & 26 & 6 & 41 & 1 & 56 & 2\\
12 & 1 & 27 & 4 & 42 & 6 & 57 & 4\\
13 & 1 & 28 & 1 & 43 & 2 & 58 & 3\\
14 & 3 & 29 & 6 & 44 & 6 & 59 & 6\\
15 & 2 & 30 & 2 & 45 & 2 & 60 & 5
\end{tabular}
\end{center}
\begin{enumerate}
\item Calculate the population mean and standard deviation:
\begin{align*}
\mu_X &= \\
\sigma_X &=
\end{align*}
\item Use a random number generator to select five samples of size $n=5$ each. Record the mean of each sample. Then copy the means from students around you until you have at least 30 sample means.
\vspace{2.5in}
\item Calculate the mean and standard deviation of this sampling distribution:
\begin{align*}
\mu_{\overline{X}} &= \\
\sigma_{\overline{X}} &=
\end{align*}
\item Repeat this process, taking five samples of size $n=10$ and recording the sample means for your samples and those of your classmates.
\vspace{2.5in}
\item Calculate the mean and standard deviation of this sampling distribution:
\begin{align*}
\mu_{\overline{X}} &= \\
\sigma_{\overline{X}} &=
\end{align*}
\item Draw a histogram for the original population, with a class width of one.
\vspace{2.5in}
\item Draw a histogram for the first sampling distribution (where $n=5$), with a class width of one half.
\vspace{2.5in}
\item Draw a histogram for the first sampling distribution (where $n=10$), with a class width of one half.
\vspace{2.5in}
\item See what observations you can make.
\end{enumerate}
\subsection{Using the Central Limit Theorem}
\begin{example}{College Age}
The mean age of college students in 2008 was $\mu=25$ years, with a standard deviation of $\sigma=6.8$ years. A simple random sample of 64 students is drawn. What is the probability that the average age of the students in the sample is greater than 26 years?\\
The sample means are normally distributed with mean \[\mu_{\overline{X}} = \mu_X = 25\] and standard deviation \[\sigma_{\overline{X}} = \dfrac{\sigma_X}{\sqrt{n}} = \dfrac{6.8}{8} = 0.85.\]
Therefore, to find the probability that the mean for this sample is greater than 26, use \texttt{normalcdf}:
\begin{center}
$P(\overline{X}>26) =$ \texttt{normalcdf(26,1000000,25,0.85)} $= 0.1197$
\end{center}
\end{example}
\vfill
\pagebreak
\begin{example}{Bull Weight}
If the mean weight of a bull is 1135 pounds, with a standard deviation of 97 pounds, would it be unusual for the mean weight of 100 head of cattle to be less than 1100 pounds?\\
The sample means are normally distributed with mean \[\mu_{\overline{X}} = \mu_X = 1135\] and standard deviation \[\sigma_{\overline{X}} = \dfrac{\sigma_X}{\sqrt{n}} = \dfrac{97}{10} = 9.7.\]
Therefore, the probability that the mean of this sample is less than 1100 pounds is
\begin{center}
$P(\overline{X} < 1100) = $ \texttt{normalcdf(-1000000,1100,1135,9.7)} $= 0.0002$
\end{center}
Yes, that would certainly be unusual.
\end{example}
This example sets up the kind of thing we'll do later: it would be unusual to get a sample like this if the claim is true about the mean weight of a bull, so if we DID get a sample like this, we might doubt that claim.
\begin{example}{Gas Mileage}
The EPA rates the mean highway gas mileage of the 2011 Ford Edge to be 27 miles per gallon. Assume the standard deviation is 3 miles per gallon. A rental car company buys 60 of these cars.
\begin{enumerate}[(a)]
\item What is the probability that the average mileage of the fleet is greater than 26.5 miles per gallon?
\begin{center}
\texttt{normalcdf(26.5,1000000,27,0.3873)} $= 0.9016$
\end{center}
\item What is the probability that the average mileage of the fleet is between 26 and 26.8 miles per gallon?
\begin{center}
\texttt{normalcdf(26,26.8,27,0.3873)} $= 0.2979$
\end{center}
\vspace{1in}
\pagebreak
\item Would it be unusual if the average mileage of the fleet were less than 26 miles per gallon?
\begin{center}
\texttt{normalcdf(-1000000,26,27,0.3873)} $= 0.0049$, so YES
\end{center}
\end{enumerate}
\end{example}
\begin{example}{NYC Rent}
The Real Estate Group NY reports that the mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is \$2631. Assume the standard deviation is \$500. A real estate firm samples 100 apartments.
\begin{enumerate}[(a)]
\item What is the probability that the sample mean rent is greater than \$2700?
\begin{center}
\texttt{normalcdf(2700,1000000,2631,50)} $= 0.0838$
\end{center}
\item What is the probability that the sample mean rent is between \$2500 and \$2600?
\begin{center}
\texttt{normalcdf(2500,2600,2631,50)} $= 0.2632$
\end{center}
\item Find the 60th percentile of the sample mean.
\begin{center}
\texttt{invNorm(0.6,2631,50)} $=$ \$2643.67
\end{center}
\item Would it be unusual if the sample mean were greater than \$2800?
\begin{center}
\texttt{normalcdf(2800,1000000,2631,50)} $= 0.0004$, so YES
\end{center}
\item Do you think it would be unusual for an individual apartment to have a rent greater than \$2800?
\begin{center}
NO
\end{center}
\end{enumerate}
\end{example}
\vfill
\pagebreak
\begin{example}{Battery Life}
A battery manufacturer claims that the lifetime of a certain type of battery has a population mean of $\mu=40$ hours and a standard deviation of $\sigma=5$ hours. Let $\overline{x}$ represent the mean lifetime of the batteries in a simple random sample of size 100.
\begin{enumerate}[(a)]
\item If the claim is true, what is $P(\overline{x} \leq 39.8)$?
\begin{center}
\texttt{normalcdf(-1000000,39.8,40,0.5)} $= 0.3446$
\end{center}
\item Based on the answer to part (a), if the claim is true, is a sample mean lifetime of 39.8 hours unusually short?
\begin{center}
Not really.
\end{center}
\item If the sample mean lifetime of the 100 batteries were 39.8 hours, would you find the manufacturer's claim to be plausible?
\begin{center}
Yeah, I think so.
\end{center}
\item If the claim is true, what is $P(\overline{x} \leq 38.5)$?
\begin{center}
\texttt{normalcdf(-1000000,38.5,40,0.5)} $= 0.0013$
\end{center}
\item Based on the answer to part (d), if the claim is true, is a sample mean lifetime of 38.5 hours unusually short?
\begin{center}
Yes.
\end{center}
\item If the sample mean lifetime of the 100 batteries were 38.5 hours, would you find the manufacturer's claim to be plausible?
\begin{center}
No.
\end{center}
\end{enumerate}
\end{example}