- 00:01
Hello.Welcome to this lesson in Mastering Statistics.Hopefully, by now, you've startedto see the usefulness of somethinglike the central limit theorem.It really allows you to solve problemsthat seem to be very complicated at first but, in fact,become quite easy.There are not too many mathematical steps involved,but we just need to know what to do and get practice with it.So let's continue that practice now.

- 00:23
47% of people in a city want to buy a bike.If we survey 61 people at random,what is the probability that less than 40% of those peoplewant to buy a bike?All right, so this is the kind of common thingthat you'll see, and at first we're given informationabout the population.We're told that 47% of people in the city want to buy a bike.

- 00:45
How could you possibly know that?Well, usually broad data like that guarding a populationmight be something that's been publishedby a larger study in the past.Maybe they studied lots and lots of people.Maybe they took a survey during the last census or somethingabout people who wanted to buy a bike.But they know somehow that 47% of the people in the citywant to buy a bike.

- 01:06
All right, so now we want to know,if I go survey people at random-- in this caseI'm surveying 61 people-- what is the probabilitythat less than 40% of those people want to buy a bike.So we have information about the population proportion, 47%.We have information about the sample proportionthat we care about because it's 40% in regards to the sample

- 01:29
size, which is 61.So let's put this information on the boardand try to get some practice with it.So the first thing we want to noticeis that the sample size is given in the problem of 61because it says, we survey 61 people.Then the population proportion is also given to us as 47%,

- 01:50
but we never write it as percentagesin our calculations.So I want to get in the habit of putting 0.47,move the decimal point two times.We're also told that the sample proportion that we'reinterested in investigating is, whatis the probability that less than 40% of that sample.So the sample proportion is 40%.But, again, we don't write it that way.

- 02:11
We put 0.40.So we have the relevant information up on the board,and before we go any further, I wantto verify that this problem can producea sampling distribution of sample proportionsthat looks normal.So you want to just verify that.And the first constrain is that n times pis greater than or equal to 5.

- 02:31
All right, so what we have is n, whichis 61, times p, which is 0.47.And when you do that calculation youget 28.7, which is greater than or equal to 5.So we put a check mark there and say, OK, weunderstand that that's right.And then let's check the other part of it which is in times

- 02:51
1 minus p.It also has to be greater than or equal to 5, so n is 61.1 minus p is 1 minus 0.47.when You do 1 minus 4.7 multipliedby 61, what you're going to get is32.3, which is also greater than or equal to 5.So we can put a check mark there.So because both of those things are true, what it basically

- 03:12
does is it ensures that we're samplingenough people to make the central limit theorem apply.The central limit theorem means that the sampling distributionwe get is going to look normal.It's going to look normal.And so the next thing we want to dois figure out what z-scores do weneed in order to find the answer to our particular problem.So the equation for a z score obviously has changed.

- 03:35
It's the sample proportion minus the population proportionover the square root of p 1 minus p over n like this.And what we know is that the sample proportionthat we're interested in investigating is 0.4.The population proportion is 0.47.

- 03:56
And on the bottom, we're going to have a square root.Here we have p, which is 0.47 and then 1 minus 0.47.And then on the bottom, we have n, which is 61.So you see, it's numbers, numbers everywhere.It's a little bit of a pain, but it's not too terribly hard.So what you're going to get on the top, 0.4 minus 0.47.

- 04:18
What you're going to get is negative 0.07,and on the bottom, you're going to get 0.0639.And when you do this division, you're going to get negative1.10.That's your z-score.So basically, you're just calculating the z-scoredifferently, but the end result of what you're trying do

- 04:41
is exactly the same thing.You know that you've created a sampling distribution of sampleproportions that looks normal, you justneed to find the point on that normal curvethat you're interested in investigating.You're trying to find an area either to the leftor to the right of that point.Now in this case, the problem actually says,what's the probability that less than 40% want to buy a bike?So in this case, because we're finding less than 40%,

