Last time we learned that a *random variable* $X$ is a random process with some numerical outcome and we focused on *discrete* random variables - i.e. random variables that generate integer values. Today, we'll continue our discussion of random variables but focus on *continuous* random variables that can, in principle, generate any real number. We'll also meet the normal distribution and explore its relationship with the binomial distribution.

This material is mostly discussed in sections 3.4, 3.5, and 4.1. There, material describing the relationship between the binomial and normal distributions is discussed in section 4.3.

Even last time, we met a few continuously distributed random variables:

- Example A: Let $X$ be the average number of points that Ohio State beats Michigan by until Michigan (someday, a long time from now) finally wins,
- Example B: Find somebody and choose $X$ to be a very precise measure of their height,
- Example C: Randomly choose a college and choose $X$ to be the average salary of all the professors.

The tricky thing is trying to figure out how to describe the distribution of a continuously distributed random variable.

The *continuous, uniform distribution* is probably the simplest example of a continuous distribution in general. Suppose I pick a *real* number between $-1$ and $1$ completely (and uniformly) at random. What does that even mean?

I suppose that the probability that the number lies in the left half of the interval (i.e. to the left of zero) should be equal to the probability that the number lies in the right half. Phrased in terms of a probability function applied to events, we might write

$$P(X<0) = P(X>0) = \frac{1}{2}.$$Pushing the previous slide a bit further, suppose we a number uniformly at random from an interval $I$. The probability of picking a number from a sub-interval should be proportional to the length of that sub-interval.

If the big interval $I$ has length $L$ and the sub-interval $J$ has length $\ell$, then I guess we should have

$$P(X \text{ is in } J) = \ell/L.$$Suppose we pick a number $X$ uniformly at random from the interval $[-10,10]$. What is the probability that the number lies in the interval $[1,3]$?

**Solution**: We simply divide the length of the sub-interval by the length of the larger interval to get

Note that we've indicated the event using the inequality $1<X<3$, as we will typically do.

A common way to visualize continuous distributions is to draw the graph of a curve in the top half of the $xy$-plane. The probability that a random value $X$ with that distribution lies in an interval is then the area under the curve and over that interval. This curve is often called the *density function* of the distribution.

In the case of the uniform distribution over an interval $I$, the "curve" is just a horizontal line segment over the interval $I$ at the height 1 over the length of $I$. In the picture below, for example, $I=[0,2]$. On the left, we see just the density function for the uniform distribution over $I$. On the right, we see the area under that density function and over the interval $[0.5,1]$. The area is $1/4$ since $$P(0.5<X<1) = 1/4.$$

This pictorial view helps us see how continuous distribution might arise as a *limit* of discrete distributions.

That brings us to the *normal distribution* - the most important distribution in elementary statistics!

Suppose we go outside, grab the first person we see and measure their height. Accoring the data in our CDC data set, the average person has a height of about 67.18 inches. I suppose that our randomly grabbed person will have a height of close to that 67.18 inches. In fact, *most* people have a height of
$$67.18 \pm 4 \text{ inches}.$$
Of course, there are people taller than 72 inches and shorter than 62 inches, but the number of folks you find of a certain height grows more sparse as we move away from the mean.

If we try to plot a curve that meets the criteria of a density function for heights, I guess it might look something like so:

Note that the area is concentrated near the mean of $67.18$. The shaded area represents 1 standard deviation (or $4.12$) away from the mean; which represents about $68\%$ of the population. There are people 2 and even 3 standard deviations away from the mean but they taper off rapidly.

There's more than one normal distribution; there's a whole family of them - each specified by a *mean* and *standard deviation*.

All normal distributions have the same basic *bell shape* and the area under every normal distribution is one.

The mean $\mu$ of a normal distribution tells you where its maximum is.

The standard deviation $\sigma$ of a normal distribution tells you how concentrated it is about its mean.

The *standard normal distribution* is the specific normal with mean $\mu=0$ and standard deviation $\sigma=1$.

The interactive graphic below shows how the mean and standard deviation determine the corresponding normal distribution.