Probability The Central Limit Theorem

Hi, welcome back to the probability lectures here on www.educator.com, my name is Will Murray.0000

This is our very last probability lecture, I’m want to say a special thank you0006

to those of you who stuck with me through all the videos.0010

Today, we are going to talk about the central limit theorem which is one of the crown jewels of probability.0014

I'm very excited to talk about the central limit theorem and show you how it plays a role in sampling.0019

We will be doing a lot of problems, solving questions about samplings.0026

I need to give you the background here.0031

It starts out just like the previous video.0033

If you watched the previous video, sampling from a normal distribution, then the first slide is going to be exactly the same.0036

You can safely skip that and then, I will show you what the difference is when we are using the central limit theorem .0044

Let us jump into that, the setting here, like I said, this is exactly the same as in the previous video, at least for the first slide.0051

The idea is that we have a population of stuff.0061

For example, we could have a whole bunch of students at a university and each student is a different height.0063

We have some distribution of heights at the university.0071

There are some population mean which means the average of all the students at the university.0074

We might or might not know that, that μ might be known or it might not be known.0080

There is some variance σ².0085

In the problems that we are going to solve today, we will need to know what the variance is.0088

That should be given to you in the problems.0092

And then, we are going to take some samples which means we are going to go out in the quad of the university.0094

We will stop some students randomly and survey them on how tall they are.0102

Or if we do not like measuring how tall people are, you can ask them how many units they are carrying0107

or how much student that they have, or what their bank balance is, or other GPA.0113

It does not really matter for the purposes of probability, what quantity we are keeping track of,0119

the important thing is that we are taking random samples.0124

The way we are going to keep track of them is, each student that we talked to counts as 1 random variable.0130

For example, if we are talking about the heights of the students then Y1 is the height of the first student,0137

Y2 is the height of the second student, and so on, until nth student.0142

If we want to survey N students then YN is the height of the last student.0148

Each one of those counts as a random variable and we will calculate the average of those samples.0154

We will talk about probability questions related to whether the average of our sample0161

is really close to the average of the entire population.0166

That is all the same as in the previous lecture.0171

What else is the same as the previous lecture is that our samples are independent.0174

There is this catch phrase that you hear in probability a lot and in statistics,0179

independent identically distributed random variables.0184

They are independent meaning that, if we meet 1 student and that student is very tall,0188

it does not really tell us that the next student is going to be tall or short because they are independent.0194

Identically distributed means they are all coming from the same population.0200

That is a buzz phrase to say independent identically distributed random variables.0206

Here is where this lecture is different from the previous lecture.0212

In the previous lecture, we have to assume that our population was normally distributed,0216

which is not really valid when you are talking about heights of students.0221

Because one thing, a student can never have a negative height so it is not really normally distributed.0225

In this lecture, using the central limit theorem, which I have not gotten to yet,0232

we do not have to assume that the population is normally distributed.0236

The beautiful thing about the central limit theorem is that the population could have any distribution at all.0241

The central limit theorem is very broad and it applies to any distribution at all.0247

In particular, when you are doing sampling, you do not have to know the general parameters of the population at all.0254

You do not need to know that your population is a normal distribution.0261

That is the key feature of the central limit theorem is that,0265

you do not need to know ahead of time what kind of distribution you are working with.0269

Let me actually tell you, what the central limit theorem is.0275

It says that, if you have independent samples from any population with mean μ and variance σ²,0279

these are IID independent identically distributed samples.0287

The conclusion of the central limit theorem says that, Y ̅ is the sample mean.0292

That means you take the samples that you collect and you take their average,0300

just of those samples, and that is a new random variable.0309

That is a function of the random variables you had before.0313

It says that, the distribution of that random variable approaches as N goes to infinity,0316

it gets closer and closer to a normal distribution with mean μ and variance σ²/N.0324

This is really one of the most extraordinary facts in all of mathematics,0335

which is that we did not assume that the original population was normally distributed.0340

But, even without that assumption, the sample mean approaches a normal distribution.0347

This is sort of why the bell curve, the normal distribution is considered the most important distribution in all probability and statistics.0356

It is because even if you do not start with the normal distribution, you could start with any distribution at all,0365

you always end up with a normal distribution, as you take samples and you look at the sample mean.0371

That is really quite extraordinary but the math is very powerful and it does work out that way.0379

Let me mention that in practice, it is kind of the rule of thumb that people use in practice0386

when applying the central limit theorem is that, it starts to kick in, remembering it applies as N goes to infinity.0398

It really starts to become useful when N is bigger than about 30.0408

N is the number of samples that you take.0416

If you take more than 30 samples, you can safely assume that your sample mean will follow a normal distribution.0418

We can invoke the central limit theorem and say that the sample mean will have a normal distribution,0430

a normal distribution with mean μ and variance σ²/N.0446

That is kind of how the central limit theorem is used in practice.0451

As long as you take at least of 30 samples then, you can say that your sample mean is going to have a normal distribution.0456

It does not even matter, what the distribution of your original population was.0466

What do we actually do with that, once we know that the sample mean has a normal distribution,0471

what are we supposed to do with that.0476

From then on, it is pretty much the same as in the previous lecture.0478

We can walk you through that, in case you did not just watch the previous video.0483

What you do with a normal distribution is you convert it to a standard normal distribution.0488