- 05:05
we're interested in finding the probabilitythat z is less than negative 1.1.We put a less than sign because the problem actuallysays less than 40%, and the 40% iscorresponding to this z-score.And when you put that into your table,look at 1.10 into the z-chart table-- negative 1.10, sorry,

- 05:25
you're going to get a probability of 0 point 1357.So this is the probability, the probability thatout of the sample size of 61, less than 40%want to buy that bike, is 0.1357.And that makes sense from our problembecause we know that the population of everybody

- 05:47
in the city, about 47% of all the people living there,say that they want to buy a bike.So the odds of us sampling some peopleand getting less than 40% have are probably fairly small, notcrazy small because the percentages aren'ttoo far apart but still a fairly small percentage.And that's why we get a probability that's

- 06:08
pretty small, close to 0, but not crazy tiny.So are our answer makes intuitive senseto what we're doing.All right, now let's work the last problem,which is a little bit more involved, nottoo terribly involved.The problem goes like this.72% of shoppers at a store are women.That information could be coming from justdemographic information that we have

- 06:30
from a long time of studying this particular store.40 shoppers are chosen at random.What is the probability that the proportionof women in the sample differs from the mean by more than 3%?We have that phrase again that we'vedone several times in the past, differs from the meanby more than something.So that means we have a mean, and then we're

- 06:50
looking 3% above the mean, 3% below the mean.What is the probability that we'reon either side of the mean by more than 3%.So the first thing you need to do is right now what you know,as you need to do with all problems.We know that the sample size is 40 because that'sa given of the problem.We know that the sample proportionis 72%, which is 0.7, because that's

- 07:13
the proportion of shoppers that are women.So we're given that.That's true.Now let's verify that we can use the central limit theorem.So n times p is 40 times 0.72.And what you get is 16.8, which is greater than or equal to 5.So that checks out.

- 07:34
And the next thing is n times 1 minusp is equal to 40-- that goes in n spot-- 1 minus 0.72.And when you do the subtraction and then the multiplication,you get 11.2, which is greater than or equal to 5.So both of the constraints are satisfied,which I already told you they would be in this class.So we have a large enough sample size

- 07:54
to guarantee that when we find the samplingdistribution of sample proportions,it's going to look normal.If we were to do all the people in that storeand survey them and batches of 40 peopleand figure out what the percentage of womenwe have and we collect all the informationand put it into a distribution, it's going to look normal.Now in this case, I think it's goingto be useful to draw a picture of that normal distribution

- 08:17
because this problem is a little bit more complicatedthan the previous problems, not difficult.It's just a little bit more involved.So here is our sampling distributionof sample proportions.We'll guarantee that it's normal because of what we set up here.This is the sampling distribution.Now the problem-- first of all, what you need to knowis that the parent population, the population of proportion,

- 08:40
is 0.72.So then we know that our sampling distribution will alsobe centered around 0.72 as much as it was for the meanwhen we're studying the mean over there.But when we're asked in the problem is--and this you can think of this as 72%--what we're asked here is, what is the probability thatthe proportion of married women in the sample differs from

- 09:01
the mean by more than 3%?That means, I need to go 3% above the mean, 3% this way.What would 3% above 72%-- it's proportion here.So if you're looking at percentages,you can just call that 75%.And then this number here is 0.75And then when you look on the low side,again, you're going it's symmetric

- 09:22
because it's 3% plus and minus, then what you have here is69%, which is 0.69.So these break points here is 3% below and 3% above,and what we're actually interested in is figuring outwhat the area of this shaded regionis plus the area of this shaded region here because it says,

- 09:43
what's the probability that in our samplethat the proportion of women will differ from the meanby more than 3%.That means on both sides of the mean wehave to take both into account.So just to start, let's focus on this side.Let's focus on this z-value here.The z-value is p hat minus p over the square root, p 1 minus

- 10:08
p over in.In this case, p hat-- because we know what our break pointsare because we're looking at the sample proportions.This is 69%, so we'll put 0.69 here.The population proportion is 0.72.And then on the bottom, we have the square rootof p, which is a 0.72, then 1 minus 0.72.

