\setcounter{ExampleCounter}{1}
This bell-shaped curve is similar to a probability distribution function (it's called a \textbf{probability density function}).

Just like a probability distribution, the density curve tells us how likely a given outcome is, based on the height of the density curve at that point.\\

For instance, the average SAT verbal score is 508, and the distribution of scores looks like the following:

\begin{center}
\begin{tikzpicture}
\begin{axis}[
  no markers, domain=-4:4, samples=100,
  axis lines*=none, xlabel=$x$,
  hide y axis,
  every axis y label/.style={at=(current axis.above origin),anchor=south},
  every axis x label/.style={at=(current axis.right of origin),anchor=west},
  height=5cm, width=12cm,
  %xtick={-3,-2,-1,0,1,2,3}, ytick=\empty,
  %xticklabels={$\mu-3\sigma$,$\mu-2\sigma$,$\mu-\sigma$,$\mu$,$\mu+\sigma$,$\mu+2\sigma$,$\mu+3\sigma$},
  xtick={},
  xticklabels={,,,,,508,,,},
  enlargelimits=false, clip=false, %axis on top,
  grid = none
  ]
  \addplot [fill=cyan!20, draw=none, domain=-4:4] {gauss(0,1)} \closedcycle;
  %\addplot [fill=cyan!20, draw=none, domain=-1:1] {gauss(0,1)} \closedcycle;
  %\addplot [fill=yellow!20, draw=none, domain=-2:-1] {gauss(0,1)} \closedcycle;
  %\addplot [fill=yellow!20, draw=none, domain=1:2] {gauss(0,1)} \closedcycle;
  %\addplot [fill=green!20, draw=none, domain=-3:-2] {gauss(0,1)} \closedcycle;
  %\addplot [fill=green!20, draw=none, domain=2:3] {gauss(0,1)} \closedcycle;
  \addplot [very thick,cyan!50!black] {gauss(0,1)};


%\draw [yshift=2.5cm, latex-latex](axis cs:-1,0) -- node [fill=white] {68\%} (axis cs:1,0);
%\draw [yshift=1.5cm, latex-latex](axis cs:-2,0) -- node [fill=white] {95\%} (axis cs:2,0);
%\draw [yshift=0.5cm, latex-latex](axis cs:-3,0) -- node [fill=white] {99.7\%} (axis cs:3,0);
\end{axis}

\end{tikzpicture}
\end{center}

This means that the most common score is 508 (and thus that's the most likely result for a randomly chosen test taker).  Not only that, but this distribution gives us a precise description of how scores are clustered around this center.\\

\begin{proc}{Normal Distribution}
Two parameters define a particular normal distribution: the mean (center) and standard deviation (spread).
\begin{center}
\includegraphics[width=0.5\textwidth]{MultNorm}
\end{center}

Recall the Empirical Rule:
\begin{center}
\begin{tikzpicture}
\begin{axis}[
  no markers, domain=-4:4, samples=100,
  axis lines*=none, xlabel=$x$,
  hide y axis,
  every axis y label/.style={at=(current axis.above origin),anchor=south},
  every axis x label/.style={at=(current axis.right of origin),anchor=west},
  height=5cm, width=12cm,
  xtick={-3,-2,-1,0,1,2,3}, ytick=\empty,
  xticklabels={$\mu-3\sigma$,$\mu-2\sigma$,$\mu-\sigma$,$\mu$,$\mu+\sigma$,$\mu+2\sigma$,$\mu+3\sigma$},
  enlargelimits=false, clip=false, %axis on top,
  grid = major
  ]
  \addplot [fill=cyan!20, draw=none, domain=-1:1] {gauss(0,1)} \closedcycle;
  \addplot [fill=yellow!20, draw=none, domain=-2:-1] {gauss(0,1)} \closedcycle;
  \addplot [fill=yellow!20, draw=none, domain=1:2] {gauss(0,1)} \closedcycle;
  \addplot [fill=green!20, draw=none, domain=-3:-2] {gauss(0,1)} \closedcycle;
  \addplot [fill=green!20, draw=none, domain=2:3] {gauss(0,1)} \closedcycle;
  \addplot [very thick,cyan!50!black] {gauss(0,1)};