A standard normal distribution, I will remind you is a normal distribution with mean 0 and variance 1.0494

It is what it means to be a standard normal distribution as mean 0 and variance 1.0507

The point of the standard normal distribution is that, you can look up probabilities for a standard normal distribution using charts.0513

Of course, there also lots of online applets that you can use,0521

a lot of computer programs will know tell you probabilities for a standard normal distribution.0525

For the videos in this lecture, I'm going to use charts.0531

If you are lucky enough to have access to the some kind of online tool or computer program,0536

that will tell you probabilities for a standard normal distribution then by all means, have at it and use that.0541

This is kind of an archaic method that I'm showing you here.0550

But still, in a lot of classroom settings people still use chart that is why I’m showing it to you.0553

That is how you can look up probabilities for a standard normal distribution.0558

This picture is one that is really useful to keep in mind.0563

This chart, the way it works is, it tells you the probability of being above a certain cutoff.0570

If you want to find the probability of being less than that cutoff, then you have to do something like subtracting from 1.0577

If you want to be between 2 cutoffs, then you have to figure out the probability of being in the tails0583

and then subtract those from 1.0589

That is a kind of computations that you have to do to use these charts.0591

That is all based on a standard normal distribution.0596

In practice, you usually do not get a standard normal distribution.0602

You usually get some kind of random normal distribution.0606

Let me show you how you convert it to a standard normal distribution.0609

I say recall because I did a whole lecture on this, earlier on in these probability lectures.0615

If this is totally new to you, what you might want to do is go back0621

and work through the lecture on the normal distribution, that we studied earlier on in this lecture series.0624

But if you already worked through that lecture, maybe you just need a quick refresher, here you go.0631

What we learned back then was that, if Y is any normal distribution then0635

what you can do is convert it to a standard normal distribution,0642

by subtracting off the mean and dividing by its standard deviation.0645

We call that new variable Z, and what we learned is that Z is a standard normal distribution.0651

The point of getting a standard normal distribution is then,0660

you can look up probabilities in terms of Z and convert them back to find probabilities in terms of Y.0662

What we learned in our central limit theorem was that, Y ̅ is essentially,0670

it approaches a normal distribution with mean μ and variance σ²/N.0679

What that means is that, if we do Y ̅ and we subtract its mean, that is Y ̅ - μ and divide by,0687

its standard deviation is always the square root of its variance.0697

I have to do √ σ ⁺N/N, I’m going to call that Z.0702

Now, you notice if I take the denominator and flip it upside down because of the fraction in the denominator,0710

then I will get σ², √σ² is just σ.0719

√N is going to flip up to the numerator, that is why I get that √N/σ × Y – μ.0726

That is where I’m getting this expression right here, that is where that comes from.0734

The variable that we just created is a standard normal variable.0741

I can use the charts to look up probabilities for that standard normal variable.0745

That is how I’m going to be solving the examples.0751

I’m going to ask you some kind of question about Y ̅,0755

and then what we will do is we will build up the standard normal variable, translate it into a question about Z.0760

And then, we will use the charts to look up probabilities on Z, that is how that is going to work.0768

Let us jump into the exercises and practice that.0775

In example 1, this is a very realistic problem.0778

Homework problems take you an average of 12 minutes each, but there is a lot of variation there,0782

there is a standard deviation of 10 minutes.0787

Maybe, if you get a real quick problem, you can quickly dispense of it in 2 minutes.0790

Or if you get a really tough one, it could take you 22 minutes or possibly even longer.0797

Your assignment is to solve 36 problems, this is going to take awhile.0802

What is the probability that it will take you more than 9 hours?0806

Let us think about that, first of all I have a conversion to solve here.0811

9 hours is 9 hours × 60 minutes per hour, that is 540 minutes.0819

If I'm going to take 540 minutes, I want to convert that into an average .0838

Remember, what I want to use is that Z is √N/σ × Y ̅ – μ.0845

I know that, that will be a standard normal variable.0858

Somehow, I got to buildup those quantities.0861

If my total time spent on the homework is going to be 9 hours, that is 540 minutes.0864

How much time will that be per problem on average?0871

My Y ̅, which is the average time per problem, if I spent 540 minutes total then0875

that would be 540 minutes divided by 36 problems.0888

I rigged up those numbers to work fairly nicely.0894

That is 15 minutes per problem.0898

I want to know the likelihood that I'm going to end up spending more than 15 minutes per problem,0909

over a 36 problem assignment.0915

I want to find the probability that my Y ̅ is bigger than 15.0922

Somehow, I want to build up this standard normal variable so I can use my normal distribution charts to solve this.0935

If Y ̅ is bigger than 15, that mean Y ̅ - μ is bigger than, what was my μ?0942

My μ is the average of all homework problems, 12 minutes on average,0951

is what it takes me to solve a homework problem.0958

15 -12, I will put in a σ and what is my σ?0962

My σ is standard deviation, that is 10.0973

The last ingredient here is √N, √N I'm going to include that.0980

What is my √N, N is the number of problems that I have.0991

N was 36, the √N is 6, this is 3 × 6/10, 18/10 is 1.8.0995

That was my variable Z, I want to find the probability that Z, my standard normal variable is going to be bigger than 1.8.1007

I'm going to look that up on the next page because I have a standard normal chart all set to go, on the next page.1021

