Hello.Welcome to this lesson in Mastering Statistics.Here, we're going to continue working with the central limittheorem as they apply to dealing with population means.And so, by this time, you should bepretty comfortable with identifying the type of problemin which you're typically going to be using the central limittheorem.You'll be given some information about some sort of population,usually, and then we'll sample x number of people.
We're trying to find the probability that that samplemean is less than or greater than a certain number.So that's the kind of problem templatethat you typically have when you're dealingwith the central limit theorem.All right.So let's go ahead and work this problem here.This problem's a little bit different, and it's not hard.I just want you to think carefullyabout what you're being asked in this problem here.
The average fetal heart rate is 140 beats per minute,with a standard deviation of 12 beats per minute.What is the probability that a fetal heartrate will differ from the mean by more than 25 beatsper minute?So, the first thing you do, you need to read the problemand understand what it's asking you.All right?So we're given-- we know what the average fetal heart
rate is.That's presumably some population of unborn babies,that we know that their fetal heartrate's140 beats per minute.We know the standard deviation of that number.So we know what the spread is of that dataover a large population of unborn babies.What is the probability that the fetal heartrate will differ from the mean by more than 25 beats
per minute?All right.So, the trick here is, usually whenyou're dealing with central limit problems,you need to identify the sample, the sample size.How many people are we studying at once for every samplethat we're doing here, so we can apply the ideaof the central limit theorem?But in this problem, it doesn't talk about the sample sizeat all.It-- in fact, it's really just giving us
information about the population and it'sasking us, what's the probability that a single heartrate differs by the mean by more than 25 beats per minute.The reason I put this problem in hereis because sometimes your teacherswill like to confuse you.And so a lot of students when they're studying central limittheorem will try to apply central limit to this,but then you can't.Because it doesn't actually talking about sample size.
It doesn't talk about sample means or anything like that.So you're scratching your head.This problem type is exactly the type of problemthat we've done in previous lessons, mastering statistics.Remember way back when we talked about the ideaof a normal distribution.Okay?So if these fetal heart rates are normally distributed,which is what we're going to assume in this problem,
then basically what you have is, what problem is givingyou is, since it's a normal distribution there,is that the fetal heart rates looklike a normal distribution?Which typically, things do look like a normal distributionin real life, a lot of time anyway.And what we're given the problem is that the average fetal heartrate, the mean of pretty much all the babies that we studied
there, is 140 beats per minute.And the standard deviation of this populationis 12 beats per minute.So this characterizes where this normal distributionis centered.It's centered at 140, and the spread of the distribution,how fat it is, is 12 beats per minute.We studied these concepts before.Now we're trying to figure out, what
is the probability that a single fetal heart rate,if you just pull it from the populationand study that one baby, unborn baby-- whatis the probability that that fetus has a heart rate thatdiffers from the mean by more than 25.When you see the key word, differs from mean,then what you need to really look at
is-- the mean is the center of this guy,so differing on the plus side of the mean will be 140 plus 25.So this number right here-- I kind of ran out of space,so I will write it down as 160-- actually what I'll do,instead of writing it in there, because I'llend up shading in there.I'll just draw a little arrow.This is 165.And of course, if you go 25 beats per minute
lower, what you get is 115.I'll just write that here.So really what we're after is whatwe're looking at the shaded region to the right,and we're looking at the shaded region to the left, right?So when you have the mean of 140,that's what the peak of this guy is.That's given in the problem.Then if you differ by 25 beats per minute higher
than the mean, it's the shaded regionif you differ by 25 beats per minute lower than the mean,it's in this shaded region.What we're really asking in this problem iswhat is the area under the probabilitydistribution of the total shaded region there?But notice there's nothing about a sample,sampling distribution, sample size.None of that stuff's listed here.This is a classic case of just using
a probability distribution, a normal probabilitydistribution.So if we wanted to figure out what the-- what this partof the curve is over here, then wewould need to go and calculate a z-score for this number here,and we would have to calculate a z-score for this numberright there.So let's first work on the first z-score, we'll call it z_1.