\draw [yshift=2.5cm, latex-latex](axis cs:-1,0) -- node [fill=white] {68\%} (axis cs:1,0);
\draw [yshift=1.5cm, latex-latex](axis cs:-2,0) -- node [fill=white] {95\%} (axis cs:2,0);
\draw [yshift=0.5cm, latex-latex](axis cs:-3,0) -- node [fill=white] {99.7\%} (axis cs:3,0);
\end{axis}

\end{tikzpicture}
\end{center}
\end{proc}

\begin{example}{The Intelligence Quotient}
IQ is normally distributed with a mean of 100 and a standard deviation of 16. Use the Empirical Rule to the find the data that is within one, two, and three standard deviations of the mean.

\marginnote{\bfseries Solution}
\begin{enumerate}
\item{68\% of the data is within one standard deviation of the mean.
\\ 
 IQ = mean $\pm (1 \cdot$ standard deviation $) = 100 \pm (1 \cdot 15) = 100 \pm 15 = (85, 115)$
 Thus, 68\% of people have an IQ between 85 and 115.
}
\item{95\% of the data is within two standard deviations of the mean.
\\ 
 IQ = mean $\pm (2 \cdot$ standard deviation $) = 100 \pm (2 \cdot 15) = 100 \pm 30 = (70, 130)$
 Thus, 95\% of people have an IQ between 70 and 130.
}
\item{99.7\% of the data is within three standard deviations of the mean.
\\ 
 IQ = mean $\pm (3 \cdot$ standard deviation $) = 100 \pm (3 \cdot 15) = 100 \pm 45 = (55, 145)$
Thus, 99.7\% of people have an IQ between 55 and 145.
}
\end{enumerate}
\begin{center}
\begin{tikzpicture}
\begin{axis}[
  no markers, domain=-4:4, samples=100,
  axis lines*=none, xlabel=$x$,
  hide y axis,
  every axis y label/.style={at=(current axis.above origin),anchor=south},
  every axis x label/.style={at=(current axis.right of origin),anchor=west},
  height=5cm, width=12cm,
  xtick={-3,-2,-1,0,1,2,3}, ytick=\empty,
  xticklabels={55,70,85,100,115,130,145},
  enlargelimits=false, clip=false, %axis on top,
  grid = major
  ]
  \addplot [fill=cyan!20, draw=none, domain=-1:1] {gauss(0,1)} \closedcycle;
  \addplot [fill=yellow!20, draw=none, domain=-2:-1] {gauss(0,1)} \closedcycle;
  \addplot [fill=yellow!20, draw=none, domain=1:2] {gauss(0,1)} \closedcycle;
  \addplot [fill=green!20, draw=none, domain=-3:-2] {gauss(0,1)} \closedcycle;
  \addplot [fill=green!20, draw=none, domain=2:3] {gauss(0,1)} \closedcycle;
  \addplot [very thick,cyan!50!black] {gauss(0,1)};


\draw [yshift=2.5cm, latex-latex](axis cs:-1,0) -- node [fill=white] {68\%} (axis cs:1,0);
\draw [yshift=1.5cm, latex-latex](axis cs:-2,0) -- node [fill=white] {95\%} (axis cs:2,0);
\draw [yshift=0.5cm, latex-latex](axis cs:-3,0) -- node [fill=white] {99.7\%} (axis cs:3,0);
\end{axis}

\end{tikzpicture}
\end{center}
\end{example}

\begin{example}{Car Sales}
Suppose you know that the prices paid for cars are normally distributed with a mean of \$17,000 and a standard deviation of \$500.  Use the 68--95--99.7 Rule to find the percentage of buyers who paid less than \$16,000.