Let me just go ahead and tell you the answer, so we can wrap it up on this page.1030

From the chart on the next page, and I will show you in a moment where that comes from.1035

The probability what was it, it was 0.0359 is what we are going to find on the next page.1040

If I want to think about in terms of percentages, that is just about 3.6%.1052

That is my probability that I'm going to spend more than 9 hours on this homework assignment.1060

It was about 3.6% chance that I'm going to spend more than 9 hours on this homework assignment.1068

Maybe, I’m worried because I have something I need to do in 9 hours.1074

I’m worried I would not get finished in time.1076

It actually looks pretty good, it looks like there is more than 96% chance that I will finish on time, that is kind of reassuring.1080

That is the answer to the problem, there is 3.6% chance that we will spend more than 9 hours on this homework assignment.1088

Let me just recap the steps there, before I show you that one missing step of looking it up on the chart.1095

We want to, first of all, convert into a standard unit here.1102

I got 9 hours and I got 12 minutes, I decided to convert the hours into minutes.1107

You can also convert the other way, if you wanted, but I think it is a little easier this way.1114

9 hours is 540 minutes, that was a total on our time that I would spend doing all the homework problems.1119

Since, I know that I have a result about averages here, I wanted to convert that into an average amount of time.1128

The average time is 540 minutes divided by 36 problems and that is just 15 minutes per problem.1138

The question is really, how likely am I to spend an average of more than 15 minutes per problem?1148

I want to find the probability that Y ̅ is bigger than 15.1154

For Y ̅ to be bigger than 15, I want to build up this expression Y ̅ - μ/σ × √N.1159

I filled in my μ is 12, μ right there is the average.1168

My σ was 10, there is σ and I got that from the problem here.1176

My √N is 6, that comes from N = 36, the number of problems that we have to solve here.1185

Then, I just simplified the numbers 15 -12 is 3, 3 × 6 is 18, 18 divided by 10 is 1.8.1193

I want to find the probability that my standard normal variable is bigger than 1.8.1199

That is what I’m going to confirm on the next page.1205

I will show you where that comes from, but we will see that it comes out to 0.0359.1207

If you think about that as a percentage, that is just about 3.6%.1214

I just want to confirm the result that we used on the previous page.1225

What we use on the previous page was, we calculated the probability that a standard normal variable,1231

because we had converted a Y ̅ to a standard normal variable, was bigger than 1.8.1236

I’m going to look up 1.8 on this chart, 1.80.1245

There is 1.8, there is 1.80, it is 0.0359 which was the number that I gave you back on the previous slide.1249

This shows you where that come from, it just come from this chart.1260

.0359, that is where I got that answer of 3.6% that we used on the previous slide.1263

That just completes that little gap that we had on the previous side.1274

That totally answers our probability of having to spend more than 9 hours on this horrible homework assignment.1278

In example 2, we have a bakery that is charting how many muffins do they start per day.1288

We have figure out a long-term average of 30 muffins per day but there is a lot of variation in there.1295

Maybe, they sell more muffins on the weekend and fewer on a weekday.1300

They figure out that, there is a standard deviation of 8 muffins.1305

What they are doing is, they are planning out the next month or so, actually 36 days.1309

They are worried about, what is the chance that they will sell more than 1000 muffins?1315

Maybe, they are worried about whether they are going to have to order some more supplies,1320

some more flours, some more eggs, or something like that.1323

Or maybe, they are worry about whether they are going to make enough money.1327

They know they need to sell 1000 muffins in the next 36 days.1330

It was the kind of calculations that a business person would make.1334

We are going to answer them using the central limit theorem.1338

Let me remind you what we have, our mean theorem is that, if we start out with Y ̅ – μ, very important distinction there.1342

Y ̅ - μ × √N/σ is a standard normal variable, that is kind of our main result for this lecture.1357

We want to figure out how we can use that here.1371

I see a Y ̅ there, that Y ̅ is the average of the number of muffins we are going to sell each day.1375

If the total muffins is going to be 1000, then that means the daily average is Y ̅ which will be 1000/36,1384

because there are 36 days.1412

It does simply a bit, I can take a 4 out of top and bottom there, simplify that down to 250/9,1413

still not the nicest fraction in the world.1421

I'm going to try to build up the standard normal variable and get an answer that I can look up easily on the normal charts.1425

Y ̅ - μ is 250/9 - μ is the overall average which we figure out is 30 muffins per day, that is -30.1434

That is a little awkward, let me go ahead and try to combine those fractions.1451

I did do this one in fractions because I rigged it up so the fraction work fairly nicely,1457

something we can work out in our heads.1462

If the fractions did not work nicely, I would probably just be going to a decimal right now.1464

But, this one works nicely.1468

250/9 - 30 is 270/9, we get -20/9, you can convert that into a decimal, if you like.1471

Let me continue to build up the standard normal variable.1486

√N × Y ̅ - μ/σ, I want to be bigger than the values that we have, because we want to sell more than 1000 muffins.1490

This should have been a greater than or equal to.1507

This should be greater than or equal to -20/9.1512

I’m multiplying by √N, the √N is 36, that is because N is 36.1517

√36 is 6, and now I have a σ.1522

Σ is our standard deviation, that is 8.1531

That is because, what we are told in the problem here.1537

I think this does simple fairly well, this is - 20/8 could simplify to 5/2, 6/2 could simplify to 3/1.1542