Remember, the equation for z-scoresis x minus the mean over the standard deviation.And in this case, x is 115.The mean is 140.And the standard deviation of this distribution is 12.And so whenever you subtract and dividelike that you get negative 2.08.So this 115 here, on this normal distribution, corresponds
to negative 2.08 on the standard normal distributionwe use for z-scores.Right?Now to calculate the second z-score, we look at this pointhere, which is 165 minus the mean, whichis 140 divided by the standard deviation, which is 12.We do the subtraction and that division, we get positive 2.08.
It makes sense that these are the same numberbut with a negative and a positive,because this is 25 above.This is 25 below.So the numbers that we chose were symmetric about the mean,so the z-scores should be symmetric about the mean.Because don't forget, for the standard normal distribution,when using a z-chart, the mean is defined to be zero.
So plus and minus you're going to get basicallyon both sides of the mean.So what we need to do is figure outthe area of this part of the curveand the area of this part of the curve and add them together.And so what we're going to end up gettingis the probability-- the answer that we're seeking
is going to be the probability that zis less than negative 2.08-- that'sgoing to cover the shaded area here--plus the probability that z is greater than positive 2.08,because this is going to give me the shaded region here.So now we start thinking about looking upz-scores in our chart.
But there's one little problem, first.And I'll show you what that is.Let's go ahead and say z less than negative 2.08.This is fine.I can look at negative 2.08 in the chart, because remember,when you look at a z-score, you'regetting the area to the left of the z-score.So this is fine.But here, your chart is not set up to give you areas
to the right for the z-score.So I need to change this, and I needto change the direction of this to the other direction.And when I do that, I need to put a negative 2.08.This is the standard way we deal with looking upthings in the z chart.We've studied that many times before.And so we end up having, basically, is2 times the probability of z less than negative 2.08,
because these are basically goingto give you the same thing.When you look this number up in the chart, and this numberin a chart, it's exactly the same thing,because the numbers are the same.That makes sense because the area hereshould exactly equal the area here,from the symmetry of the problem.So at the end of the day, the probabilityis going to be 2 times-- and if I look at negative 2.08
in my z chart table, I'm going to get 0.0188.And so the probability, when I multiply by 2, is 0.0376.0.0376.And it's a fairly low probability.It's not minuscule, but it's pretty low.And that kind of makes sense because if the mean is here
at 140, and the standard deviation of this curveis only 12, that means most of my datais going to be occupying between plus or minus 12 beatsper minute about that mean.But what I've asked in this problemis what's the probability of getting a heartrate greater than 25 from the mean and also greater than 25in the other direction less than the mean.
So I'm basically looking at the outlier data,so it makes sense that this probability that youget by looking farther away from the mean greaterthan the standard deviation is pretty low.But the thing I really want to show you hereis that we've been studying the central limittheorem for a long time, and a lot of times on a test or exam,teachers will throw problem like this that doesn't evenuse the central limit theorem.
You need to be aware of all of the stuffthat you've learned in the past and learn howto identify what to use when.In this particular problem, there's no mention of a sample.There's no mention of sample size.It simply gives us information about a population.And it says what's the probabilitythat the heart rate differs from the meanby more than 25 beats per minute?It's taking a single guy from the population--
what's the probability that he's going to do something.That's just using the standard normal distributionthat we've done in the past.So you don't even need to use the central limit theorem.In order to use the central limit theorem,there's got to be some kind of sampling going on.Some type of sampling, which is what willhappen in this problem here.And you could compare how this isworded to the problem we just worked here.
The average person plays video games 7.5 hours per week,with a standard deviation of 3 hours.If I choose 110 people at random,what is the probability that their meantime of game playis more than eight hours per week?This is much more in tune with what
we know a central limit theorem type of problem is going to be,because we're given information about a population.And this information, the 7.5 hours per week gameplay with a standard deviation of 3 hours about the mean--that information can come from some previous study,some previous very large survey, itcould come from something else.But in any case, it's given to you as fact in this problem.That's information about the population.
