Section 9.2 The Normal Distribution
Definition 9.2.1. The Normal Distribution.
Given two parameters \(\mu\) and \(\sigma\text{,}\) a random variable X over \(R = (-\infty,\infty)\) has a normal distribution provided it has a probability function given by
\begin{equation*}f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{ -\left ( \frac{x-\mu}{\sigma} \right ) ^2 / 2}\end{equation*}
The normal distribution is also sometimes referred to as the Gaussian Distribution (often by Physicists) or the Bell Curve (often by social scientists).
Theorem 9.2.2. Standard Normal Distribution.
If \(\mu = 0\) and \(\sigma = 1\text{,}\) then we say X has a standard normal distribution and often use Z as the variable name and will use \(\Phi(z)\) for the standard normal distribution function. In this case, the density function reduces to
\begin{equation*}f(z) = \frac{1}{\sqrt{2 \pi}} e^{ -z^2 / 2}\end{equation*}
Proof.
Convert to "standard units" using the conversion
\begin{equation*}z = \frac{x-\mu}{\sigma} = \frac{x-0}{1} = x.\end{equation*}
Notice that the work needed to complete the integrals over the entire domain above was pretty serious. To determine probabilities for a given interval is however not possible in general and therefore approximations are needed. When using TI graphing calculators, you can use
\begin{equation*}P( a \lt x \lt b ) = \text{normalcdf}(a,b,\mu, \sigma).\end{equation*}
Or you can use the calculator below.
Note that when no mean or standard deviation for normalcdf is provided, the calculator presumes standard normal.
Note that you can use a graphing calculator's normalcdf(a,b) = \(\Phi(b) - \Phi(a)\) to compute probabilities in the standard normal distrubution. If you have a normal distribution other than the standard and don't want to convert to standard, then the graphing calculator usage is normalcdf(a,b,\(\mu,\sigma\)).
Checkpoint 9.2.3. WebWork - Computing Standard Normal Probabilities.
Checkpoint 9.2.4. WeBWorK - Computing Normal Probabilities when non-Standard.
Theorem 9.2.5. Verifying the normal probability function.
\begin{equation*}\int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2 \pi}} e^{ -\left ( \frac{x-\mu}{\sigma} \right ) ^2 / 2} dx = 1\end{equation*}
Proof.
Note that you can convert the integral above to standard units5.6.1 so that it is sufficient to show
\begin{equation*}I = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{ -\frac{z^2}{2} } dz = 1\end{equation*}
Toward this end, consider \(I^2\) and change the variables to get
\begin{align*}I^2 & = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{ -\frac{u^2}{2} } du \cdot \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{ -\frac{v^2}{2} } dv\\& = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{ -\frac{u^2+v^2}{2} } du dv\end{align*}
Converting to polar coordinates using
\begin{equation*}du dv = r dr d\theta \end{equation*}
and
\begin{equation*}u^2 + v^2 = r^2\end{equation*}
gives
\begin{align*}I^2 & = \frac{1}{2 \pi} \int_0^{2 \pi} \int_0^{\infty} e^{ -\frac{r^2}{2} } r dr d\theta\\& = \frac{1}{2 \pi} \int_0^{2 \pi} -e^{ -\frac{r^2}{2} } \big |_0^{\infty} d\theta\\& = \frac{1}{2 \pi} \int_0^{2 \pi} 1 \cdot d\theta\\& = \frac{1}{2 \pi} \theta \big |_0^{2 \pi} = 1\end{align*}
as desired.
Theorem 9.2.6. Verifying the normal probability mean.
\begin{equation*}E[X] = \int_{-\infty}^{\infty} x \cdot \frac{1}{\sigma \sqrt{2 \pi}} e^{ - \left ( \frac{x-\mu}{\sigma} \right ) ^2 / 2} dx = \mu\end{equation*}
Proof.
The definition of expected value for a continuous variable in this case gives an integral to evaluate since X is continuous. In that integral, it is useful to use a standard change of variables as in basic integral calculus to convert the integral to something easier to evaluate. In this case, you will want to convert the X variable to the standard units variable Z so that
\begin{equation*}z = \frac{x-\mu}{\sigma}\end{equation*}
or by solving for x
\begin{equation*}x = \mu + z \sigma.\end{equation*}
So, it follows that
\begin{align*}E[X] &= \int_{-\infty}^{\infty} x \cdot \frac{1}{\sigma \sqrt{2 \pi}} e^{ - \left ( \frac{x-\mu}{\sigma} \right ) ^2 / 2} dx \\&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} (\mu + z\sigma) \cdot e^{ -z^2 / 2} dz\\&= \mu \cdot \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{ -z^2 / 2} dz + \sigma \cdot \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} z \cdot e^{ -z^2 / 2} dz\\&= \mu \cdot 1 + \sigma \cdot 0\\& = \mu\end{align*}
and therefore the use of \(\mu\) as the parameter in the normal distribution probability function is warranted.
Theorem 9.2.7. Verifying the normal probability variance.