3/9 could simplify to 1/3, we just get -5/3.1557

Now, I'm going to convert it into a decimal.1563

The point of this was that, this was a standard normal variable.1566

I want that Z to be bigger than or equal to -5/3 which is as a decimal is -1.67.1573

2/3 is about 0.67, most technically that is an approximation.1580

I do not want any pure mathematicians to complain about that.1587

It is -1.67, and now I want to figure out the probability that Z will be bigger than -1.67.1591

Let me draw what I'm going to be looking for, then we will use a chart on the next page to actually calculate that.1602

-1.67 was down here somewhere.1610

I’m looking for the probability that Z is bigger than that.1617

I’m looking for all that probability.1619

The way the chart works is it will tell me probabilities of being bigger than a certain cutoff.1622

What I'm going to do is find the probability that Z is bigger than 1.67 and then subtract that.1630

The probability that Z is bigger than -1.67 is going to be 1 - the probability that Z is bigger than +1.67.1639

This is what we are looking for, this probability that Z is bigger than -1.67.1660

But I can figure it out as 1- this probability, the probability that Z is bigger than 1.67.1670

That is how I'm going to calculate that out.1681

The rest of it is simply a matter of looking it up on the chart, because that is the form that I can look things up on the chart.1684

I have already done this, if I look it up on the chart on the next page.1690

Let me tell you what the answer comes out to be.1696

It is 1- 0.0475 and that comes from the chart on the next page.1699

And then, 1- 0.0475 is 0.9525, and that is approximately 95%.1708

It is in fact very likely that, this bakery is going to sell more than 1000 muffins in the next 36 days.1721

If they are planning on buying supplies, buying flour, eggs for their muffins,1730

then they better go out and buy more supplies because it is very likely that they will sell more than 1000 muffins.1735

If they are worried about revenue then things are looking pretty good,1742

because there was a good chance that they will sell more than 1000 muffins.1745

Let me recap the steps here.1750

There is one missing step which is the chart, which I will fill in on the next slide.1751

In the meantime, the total muffins, we want that to be bigger than 1000,1757

which means the daily average should be bigger than 1000/36, which is more than 250/9.1762

I was just reducing the fractions there.1770

I rigged this one up to give us nice fractions.1772

And then, I kind of built up this standard normal variable.1775

I subtract a μ, the μ was the average of 30 muffins per day, that comes from there.1779

That is where that 30, and subtracted it and I got a negative number.1787

It is significant that it is negative there.1792

We do want to keep track of the negative sign.1795

And then, I multiply by √N which was, there is my N is 36.1798

√N is my 6 right there, divided by σ which is the standard deviation, 8 muffins right there, there is my 8.1804

And then, I just did some simplifying fractions there, got down to -5/3.1813

And I convert that back into a decimal which is -1.67.1820

In order to find the probability of Z being bigger than -1.67, I flipped it around and I calculated that the probability of being in this tail.1827

That is the probability of being in the tail, the probability that Z is bigger than 1.67.1840

For that, I’m going to use the chart on the next page.1847

I hope I have been reading the chart correctly.1849

When we look on the next page, it really will be 0.0475.1852

It will work out then to be 1- that is 95% chance that this bakery will sell more than 1000 muffins.1856

That is just filling that one missing step from the chart.1867

This is the normal distribution chart, this will tell you the probability that in normal variable,1871

we will end up being bigger than a particular cutoff.1878

In this case, our cutoff is 1.67, we are using that to solve the problem on the previous slide.1881

The probability that Z is bigger than 1.67, here is 1.6 and the second decimal place is 7, it is right there.1887

I hope this works out, 1.6 and 0.0475, that is what we use before.1899

It is 0.0475 and that was the answer that we plugged into our calculations on the previous side.1907

Just take this answer, drop into the calculations on the previous side.1918

And then, we did some more work and we got our answer to be 95%.1923

That fills in the one missing step from the previous slide.1932

It is just a matter of taking this 1.67 and matching up 1.6 and 0.07, and finding the right probability.1938

Of course, if you are using electronic tools, you probably do not need to use this chart.1948

You can just ask what is the probability that a standard normal variable will be bigger than 1.67,1953

and it should just spit out the answer for you.1959

In example 3, we have a technician fixing a soda machine.1963

It is supposed to give a certain amount of soda and she wants to check out whether it is dispensing the right amount of soda.1969

She takes 100 samples and the standard deviation in this machine is 2.5 ml.1976

We want to find the chance that her sample mean, that is the Y ̅,1986

that is the mean of her 100 samples is within 0.5 ml of the true average amount.1990

That is the true population mean, that is the μ of soda dispensed.1997

Let me setup that one out for you.2002

The whole point of this lecture is that, we look at Y ̅ - μ and we convert that to a standard normal variable.2005

The way we do that is by multiplying by √N/σ.2016

That turned out to be a standard normal variable meaning it has mean 0 and standard of deviation 1.2022

What we want to is figure out Y ̅ – μ.2032

In this case, we want the sample mean to be within 5 ml of the true mean.2038

Within 5 ml means it could go 0.5 ml either way.2044

I’m going to put absolute values here and set it less than or equal to 0.5 ml.2051

