CARNEGIE MELLON UNIVERSITY
Department of Mathematical Sciences
21-256 Normal Distributions

The probability density function corresponding to a normal distribution has the form

$$f(x)=\frac{1}{\sigma \sqrt{2\pi}} e^{-( (x-\mu)/\sigma )^2/2}$$

The mean, or average of this p.d.f. is $\mu$, the median is also $\mu$, and the standard deviation is $\sigma$, where in general for any p.d.f. $f$,

$$\sigma = \left[ \int_{-\infty}^{\infty} (x-\mu)^2 f(x) \ dx \right] ^2$$

and $\sigma$ measures how spread out the p.d.f is about the mean.

The probability of a continuous random variable, $X$, with a normal distribution is given by the integral

$$P(x_1 \leq X \leq x_2)=\frac{1}{\sigma \sqrt{2\pi}} \int_{x_1}^{x_2} e^{-( (x-\mu)/\sigma )^2/2} \ dx$$

This corresponds to the area under the graph of the p.d.f for $x_1 \leq X \leq x_2$. The integral can't be computed exactly, so we look to approximate this integral. Many scientific calculators and computer can compute the integral, and also integral tables can be used to get a reasonable approximation.

The first thing we do is we make a substitution, $z=(x-\mu)/\sigma$, which converts the integral to

$$\frac{1}{\sqrt{2\pi}} \int_{z_1}^{z_2} e^{-z^2/2} \ dz$$

where

$$z_1=(x_1-\mu)/\sigma, \ z_2=(x_2-\mu)/\sigma$$

This substitution may also be considered as the probability $P(z_1 \leq Z \leq z_2)$ for the continuous random variable $Z$ with p.d.f.

$$f(z)=\frac{1}{\sqrt{2\pi}} e^{-z^2/2}$$

We introduce the function $\Phi(z)$ to assist us in computing this integral.

$$\Phi(z)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{z} e^{-y^2/2} \ dy$$

Some properties of this function include

• $0 \leq \Phi(z) \leq 1$
• $\Phi$ is increasing.
• $\Phi(-\infty)=0$, $\Phi(0)=1/2$, $\Phi(\infty)=1$
• $P(z_1 \leq Z \leq z_2)= \Phi(z_2)-\Phi(z_1)$
• $\Phi(-z)=1-\Phi(z)$

Example: Suppose heights of students are normally distributed with mean $\mu=64$ and standard deviation $\sigma=4$. Find the probability that a student's height is between $62$ and $70$ inches.

Solution: We'll need to compute

$$P(62 \leq X \leq 70)=\frac{1}{4 \sqrt{2\pi}} \int_{62}^{70} e^{-( (x-64)/4 )^2/2} \ dx$$

which after transforming to the $Z$ variable becomes

$$P(-1/2 \leq Z \leq 3/2)=\frac{1}{ \sqrt{2\pi}} \int_{-1/2}^{3/2} e^{-y^2/2} \ dy$$

which requires that we compute

$$\Phi(3/2)-\Phi(-1/2)$$

Since $\Phi(-1/2)=1-\Phi(1/2)$, we need to compute

$$\Phi(3/2)+\Phi(1/2)-1$$

Looking up these values in a table, we obtain

$$\Phi(3/2)+\Phi(1/2)-1= .9332+.6915-1=0.6247$$