Using a similar change of variable as in the previous theorem9.2.6,
\begin{equation*}E[(X-\mu)^2] = \int_{-\infty}^{\infty} (x-\mu)^2 \cdot \frac{1}{\sigma \sqrt{2 \pi}} e^{ - \left ( \frac{x-\mu}{\sigma} \right ) ^2 / 2} dx = \sigma^2\end{equation*}
Proof.
\begin{align*}E[(X-\mu)^2] & = \int_{-\infty}^{\infty} (x-\mu)^2 \cdot \frac{1}{\sigma \sqrt{2 \pi}} e^{ - \left ( \frac{x-\mu}{\sigma} \right ) ^2 / 2} dx\\& = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \sigma^2 z^2 \cdot e^{ -z^2 / 2} dz\\& = \frac{\sigma^2}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} z \cdot z e^{ -z^2 / 2} dz\\& = \frac{\sigma^2}{\sqrt{2 \pi}} \cdot \big [ -z e^{-z^2 / 2} \big |_{-\infty}^{\infty} + \int_{-\infty}^{\infty} e^{ -z^2 / 2} dz \big ]\\& = \frac{\sigma^2}{\sqrt{2 \pi}} \cdot \big [ 0 + \sqrt{2 \pi} \big ]\\& = \sigma^2 \end{align*}
using integration by parts. So, the use of \(\sigma\) as the other parameter in the normal probability function is also warranted.
Using simple calculus on the normal probability function provides a few tips regarding how to best sketch a nice graph.
Theorem 9.2.8. Normal Distribution Maximum.
The maximum of the normal distribution probability function occurs when \(x = \mu\)
Proof.
Take the derivative of the probability function to get
\begin{equation*}f'(x) = \frac{\sqrt{2} {\left(\mu - x\right)} e^{\left(-\frac{{\left(\mu - x\right)}^{2}}{2 \, \sigma^{2}}\right)}}{2 \, \sqrt{\pi} \sigma^{3}}\end{equation*}
which is zero only when \(x = \mu\text{.}\)
It is easy to see by evaluating to the left and right of this value that this critical value yields a maximum.
Points of Inflection for the normal distribution probability function occurs when \(x = \mu + \sigma\) and \(x = \mu - \sigma\text{.}\)Theorem 9.2.9. Normal Distribution Points of Inflection.
Proof.
Take the second derivative of the probability function to get
\begin{equation*}f''(x) = \frac{\sqrt{2} {\left(\mu + \sigma - x\right)} {\left(\mu - \sigma - x\right)} e^{\left(-\frac{\mu^{2}}{2 \, \sigma^{2}} + \frac{\mu x}{\sigma^{2}} - \frac{x^{2}}{2 \, \sigma^{2}}\right)}}{2 \, \sqrt{\pi} \sigma^{5}}\end{equation*}
which is zero only when \(x = \mu \pm \sigma\text{.}\)
It is easy to see by evaluating to the left and right of this value that these critical values yield points of inflection.
When using a graphing calculator's normalcdf(a,b,\(\mu,\sigma\)), pay attention to the the order of terms. For normal distributions, the calculator function always requires an interval. If you are looking for a one-sided probability, such as \(P(X \gt 4)\) for a problem with (say) mean \(\mu = 2\) and \(\sigma = 3\text{,}\) you can replace the infinite upper limit with "large" finite endpoint. Providing something that is more than 10 standard deviations above the mean is for all practical purposes infinity with respect to calculations. So, in this case \(P(X \gt 4)\) can be approximated by normalcdf(4,32,2,3). If you are brave, you can go even higher and use normalcdf(4,100000,2,3) if desired.
Checkpoint 9.2.10. WebWork - Computing Regular Normal Probabilities.
In the standard normal distribution, we have considered the case where you get the probability when given an interval. However, what about the reverse problem of finding an interval that would result in a given probability? That is, to solve for example the problem
\begin{equation*}\Phi(b) - \Phi(a) = P(a < z < b) = 0.6217.\end{equation*}
To deal with this we need an "inverse function" \(\Phi^{-1}\text{.}\) Toward that end, consider the simpler problem of solving
\begin{equation*}\Phi(z_0) = P(z < z_0) = \text{some given probability value} = \alpha.\end{equation*}
Then, technically the answer would be
\begin{equation*}z_0 = \Phi^{-1}(\alpha).\end{equation*}
Since integrating the normal probability function is impossible you can expect that finding a nice formula for the inverse of that integration might also be challenging and that is certainly the case. However, you have two options:
- Guessing \(z_0\) until you get a probabilty that is close enough to \(\alpha\text{.}\)
- Use a calculator that has the inverse built in!
For most graphing calculators, there is a function called "invNorm" and that is a way to compute values involving \(\Phi^{-1}\text{.}\) Indeed, for example if you wanted to solve
\begin{equation*}\Phi(z_0) = P(z < z_0) = 0.813\end{equation*}
then just use invNorm(\(0.813\)) to get \(z_0 \approx -0.2198\text{.}\)
Checkpoint 9.2.11. WebWork - Inverse Standard Normal Calculations.
While we are at it, can we "go backwards" and figure out the mean and variance if given some probabilities. This requires some problem solving skills and enough provided information to figure things out. For the exercise below, what does it mean to talk about the "middle" percentage of the area? That gives one the mean and also describes a probability of the sort
\begin{equation*}P(\mu - z_0 \sigma < X < \mu + z_0 \sigma),\end{equation*}
where \(z_0\) is a z-value obtained by using the inverse normal distribution function. It might help to draw a picture first of the area under the normal curve described in any such exercise.