I’m going to try to build up my standard normal variable.2061

I'm going to build up by putting √N here and dividing by σ.2065

Of course, I got to do that on the other side, as well, √N and divided by σ.2071

The point of that is, that gives me a standard normal variable or actually there is a mass of values their,2078

I put the absolute values on Z as well.2084

That means, I want the absolute value of Z less than or equal to, let me fill in what I can here.2087

0.5 √N, N is 100 because we are taking 100 samples here.2093

That is × 10, √100 is 10, I rigged that up to make the numbers easy.2101

What is my σ, σ is 2.5 here.2108

10 divided by 2.5 is 4, this is 0.5 × 4 which is 2.2117

That worked out really nicely, is not it.2125

I want the probability that the absolute value of Z will be less than 2.2127

It is the same as the probability that Z is between -2 and 2.2134

Let me draw a little picture, it is always useful when you are working with these normal distributions2146

to draw a picture and figure out what it is you are actually calculating.2150

I want Z to be between -2 and 2, there is -2 and 2.2155

I want to be in between there.2162

The way my chart is set up, your chart might be different, but the way my chart works is,2164

it will tell you the probability of Z being in the positive tail.2170

It will tell you that area right there.2175

In order to find the probability in the middle, what I'm going to do is2179

calculate that probability in the tail and then subtract off 2 tails, that will give me the probability in the middle.2184

I will do 1-2 × the probability that Z is bigger than 2.2192

That should give me the probability of being between -2 and 2.2199

That is something now that I can look up on my chart,2205

or if you do not like charts and you got access to some kind of electronic tool, you can look that up more quickly.2209

Just drop the number 2 into your tool.2216

I’m going to use the chart and it is on the next page where I set up the charts.2219

Let me go ahead and tell you the numerical answer now, and then let me just confirm that on the next page.2224

I found the probability of Z being bigger than 2 from the chart.2229

It was 0.0228, and that is 1- 2 × 0.0228 is 0.0456.2234

That is 0.9544, and that is approximately 95%, just slightly over 95%.2250

That is my answer, that is the probability that this technician’s sample mean will be within 0.5 ml of the true mean.2271

Let me recap the steps there.2286

There is one missing step which is the step from the chart, which will confirm that on the next slide.2288

Just while we have the slide in front of us, let me recap the steps.2295

I set up my standard normal variable, I know that Y ̅ - μ is always × √N/σ is a standard normal.2299

I wanted Y ̅ - μ to be within 0.5 of each other.2311

Y ̅ μ should be within 0.5 of each other.2317

When we say two things are within a certain distance from each other, that really means that you want to bound their absolute value.2320

The distance from A to B is the absolute value of A – B.2329

We want the absolute value of Y ̅ - μ to be less than 0.5.2333

I just tacked on these other quantities, √N/σ.2338

The point of that was that gave me a standard normal variable, I can call that Z.2343

And then, I wanted to fill in what my √N and σ were.2350

N was 100, √N gave me 10, that is where that came from.2354

The σ was the standard deviation given there, that is my σ 2.5.2360

The numbers simplified nicely, that was me being clever, setting up nice numbers there.2367

It simplified down to 2, but now you have to think a bit more because you want the probability that Z is less than 2.2374

Absolute value, that means Z is between -2 and 2, that is the middle region here.2383

The way my chart works, it will tell you the tail region, that would not you the middle region directly.2391

What I did was, I said you could solve this by doing 1-2 tails, because there is a lower tail and there is an upper tail there.2397

I'm going to look up on the chart on the next page and we will see that the probability of Z being bigger than to is 0.0228.2407

It is 1-2 × that, 2 × that is 0.0456.2416

We finally get 0.9544, get a 95% chance that we will be within 0.5 ml.2421

By the way, example 4 is a follow up to example 3.2430

I want to make sure that you understand example 3.2435

It there is any steps in here that you are a little fuzzy on, just watch the video again and2437

just make sure that you are very clear on all the steps here,2442

because in example 4, we are going to tweak the numbers a little bit.2445

It really will help if example 3 is already very solid for you.2449

One missing step here is where that number comes from.2453

Let us fill in that step, this is still example 3 and we had one missing step2457

which was the probability that Z was bigger than 2, that was from the previous side.2464

In order to solve that, I see that there is 2.0 right there.2471

2.00 gives me 0. 0228 is what I used on the previous side.2477

I use that number on the previous side and you can catch up with all the rest of the arithmetic on the previous side.2492

If you do not remember how that worked out but we did some calculations with that.2501

We came up with an answer of 95% for this soda dispensing machine.2505

That wraps up example 3, we are going to use the same scenario for example 4.2514

I do want to make sure that you understand example 3, before you go ahead and try example 4.2519

In example 4, we have the same technician from example 3.2528

Remember, this technician is taking samples from a soda dispensing machine.2532

She wants to guarantee with probability 95% that her sample mean will be with 0.4ml of the true average.2537

This looks a lot like example 3, the difference, I will just remind you was with examples 3,2547

we had not 0.4 but a 0.5 ml tolerance.2555

Here we are restricting that to 0.4 ml tolerance.2562

We are deciding that 0.5 is not close enough, I want to get a 0.4 ml tolerance.2566

I still want to keep the probability at 95%.2572

That means, I have to change something else.2576

Since, I want a more accurate answer, that means I'm going to need to take more samples than I did before.2580

