So, remember that for a die, the expected value of a die roll is 3.5.

If you don't remember,

we went through this calculation a couple lectures ago when we covered the variants.

The variance of a die roll is 2.92.

And the standard error is square root 2.92 divided by the number of

die rolls going into the average so that's 1.71 divided by square root n.

And so the standardized mean would just be the average of the die rolls minus 3.5

divided by 1.71 divided by square root of n.

On the next slide, basically what I've done is I rolled the die,

let's say one time, and I did that over and over and over again.

So an average of one.

And I standardized the average of 1, right?

So in case, for an average of 1, it would be a die roll minus 3.5 divided by 1.71.

And so now we have a distribution that's centered at zero and has variance one.

And I plotted the standard normal density in gray in the back.

I plotted the histogram of my die rolls.

And of course it can only take six possible values, and

you see those six spikes at one to six, and it's not perfectly discrete because

the software I'm using to plot the histogram assumes the data is continuous.

So, you see the six spikes basically from the fact that one die roll,

if we plot a histogram of a bunch of one die rolls, they are going to look like

a bunch of spikes, one to six, and in this case cuz it was normalized, they're gonna

look like the numbers one to six, where you subtract off 3.5 and divide by 1.71.

Okay now imagine if I just took two die rolls.

I took a die, rolled it once, I rolled it a second time, I got my average.

I subtracted off 3.5 and I dived by 1.71 divided by square root 2.

And I repeated that process over, and over, and over again and

I plotted histogram of the result of a lot of averages of two die rolls.

And so the histogram is going to give me a good sense of what the distribution of

the average of two die rolls what the standardized average of two die rolls is.

And in the background we have the normal distribution, and

on top of it we have the distribution of the average of two die rolls.

And I think that you'll agree that it's already, just by two die rolls,

looking pretty good.

Amazingly good.

And now imagine if you had six die rolls.

So, I rolled the dice six times, right?

Took the average, subtracted 3.5,

and then divided by 1.7 when divided by square root 6, right?

And then I did that process over and over and over again, and

I got lots of normalized averages of six die rolls, and I plotted a histogram,

and you can't even see the standard normal distribution in the background.

Because the distribution of the average of six die rolls looks so similar.

So if I was on my desert island and if I needed a standard normal, I think I could

probably get away with six die rolls and taking an average subtracting by 3.5 and

dividing by 1.71 divided by square root 6.

And the reason why I bring this up is because,

it's an interesting fact that the famous statistician Francis Galton,

who was quite a character, you should look up Francis Galton if you get a chance,

he was Charles Darwin's cousin, and he's a very brilliant guy.

He needed standard normals, but

was trying to get to simulate standard normals prior to having computers.

So how did he do it?

Well, he basically rolled dice and applied the central limit theorem

to get standard normals, which is really quite clever.

And he had, because it was a pain in the butt, and he wasn't on a desert island.

So he had time constraints.

He actually invented dice that made it a little bit easier for him to do.

I think he took standard dice and kind of wrote on the corners and stuff like that.

So there was more values than one to six.

But the basic idea is that the distribution of averages

looks like that of a normal distribution.

And just about in any sense,

regardless of what the underlying distribution of the date is.

There's some assumptions there that we assume that the variance was finite and

some other things like that.

But, for the purpose of this class,

it's basically any distribution that we can probably think of.

So let's take another version of the central limit theorem, or

another instance of the central limit theorem, from flipping coins.

So now, instead of a die I have coin.

And I want to evaluate the average of a bunch of coin flips.

So let's let Xi be zero or one result of the ith coin flip of a possibly unfair

coin where the p is the true success probability of the coin.

And the sample proportions say p hat is just the average of the coin flips.

Right?

The p hat is, in this case,

the percentage of ones and the average of the x i is the same thing, of course.

So remember that the expected value of the x i's is p.