Today, we'll learn more about the normal distribution and see a tool that's great for ensuring that you get enough precision for your homework answers.

We've learned a lot over the couple of days about the normal distribution:

Key question now is - given a normally distributed random variable $X$ with mean $\mu$ and standard deviation $\sigma$, how do we compute quantities like

\begin{align} &P(X < x_0), \text{ or} \\ &P(X > x_0), \text{ or} \\ &P(x_1< X < x_0) \end{align}Last time we learned how to do this using our normal table. Today, we'll learn how to use a more flexible and precise tool on the computer.

Suppose that $X$ is normally distributed with mean $42.3$ and standard deviation $3.4$. Compute,

\begin{align} &P(X < 45), \\ &P(X > 45), \text{ and} \\ &P(40 < X < 45) \end{align}**Step 1**: The first step is the same as before - we compute the $Z$-scores for the values $40$ and $45$. Once we have the $Z$-scores, we can compute those values to the *standard* normal distribution. Recall that the formula for a $Z$-score is

Thus, for this example we have

$$z_1 = \frac{40-42.3}{3.4} = -0.67647 \: \text{ and } \: z_2 = \frac{45-42.3}{3.4} = 0.79412.$$For **Step 2**, we just plug the $Z$-score into the simple calculator below. We've got a more detailed version on our class webpage.

Of course, our driving mission is to understand *data*, be it *categorical* or *numerical*. Don't forget that data drives the following histogram, that in turn drives the approximating normal curve.

Suppose that NBA players have an average height of $6.5277$ feet with a standard deviation of $0.285$ feet. Supposing their heights are normally distributed, find the proportion of NBA players taller than $6'9''$.

**Solution**: First, let's express $6'9''$ in terms of feet:
$$6'9'' = \left(6 +9/12\right)\text{ ft} = 6.75 \text{ ft}.$$
We then compute the $Z$-score:

Finally, we plug that number into our calculator to get $$P(Z<0.78)\approx0.782305.$$ Thus, our answer is $1-0.782305 = 0.21769$.

Why do we care so much about normal distributions?

Because of the Central Limit Theorem, of course!!

The *central limit theorem* is the theoretical explanation of why the normal distribution appears as the limit of binomials above and, therefore, so often in practice. Suppose that $X$ is a random variable which we evaluate a bunch of times to produce a sequence of numbers:
$$X_1, X_2, \ldots, X_n.$$
We then compute the average of those values to produce a new value $\bar{X}$ defined by
$$\bar{X} = \frac{X_1 + X_2 + \cdots + X_n}{n}.$$
The central limit theorem asserts that the random variable $\bar{X}$ is normally distributed. Furthermore, if $X$ has mean $\mu$ and standard deviation $\sigma$, then the mean and standard deviation of $\bar X$ are $\mu$ and $\sigma/\sqrt{n}$.

Note that all of this is true regardless of the distribution of $X$!

The interactive tool below allows you to play with the parameters defining a binomial distribution and shows how the corresponding normal fits in.

Here's an interactive illustration of how a Normal distribution models data:

Note that the image is *data driven*. That is,

- The data comes first,
- then comes the histogram for the data,
- and finally, the curve that models data.

More specifically, the curve on the previous slide is *the* normal curve whose mean and standard deviation is the same as the data.