We are going to try and solve that together.2587

First thing is to figure out with this probability 95%, what is that mean?2591

Let me try to illustrate that graphically.2597

I want to find some cutoff that gives me 95% of the probability in the middle.2602

I will put a value of Z there, that is –Z.2611

I want to get 95% of the probability in between these cutoffs, for the standard normal variable.2614

This is supposed to be 95% here and I want to figure out what value of Z will give me that.2621

The way to find that, since I have a chart that will tell me how much probability is in the tail of the normal distribution,2631

what I can do is say that these tails, I got two tails here,2644

the area in 1 tail should be 1- 95%, 1 - 0.95/2, which is 0.05/2 which is 0.025.2649

I want to find a Z value, since that the probability of Z being bigger than that cutoff z is 0.025.2668

I will check this on the next slide, you will see, we will look it up together.2680

I do not want to break my flow for this slide, I will tell you right now that it comes out to be Z = 1.96.2684

I want to make sure that was the right value that I used.2694

Yes 1.96, that is the cutoff Z value that we are looking for.2696

Let me go back and show you how that factors in with all the other numbers in the problem.2702

I will just remind you that Z is our standard normal variable.2706

The way we get it is we do √N/σ × Y ̅ – μ.2711

Let us work that, I wanted out to have Y ̅ – μ.2720

I wanted those two quantities, Y ̅ is the sample mean, that is Y ̅ right there.2726

The true average of this machine is μ, I do not know what that is, by the way.2733

I want this to be within 0.4 of each other.2739

I want the difference between those two in absolute value to be less than 0.4.2741

When I want to do is build up this standard normal variable.2755

I’m going to multiply on √N/σ here.2760

I will multiply on √N/σ, as well here.2764

The point of that is this gives me my Z value, my standard normal variable.2771

My Z is less than or equal to √N × 0.4.2779

I know what σ is, my σ was given to me in example 3, that was 2.5.2786

That is 2.5 from example 3 was where that was given to us.2793

Let me fill that in, example 3 was where that information came from.2799

The N in example 3, N was 100 because we took 100 samples.2805

We cannot use that anymore because now we are trying to go for more accurate estimation.2810

We are going to have to increase the number of samples.2817

I know it is going to be increased because we have to get a more accurate answer.2820

What we are really going to do is solve this for N.2823

We are going to use this value of Z that we figured out over here, 1.96, and then we will solve for N.2831

1.96 is less than or equal to √N × 0.4/2.5.2841

Now, I’m just going to manipulate the algebra a little bit until I can get a value for N.2852

I do not think this one is going to work out particularly nicely, I did not rigged the numbers for this one very well.2860

2.5, I will multiply by both sides, I will divide both sides by 0.4, that should still be less than √N.2865

N should be greater than or equal to 1.96 × 2.5 divided by 0.4.2876

Since, I changed from √N to N, I'm squaring both sides.2888

We should square that expression.2893

That is not a number you will have work out in your head.2897

I did that on a calculator and I will show you what I got there.2901

I got 150.063, just slightly over 150 there.2905

N is the number of samples that we are going to take.2916

It is got to be a whole number, because you cannot take half of the sample.2919

In order to make this work, I need a whole number bigger than 150.6063.2923

N = 151, I will round that up.2931

You always round up, if you are talking about the number of samples.2936

The samples is enough, I solved that except for one detail of showing you on the chart where that Z value came from.2940

I will go back over the steps here and then we will jump forward to the chart.2952

I will show you where that Z value came from.2957

To go back to be getting here.2960

I start out with this probability of 95%.2964

I’m going for a 95% probability here.2967

In order to get 95% in the middle, that means my values on the tail, there is two tails, they are going to split the left over.2972

The left over probability is 1 - 0.95/2.2982

1-0.95 is 0.05, .05/2 is 0.025.2988

I’m looking for cutoff value of z, that when I find the probability bigger than that, it is 0.025.2994

It will come from the chart on the next slide.3003

L will fill the part in, if you are willing to be a little patient with me, or you can skip ahead and see the chart on the next slide.3007

Then, I would just going to hang onto that Z for a little while.3015

I'm going to go back and I'm going to build up my standard normal variable here, that comes from this formula here.3018

That is supposed to be a standard normal distribution.3023

I started off with Y - μ and their absolute value, that comes from the word within here.3029

I want two things to be within 0.4 ml.3035

I want the absolute value of less than 0.4.3039

And then, I built up my √N/σ.3042

I do not know what the √N is because I have not been told how many samples I want at this point,3046

that is what I'm solving for.3051

The N is the variable but my σ, I was given the standard deviation in example 3.3054

I dropped that in, it is 2.5.3059

I fill in my Z value, that comes from over here.3062

That is where that Z value came from.3066

I just take this equation and I solve it for N.3071

That is just a matter of manipulating the algebra around, squaring both sides because I had a √N.3075

I get N = 150.063, and I need it to be a whole number.3080

To be safe, I round it up.3089

Because 150 samples would not be quite enough, I have to go for 151 samples.3092

And then, I know with probability 95% that my sample mean will be close to the true mean.3098

One missing step here is where that 1.96 came from and it comes from this 0.025.3106

But, I'm using the chart that we will see on the next slide.3114

I want to fill in that one missing step from example 4.3118

We want to find a Z for which the probability of Z being bigger than that Z is 0.025.3122

