# 16.1 - The Distribution and Its Characteristics

16.1 - The Distribution and Its Characteristics
Normal Distribution

The continuous random variable $$X$$ follows a normal distribution if its probability density function is defined as:

$$f(x)=\dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2} \left(\dfrac{x-\mu}{\sigma}\right)^2\right\}$$

for $$-\infty<x<\infty$$, $$-\infty<\mu<\infty$$, and $$0<\sigma<\infty$$. The mean of $$X$$ is $$\mu$$ and the variance of $$X$$ is $$\sigma^2$$. We say $$X\sim N(\mu, \sigma^2)$$.

With a first exposure to the normal distribution, the probability density function in its own right is probably not particularly enlightening. Let's take a look at an example of a normal curve, and then follow the example with a list of the characteristics of a typical normal curve.

## Example 16-1

Let $$X$$ denote the IQ (as determined by the Stanford-Binet Intelligence Quotient Test) of a randomly selected American. It has long been known that $$X$$ follows a normal distribution with mean 100 and standard deviation of 16. That is, $$X\sim N(100, 16^2)$$. Draw a picture of the normal curve, that is, the distribution, of $$X$$.

Note that when drawing the above curve, I said "now what a standard normal curve looks like... it looks something like this." It turns out that the term "standard normal curve" actually has a specific meaning in the study of probability. As we'll soon see, it represents the case in which the mean $$\mu$$ equals 0 and the standard deviation σ equals 1. So as not to cause confusion, I wish I had said "now what a typical normal curve looks like...." Anyway, on to the characteristics of all normal curves!

## Characteristics of a Normal Curve

It is the following known characteristics of the normal curve that directed me in drawing the curve as I did so above.

1. All normal curves are bell-shaped with points of inflection at $$\mu\pm \sigma$$.

Proof

The proof is left for you as an exercise

2. All normal curves are symmetric about the mean $$\mu$$.

Proof

All normal curves are symmetric about the mean $$\mu$$, because $$f(\mu+x)=f(\mu-x)$$ for all $$x$$. That is:

$$f(\mu+x)=\dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2} \left(\dfrac{x+\mu-\mu}{\sigma}\right)^2\right\} =\dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2} \left(\dfrac{x}{\sigma}\right)^2\right\}$$

equals:

$$f(\mu-x)=\dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2} \left(\dfrac{\mu-x-\mu}{\sigma}\right)^2\right\} =\dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2} \left(\dfrac{-x}{\sigma}\right)^2\right\}=\dfrac{1}{\sigma \sqrt{2\pi}} \text{exp}\left\{-\dfrac{1}{2} \left(\dfrac{x}{\sigma}\right)^2\right\}$$

Therefore, by the definition of symmetry, the normal curve is symmetric about the mean $$\mu$$.

3. The area under an entire normal curve is 1.

Proof
We prove this later on the Normal Properties page.
4. All normal curves are positive for all $$x$$. That is, $$f(x)>0$$ for all $$x$$.

Proof

The standard deviation $$\sigma$$ is defined to be positive. The square root of $$2\pi$$ is positive. And, the natural exponential function is positive. When you multiply positive terms together, you, of course, get a positive number.

5. The limit of $$f(x)$$ as $$x$$ goes to infinity is 0, and the limit of $$f(x)$$ as $$x$$ goes to negative infinity is 0. That is:

$$\lim\limits_{x\to \infty} f(x)=0$$ and $$\lim\limits_{x\to -\infty} f(x)=0$$

Proof

The function $$f(x)$$ depends on $$x$$ only through the natural exponential function $$\exp[-x^2]$$, which is known to approach 0 as $$x$$ approaches infinity or negative infinity.

6. The height of any normal curve is maximized at $$x=\mu$$.

Proof

Using what we know from our calculus studies, to find the point at which the maximum occurs, we must differentiate $$f(x)$$ with respect to $$x$$ and solve for $$x$$ to find the maximum. Because our $$f(x)$$ contains the natural exponential function, however, it is easier to take the derivative of the natural log of $$f(x)$$ with respect to $$x$$ and solve for $$x$$ to find the maximum. [The maximum of $$f(x)$$ is the same as the maximum of the natural log of $$f(x)$$, because $$\log_e(x)$$ is an increasing function of $$x$$. That is, $$x_1<x_2$$ implies that $$\log_e(x_1)<\log_e(x_2)$$. Therefore, $$f(x_1)<f(x_2)$$ implies $$\log_e(f(x_1))<\log_e(f(x_2))$$.] That said, taking the natural log of $$f(x)$$, we get:

$$\text{log}_e (f(x))=\text{log}\left(\dfrac{1}{\sigma \sqrt{2\pi}} \right)-\dfrac{1}{2\sigma^2}(x-\mu)^2$$

Taking the derivative of $$\log_e(f(x))$$ with respect to $$x$$, we get:

$$\dfrac{d\text{log}f(x)}{dx}=-\dfrac{1}{2\sigma^2}\cdot 2(x-\mu)$$

Now, setting the derivative of $$\log_e(f(x))$$ to 0:

$$\dfrac{d\text{log}f(x)}{dx}=-\dfrac{1}{2\sigma^2}\cdot 2(x-\mu) \stackrel{\equiv}{\scriptscriptstyle{SET}} 0$$

and solving for $$x$$, we get that $$x=\mu$$. Taking the second derivative of $$\log_e(f(x))$$ with respect to $$x$$, we get:

$$\dfrac{d^2\text{log}f(x)}{dx^2}=-\dfrac{1}{\sigma^2}$$

Because the second derivative of $$\log_e(f(x))$$ is negative (for all $$x$$, in fact), the point $$x=\mu$$ is deemed a local maximum.

7. The shape of any normal curve depends on its mean $$\mu$$ and standard deviation $$\sigma$$.

Proof

Given that the curve $$f(x)$$ depends only on $$x$$ and the two parameters $$\mu$$ and $$\sigma$$, the claimed characteristic is quite obvious. An example is perhaps more interesting than the proof. Here is a picture of three superimposed normal curves —one of a $$N(0,9)$$ curve, one of a $$N(0, 16)$$ curve, and one of a $$N(1, 9)$$ curve:

As claimed, the shapes of the three curves differ, as the means $$\mu$$ and standard deviations $$\sigma$$ differ.

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