- 10:33
And on the bottom we have a sample size, which we alreadyknow is 40 from the problem.All right, so when you do this on the top,you're going to get negative 0.03.And when you do this on the bottom, do the subtraction.Multiply it by 0.72.Divide by 40, and then take the square root.You get 0.071.So what you will then have is the z-sore

- 10:54
for this part of the curve here, this negative 0.423.Now let's focus for just a second.We've figured out that this particular proportioncorresponds to a z-score of negative 0.423,so let's focus firstly on just finding the area here.We'll worry about that later.Let's work on this one now.

- 11:15
So that first probability, the probabilitythat z is less than negative 0.423--remember, we put a less-than sign herebecause we're wanting the area to the left.And since we have a less-than sign,we can just look at this value up in the z-chart right away.So we're going to look negative 0.423in the z-chart for the standard normal distribution,

- 11:35
we get 0.3372.What we figured out is that the probabilityof this left-hand side here is 0.3372.Now we could do the same calculation.We could find the z-score here, and then wecan figure out the area to the right.That's totally fine to do.But then you realize that this is a symmetric distribution.A normal distribution is symmetric about the center

- 11:58
here.So if we know what the area of this part is,and since it's 3% plus of minus, weknow that it's exactly equal to whatever the area of this is.So the final probability is just going to be 2 times 0.3372.And when you do that math, you get 0.6744, 0.6744.

- 12:23
So it's a little bit better than 50%, getting close to 75%as far as a probability goes there.So this problem was a little bit different because of the wayit was phrased, but the same sort of stepswere basically executed.First, you write down what you.You verify that the central limit theorem will apply.Once you verify that it applies, youknow the shape of the sampling distribution.

- 12:45
And so we write that information down.We look plus and minus 3% from the mean,and we calculate using the new revisedequation for the z-score-- what the z-score here is.We get an answer, negative 0.423, and thenwe look that up in the table, and we figure outthis probability of this area is this guy.We just multiply by 2.That's a nice thing about the normal distribution.

- 13:05
A lot of times, if you're looking symmetricallyon both sides, you can just calculate 1/2 and multiply by 2if it's symmetrically oriented, if your problem issymmetric like this.So that about concludes what we'regoing to use the central limit theorem for with regardto proportions.It's a common, common thing that youwould be using in either manufacturing, engineering,

- 13:27
even politics with surveying, you know, who you're voting forand things like that.So we've all been exposed to these thingsin terms of what we see on television and thingslike that.So we're kind of scratching the surface.We still have a lot more to cover of courseto truly understand what a lot of statisticsthat we have been bombarded with really means, of course.But this is just a good first start.And you can really see that the central limit theorem is--

- 13:50
it's pretty amazing actually that you can havea distribution that you don't know anything about,but if you sample it right, then the sampling distributionyou get actually looks normal, which incredibly helpsour calculations along the way.So we've wrapped up the central limit theorem here.We're going to use this to expand the topics that we'regoing to learn about in this courseand to a very important topic, which will start

- 14:12
in the next section, which will be called confidence intervals.Confidence intervals may be something you've heard about.Maybe you've kind of, at least, heard about itin general language, or maybe not.But in any case, you've all been exposedto confidence intervals.But all these surveys and things that we're alwaysbombarded with, so we're going to learn about those offin the next section.We're going to work lots of problemsand really understand confidence intervals.

- 14:34
And it dovetails in with what we have justlearned with the central limit theorem in statistics.

### Video Info

**Series Name:** Mastering Statistics, Vol 3

**Episode:** 8

**Publisher:** Math Tutor DVD LLC.

**Publication Year:** 2014

**Video Type:**Tutorial

**Methods:** Normal distribution

**Keywords:** mathematical computing; mathematics; proportion

### Segment Info

**Segment Num.:** 1

**Persons Discussed:**

**Events Discussed:**

**Keywords:**

## Abstract

Jason Gibson provides sample problems to show how to apply the central limit theorem to population proportions. He explains two examples using hypothetical data.