We work that out from the previous slide, that was what we are looking for.3130

Let me draw a little picture of what we are dealing with.3139

We want to find a Z, says that tail probability there is 0.025 which means I have to look for 0.025 in my chart.3143

I’m going to start here, they are getting smaller 0.7, 0.6, 0.5, 0.4, 0.3, 0.02.3158

It is getting close, 0.027, 0.026, 0.0256, 0.0250, there it is right there.3172

I look at where row and column that happened in, 1.96 and 0.06, that tells me that my Z value is 1.96.3182

You might also have electronic applets, you do not have to do this kind of old fashioned method of looking up on charts.3197

That is totally fine with me, if you are okay using electronic tools in your class.3205

You can jump from 0.025 to 1.96, that is totally fine with me.3211

This 1.96, we went on and use that in our calculations on the previous side.3217

Somehow, that work out on the previous side to tell us that we need N = 151 samples,3228

in order to guarantee a particular accuracy of our sample mean.3237

Just to recap there, most of the work was done on the previous side.3245

I figure out on the previous side that I was looking for cutoff value of Z,3249

such that the probability of being bigger than that was 0.025.3253

We just look through this chart until I found 0.025, found that in the 1.9 row and the 0.06 column.3257

I put those together and I get 1.96.3265

The rest of it goes back to the previous side where we threw that 1.96 in a bunch of calculations3268

and came back to N = 151 samples.3275

In our final example here, we have a restaurant that is worried about how much money it is going to make tonight.3282

It has done some studies and it is found that its customers spend an average of $30.00 per customer.3294

But, they have a standard deviation of $10.00 which means,3301

maybe if somebody just has an appetizer and a drink, maybe they will spend $20.00.3303

Maybe, if they really go for the full menu and have drinks and desserts, and a few different extras,3308

then they are going to end up spending $40.00 or even more.3317

The restaurant, their average is $30.00 and they have 25 reservations tonight.3320

I guess the reservation only restaurant, you cannot just walk in here, you have to have a reservation.3326

They are expecting 25 customers tonight.3331

They want to know the chance that their total revenue tonight, we are not worried about profit,3334

we are not worried about what we are spending on supplies, total revenue will be between $725 and $800.3338

Let me show you how this turns into a central limit theorem problem.3348

Because it is not totally obvious right now, we are talking about total revenue.3353

Let me show you here, let me remind you of what we are given at the beginning of this lecture.3356

We are given that Y ̅ is sample mean – μ the global mean, divided by σ the standard deviation, and multiplied by √N.3361

N is the size of the sample.3378

The point of that was that would give you a standard normal variable.3381

We are going to call it Z and that is a standard normal variable.3386

In turn, the point of a standard normal variable is it is very easy to calculate probabilities.3392

The way I’m doing it is I'm using charts.3397

You might use charts for your class or you might have more sophisticated electronic tools, and that is okay with me.3400

What is this have to do with this restaurant?3407

They want to make between $725 and $800 total.3409

If the total is going to be between 725, I said they want to make between that.3416

Of course, they would be happy if they made more.3425

They are worried about, they want calculate how likely is it that they will make between 725 and 800 total.3428

What we have here is a result that has to do with the mean, the sample mean.3434

How do we convert that into a mean?3440

We just divide by the customers, the number of customers,3442

and convert that into an average amount that each customer would spend.3448

The mean, the average, Y ̅ would have to be between 725 divided by 25 and 800/253453

because that is how much the average customer would have to spend, in order to get the total between 725 and 800.3468

I just did a little arithmetic here, I rigged this up so that the numbers came out fairly nicely.3476

800/25 is 32 and 725/25 is 28, 725/25 is 29.3482

Y ̅ would have to be between 29 and 32.3495

What that means is, all the customers that come in tonight,3501

they would have to spend an average of between $29.00 and $32.00.3504

It does not mean they will have to spend between that,3510

but it means you can still have some big spenders to come in and drop 50 bucks on a meal.3512

You can still have some cheapskates to just buy an appetizer and then slid out of there after spending $10.00.3517

But on average, it has to come out between $29 and $32.00 per customer for tonight's customers.3523

What I would like to do is kind of buildup that standard normal variable.3534

Y ̅ - μ would be between, my μ is my global average.3540

That is right there, that is the 30, that is how much a customer spend on the average in the long term.3547

That is 29 -30 and 32 -30, let me go ahead and divide by σ.3554

My σ is the standard deviation, there it is $10.00 right there.3569

I also need to multiply by √N, I did not really leave myself enough space to do that.3577

I will give myself another line there.3583

Y ̅ - μ/σ × √N, this not absolute value, this is not one of those within problems like the previous one.3584

We have to be careful about what is positive, what is negative, no absolute value here.3595

√N, N is 25, that is the number of customers we are going to be working with tonight.3599

√N is 5, 5 × 29 -30/10 and 5 × 32 -30/10.3604

That simplifies fairly nicely, 5/10 is ½, ½ × -1 is - ½.3619

I will write that as -0.5.3628

And then, the point of this was we are building up a standard normal variable.3631

That is my Z right there, this is between -0.5 and 5/10 is still ½, 32 -30 is 2, 2 × ½ is just 1.0.3635

I have a standard normal variable and I want to find the probability that it is between -1/2 and +1.3651

Let me draw a graph of what I’m looking for.3659

Possibly, if you have the right electronic tool, you can jump to the answer at this point.3663