\begin{center}
\begin{tikzpicture}
\begin{axis}[
  no markers, domain=-4:4, samples=100,
  axis lines*=none, xlabel=$x$,
  hide y axis,
  every axis y label/.style={at=(current axis.above origin),anchor=south},
  every axis x label/.style={at=(current axis.right of origin),anchor=west},
  height=5cm, width=12cm,
  xtick={-3,-2,-1,0,1,2,3}, ytick=\empty,
  xticklabels={{\$15,500},{\$16,000},{\$16,500},{\$17,000},{\$17,500},{\$18,000},{\$18,500}},
  enlargelimits=false, clip=false, %axis on top,
  grid = major
  ]
  \addplot [fill=cyan!20, draw=none, domain=-1:1] {gauss(0,1)} \closedcycle;
  \addplot [fill=yellow!20, draw=none, domain=-2:-1] {gauss(0,1)} \closedcycle;
  \addplot [fill=yellow!20, draw=none, domain=1:2] {gauss(0,1)} \closedcycle;
  \addplot [fill=green!20, draw=none, domain=-3:-2] {gauss(0,1)} \closedcycle;
  \addplot [fill=green!20, draw=none, domain=2:3] {gauss(0,1)} \closedcycle;
  \addplot [very thick,cyan!50!black] {gauss(0,1)};


\draw [yshift=2.5cm, latex-latex](axis cs:-1,0) -- node [fill=white] {68\%} (axis cs:1,0);
\draw [yshift=1.5cm, latex-latex](axis cs:-2,0) -- node [fill=white] {95\%} (axis cs:2,0);
\draw [yshift=0.5cm, latex-latex](axis cs:-3,0) -- node [fill=white] {99.7\%} (axis cs:3,0);
\end{axis}

\end{tikzpicture}
\end{center}

Since 95\% is between \$16,000 and \$18,000, 5\% is outside this range, so 2.5\% is below \$16,000.  
\end{example}
\vfill
\pagebreak

\subsection{z Table}
What if we want to know about points that don't happen to be exactly one, two, or three standard deviations away from the mean?

\paragraph{Recall:} z-score is how many standard deviations a data point is above or below the mean.

\paragraph{Now:} we use z-scores to describe the probability that a randomly chosen data point falls into a certain range.

\paragraph{Area under normal curve:} the area under the curve between $a$ and $b$ measures the proportion of data points between $a$ and $b$ (or the probability that a randomly chosen data point is in that range).

\paragraph{Standard normal distribution:} mean of 0, standard deviation of 1.  A z score converts a non-standard normal distribution to a standard one.\\

We can use a table like the one below to find the area under a curve in given ranges.
\begin{center}
\includegraphics[width=0.9\textwidth]{zTable}
\end{center}
\vfill
\pagebreak

More specifically, the table gives the proportion of values \textbf{below} any given $z$ score:
\begin{center}
\begin{tikzpicture}
\begin{axis}[
  no markers, domain=-4:4, samples=100,
  axis lines*=none, xlabel=$x$,
  hide y axis,
  every axis y label/.style={at=(current axis.above origin),anchor=south},
  every axis x label/.style={at=(current axis.right of origin),anchor=west},
  height=5cm, width=12cm,
  xtick={-1,0}, ytick=\empty,
  xticklabels={$z$,0},
  enlargelimits=false, clip=false, %axis on top,
  %grid = major
  ]
  \addplot [fill=cyan!40, draw=none, domain=-3:-1] {gauss(0,1)} \closedcycle;
  %\addplot [fill=yellow!20, draw=none, domain=-2:-1] {gauss(0,1)} \closedcycle;
  %\addplot [fill=yellow!20, draw=none, domain=1:2] {gauss(0,1)} \closedcycle;
  %\addplot [fill=green!20, draw=none, domain=-3:-2] {gauss(0,1)} \closedcycle;
  %\addplot [fill=green!20, draw=none, domain=2:3] {gauss(0,1)} \closedcycle;
  \addplot [very thick,cyan!40!black] {gauss(0,1)};


%\draw [yshift=2.5cm, latex-latex](axis cs:-1,0) -- node [fill=white] {68\%} (axis cs:1,0);
\end{axis}

\end{tikzpicture}
\end{center}

\paragraph{Reading the z table:} How do we read this?