Just drop these numbers in your electronic tool, but let me show you how you can figure out using your charts.3668

There is -0.5 and there is 1.0.3675

I want to find the probability of being in between those two.3680

What my chart will do is, it will tell me the probability of being in a tail.3685

It will tell me that probability right there.3690

It will also tell me that probability right there, but those are not the same because 0.5 and 1.0 are not symmetric.3693

This is the probability that Z is greater than 1.0.3701

This is the probability that Z is less than -0.5, but it is also the same as Z being bigger than 0.5.3707

What I really want here is, my probability that Z is between -0.5, 0.5, and 1.0.3716

Let me write that a little more clearly, - 0.5 and 1.03730

What I can do is, I can subtract off the two tails to get the probability.3741

That is 1- the probability that Z is bigger than 0.5 - the probability that Z is bigger than 1.0.3746

We have some other problems that is like this where we subtracted off the two tails.3758

Those are symmetric ones, they started out with absolute values.3761

We can just find one tail and then multiply it by 2.3765

But these tails are not symmetric, I would have to do two separate calculations and3769

look up two separate numbers on a chart there.3775

Let me say that I will do the steps on the chart on the next page.3780

I will just tell you what the answers are for now, and then I will prove to you by showing you the chart on the next page.3785

The probability that Z is bigger than 0.5, I look that up on my chart, I got 0.3085.3793

The probability that Z is bigger than 1.0 was 0.1587.3804

Now, it is just 1 - 0.3085 0.1587.3812

I was lazy, I threw that into a calculator and what I got was a 0.5328,3817

could have done that by hand, that would have been that bad.3827

If you want to estimate that, that would be just 53%.3832

That is the probability that this restaurant, their total profit for tonight is between $725 and $800 tonight.3840

There is that one missing step from the chart.3853

We will confirm that on the next slide but before I turn the page from the slide, let me show you the steps here.3856

I want to find the probability that my total was between 725 and 800.3862

What I really know is a result about the average that each customer tonight is going to spend.3867

I want to convert that total into an average.3875

I just divided it by the number of customers.3878

The total divided by the number of customers gives me the average.3883

725/25 gives 29, 800/25 gives me 32.63887

And then, I start to build up this formula for a standard normal variable.3894

I subtracted μ from both sides, that μ was the average that all the customers in the world spend at this restaurant.3899

I subtracted 30 from both sides and then I divided by σ, where is my σ, there is my σ right there, the standard deviation.3908

I divided by σ and then I multiply by √N on the next line.3919

N was the number of customers, there is 25 of them, I’m going to multiply both sides by 5.3926

5 is the √25.3934

I simplified the numbers here, they have worked out pretty nicely.3936

They are rigged to work nicely, simplified down to -0.5 and +1.0.3940

We are really looking for the probability between -0.5 and 1.0 for a standard normal variable.3947

The way my chart works and some people's chart work a little differently,3959

but the way my chart works is it will tell you these tail probabilities.3959

It tells you the positive tails but you can work out the negative tails the same way.3964

What I will do is, I will look up the two different tails there and subtract them off from 1.3969

That will give me this probability in the middle, that I'm really looking for.3976

I will confirm those on the next page with the chart, it is 0.3085 and 0.1587.3982

Once I look those up, I can drop them back into the calculation and just reduce it down to 0.5328 which is about 53%.3989

That is my probability that my restaurant is going to make between $725 and $800 tonight.3999

The one missing piece of the puzzle from the previous slide,4009

we are still answering example 5 now, is to find those two probabilities.4012

We were finding the probabilities of being less than - 0.5 or being bigger than 1.0.4020

We actually want to find the probability of being between those cutoffs.4029

This chart will tell you the probability of being in the tails.4033

The probability of Z being less than -0.5 is the same as being bigger than 0.5.4039

It should be in this chart somewhere, here it is 0.5 and 0.00 is 0.3085.4048

The probability of Z being bigger than 1.0, here is 1.0, it is 0.1587.4060

I took those numbers and I do not think I'm going to rehash all the calculations that I did on the previous side.4078

I will just say, you plug those numbers in to the appropriate place on the previous side.4084

You can go back and watch it, if you do not remember how it works out.4089

It would be good if I can spell previous though.4094

We work through some calculations and we came up with a 53% probability4101

that this restaurant is going to make between $725 and $800 in their nightly revenue.4107

That wraps up example 5, most of the work was done on the previous side.4118

You can go back and check it out, just the missing step on the previous side was4125

where these two numbers came from, the 0.3085 and 0.1587.4129

Where they came from was by looking up 0.5 and 1.0, and then getting these two numbers from my standard normal chart.4136

If you do not like using charts, if you have an electronic way of getting probabilities4146

for a standard normal distribution, then by all means use that.4151

It is definitely something quicker than this slightly archaic method.4156

In some probability classes, they are still using charts so I want to show you that way.4160

That wraps up this lecture on the central limit theorem and4166

that is the last lecture in the probability series here on www.educator.com.4169

My name is Will Murray, I have really enjoyed making these lectures.4174

I hope you have learned something about probability.4178

I hope you have been working through the examples and learning something along with me.4180

I thank you very much for sticking with me through these probability lectures.4186

I hope you are enjoying your probability class and all your math classes, bye now.4190