For instance, to find the proportion to the left of $z=-1.73$, go down to the $-1.7$ row and over to the 0.03 column and read the proportion: \[0.0418\]

\begin{center}
\includegraphics[width=0.9\textwidth]{zTableEx}
\end{center}
\vfill
\pagebreak

\begin{example}{Using Z Table}
Find the area under the standard normal curve that is
\begin{enumerate}[(a)]
\item to the left of $z=0.47$.
\[0.6808\]
\item to the right of $z=-1.24$.
\[0.8925\]
\item between $z=0.86$ and $z=1.15$.
\[0.0698\]
\item outside the interval between $z=-0.44$ and $z=2.10$.
\[0.3479\]
\end{enumerate}
\end{example}
\vspace{1.5in}

\begin{example}{Using Z Table}
A normal distribution has mean $\mu=20$ and standard deviation $\sigma=4$.
\begin{enumerate}[(a)]
\item What proportion of the population is less than 18?
\[z_{18} = \dfrac{18-20}{4} = -0.5 \longrightarrow P(z < -0.5) = 0.3085\]
\item What is the probability that a randomly chosen value will be greater than 25?
\[z_{25} = \dfrac{25-20}{4} = 1.25 \longrightarrow P(z > 1.25) = 1-P(z \leq 1.25) = 0.1056\]
\end{enumerate}
\end{example}
\vfill
\pagebreak

\paragraph{Note:} the normal distribution describes a \textbf{continuous} random variable, where we never talk about the probability that $X$ \emph{equals} a given value.  This is because this probability is technically zero.  This doesn't affect our problems much, except that we can talk interchangeably about 
\begin{center}
$P(X \leq x)$ \hspace{0.2in} or \hspace{0.2in} $P(X < x)$.
\end{center}

\subsection{Using Your Calculator}
Here again, there's an easier way, using your calculator.  There's a built-in function called \texttt{normalcdf} that can calculate the proportion of the data in any given range for a normal distribution with any mean and standard deviation.

\begin{enumerate}
\item Press \includegraphics[height=0.3in]{Calc2ndVars}, then scroll to the second option: \texttt{2: normalcdf(}
\item You might see the following menu:
\begin{center}
\includegraphics[width=3in]{CalcNormalcdfMenu}
\end{center}
\item Enter values for the lower and upper bounds that you're interested in, as well as the mean and standard deviation of the given data set, and press \texttt{Paste}:
\begin{center}
\includegraphics[width=3in]{CalcNormalcdfEx}
\end{center}
If you don't get the menu from the previous step, just enter the information the way it is shown here, as
\begin{center}
\texttt{normalcdf(lower,upper,mean,stdev)}
\end{center}
\end{enumerate}
\vfill
\pagebreak

\begin{example}{Pregnancy Lengths}
The average length of a pregnancy is 272 days and the standard deviation is 9 days.  Find the probability that
\begin{enumerate}[(a)]
\item a randomly chosen pregnancy will last less than 252 days.
\begin{center}
\texttt{normalcdf(-1000000,252,272,9)} $= 0.0132$
\end{center} 
\paragraph{Note:} ``below 252'': between negative infinity and 252, so we put some large negative number there (some use \texttt{1E99}).

\item a randomly chosen pregnancy will last more than 252 days. 
\begin{center}
\texttt{normalcdf(252,1000000,272,9)} $= 0.9868$
\end{center}  
\paragraph{Note:} we also could have used the answer from part (a) to solve this one.

\item a randomly chosen pregnancy will last between 252 and 298 days. 
\begin{center}
\texttt{normalcdf(252,298,272,9)} $= 0.9849$
\end{center}
\end{enumerate}
\end{example}

\begin{example}{Blood Pressure}
The Centers for Disease Control and Prevention reported that diastolic blood pressures of adult women in the US are approximately normally distributed with mean 80.5 and standard deviation 9.9.
\begin{enumerate}[(a)]
\item What proportion of women have blood pressures lower than 70?
\begin{center}
\texttt{normalcdf(-1000000,70,80.5,9.9)} $= 0.1446$
\end{center}

\item What is the probability that a randomly chosen woman would have blood pressure between 75 and 90?
\begin{center}
\texttt{normalcdf(75,90,80.5,9.9)} $= 0.5438$
\end{center}

\item A diastolic blood pressure greater than 90 is classified as hypertension (high blood pressure).  What proportion of women have hypertension?
\begin{center}
\texttt{normalcdf(90,1000000,80.5,9.9)} $= 0.1685$
\end{center}

\end{enumerate}

\end{example}
\vfill
\pagebreak

\subsection{Working Backwards: Percentiles}
What if we turn the question around?  Instead of asking what percentage of people fall into a certain range, we could ask what range corresponds to a given percentage.  Of course, we've already done this, and we called it finding percentiles.

With a normal distribution, we can do this either with the z table or with the calculator.  We'll do an example of each, but after this, we'll stick with the calculator method.

\begin{example}{IQ Scores}
IQ scores have a mean of 100 and a standard deviation of 15.
\begin{enumerate}[(a)]
\item Find the 90th percentile using the z table. 

This means to find the point with 90\% of the data below it.  To use the table, locate a proportion as close as possible to 0.9000.  The closest we can get is 0.8997, but that's good enough.
\begin{center}
\includegraphics[width=0.9\textwidth]{zTableEx2}
\end{center}
This proportion belongs to a z score of 1.28; what data value is this?

Work backwards: \[1.28 = \dfrac{x-100}{15} \longrightarrow (1.28)(15) = x-100 \longrightarrow x = 119\]

\item Find the value with 20\% of the data above it, using the z table.

If 20\% of the data is above a certain point, 80\% must be below it.  Repeat the process from part (a) with 0.8000:
\begin{center}
\includegraphics[width=0.9\textwidth]{zTableEx3}
\end{center}
\[z=0.84=\dfrac{x-100}{15} \longrightarrow (0.84)(15) = x-100 \longrightarrow x = 113\]
\end{enumerate}
\end{example}
\vfill
\pagebreak

\begin{example}{Cherry Trees}
Cherry trees in a certain orchard have heights that are normally distributed with mean $\mu=112$ inches and standard deviation $\sigma=14$ inches.
\begin{enumerate}[(a)]
\item What proportion of trees are more than 120 inches tall?
\begin{center}
\texttt{normalcdf(120,1000000,112,14)} $= 0.2843$
\end{center}

\item What is the probability that a randomly chosen tree is either less than 100 inches tall or more than 125 inches tall?
\begin{center}
\texttt{normalcdf(-1000000,100,112,14)} $+$ \texttt{normalcdf(125,1000000,112,14)} $= 0.3722$
\end{center}

\item Find the 27th percentile of the tree heights.

Press \includegraphics[height=0.3in]{Calc2ndVars} and select \texttt{3: invNorm(} to pull up the following menu:
\begin{center}
\includegraphics[width=3in]{CalcInvNormEx}
\end{center}

Enter the desired proportion (0.27 in this case for 27\%), as well as the mean and standard deviation of the data set and press paste:
\begin{center}
\includegraphics[width=3in]{CalcInvNormEx2}
\end{center}
\vfill
\pagebreak

If the menu didn't show up for you, type it in as shown:
\begin{center}
\texttt{invNorm(proportion,mean,stdev)}
\end{center}

In this case, the answer is
\begin{center}
\texttt{invNorm(0.27,112,14)} $= 103.4$
\end{center}
Thus, a tree that is 103.4 inches tall will be in the 27th percentile.
\end{enumerate}

\end{example}

\begin{example}{Pregnancy Percentiles}
Recall that the average length of a pregnancy is 272 days and the standard deviation is 9 days.  Find the 65th percentile of pregnancy lengths.

\begin{center}
\texttt{invNorm(0.65,272,9)} $= 275.5$
\end{center}
\end{example}