Then, it says, if I choose 110 people at random.That's your clue right there.You sample 110 people at random.And I'm asking a question about what those 110 people basicallydid as an average.So, in fact, it actually says, what'sthe probability that their mean of the sample whichmeans the sample mean is more than eight hours a week.
So you know that you've taken a sample,and you're asking a question about what that sample mean isdoing.That's a central limit problem, because we're doing sampling.So the bottom line here-- in this case,since we know I've chosen 110 people, I know that the samplesize is greater than 30, so I know that the samplingdistribution of sample means, that'sbasically implied in this problem,is going to yield a normal distribution.
Because notice that in the original problem statement,I don't actually tell you that the regional populationof game players.I give you a mean and a standard deviation,but I never actually told you that that original populationwas normal.You don't know if that original population is normal or not,because you don't know how that study was conducted.You're not given any information.It could be skewed.
It could be dependent on age groups.Of course it's going to be dependent on age groups.I mean, younger kids probably play more gamesthan older people, for instance.You don't know anything about this,but you know that the average of the population is 7.5,and the standard deviation's three.But since I'm sampling, with a sample size greater than 30,that I know that when I get my sampling distribution of sample
means, which we've talked about over and over again,that's going to look normal.That's the outcome of the central limit theorem.And I also know, because of the central limit theorem,right, that the mean of the sample means,right, is going to be equal to the mean of the population.In this case, 7.5.
So here, I've taken a bunch of samples of 110 people, 110people, 110 people, 110 people.I've calculated their sample means.I take all of those sample means,and I average them together, which is what this quantity is,and the central limit theorem tells methat that number is the same as the population mean, right?So I can write this down as fact.And then I might want to write down
that the standard deviation of that sampling distributionis given to us by the standard deviation of populationsdivided by the square root of the sample size.We know the square root of the population, in this case,it's three hours.We know the sample size is 110, so wedo the square root of 110.So we do 3 divided by square root of 110, I get 0.286.
So this is the standard deviationof the sampling distribution.We keep saying, hey, if you samplethe sampling distribution of sample means,right, with a sample size greater than 30,then you always get a normal distribution.This information tells me that that normal distributionthat you arrived at from the sample means, has a mean of 7.5with a standard deviation of this.
So if you were to draw a picture of that,I'll just draw it over here, right, then whatyou will have is, here's a normal distribution,and we know that this normal distribution is centeredabout this guy, and the standard deviationof this normal distribution of sample means is.
Never, never, never forget now whatwe're talking about with these variableshere that is the sampling distribution of samplemeans, that is, the distribution we get after we've donethe sampling, after we calculate all the sample means,and we arrange it in a distribution.Makes total sense if I'm sampling 110 people at a time,and I sample everybody appropriatelyat that large sample size, that I'm probably--
that I'm going to get a mean of the sample meanthat's going to match my population mean.I've got a pretty narrow standard deviationof this curve, but I've sampled so many people at once,the odds are that I'm getting a pretty good estimateof the population-- of the population mean like that.That's why this number is so low, because the samplesize is so high.See, as this number goes up, as I sample more people,
the standard deviation of sample meansgoes down, because of course if you sample more people,you expect to get a better, more refined normal distributionthat's more centered about the mean.Okay?So that's kind of a little bit of an aside.Basically, you want to calculate these two numbers.And then you want to use the z-tableto arrive at the answer.So in this case, we know that z-score for sampling
distributions is this guy.We've used it several times.So we know this number, 7.5, we know this number, 0.286.What goes over here?Well, we're asked in the problem is,it says, if I choose 110 people at random,what is the probability that their mean time of game playis more than eight hours per week?
So what I've got here, is a situationwhere I've got 7.5 hours.That's the mean.What I'm being asked here is, over here to the right,somewhere over here, is going to be eight.And I'm asking, what is the shaded region here?Because I'm asking, what is the probabilitythat the sample mean, if I choose 110 people,is greater than eight?So here I'm looking for anything greater than eight,
so I've got that region shaded.So what goes here is the sample mean I'm caring about.In this case, it's 8, minus this guy, 7.5,divided by this guy 0.286.So when I do all this stuff, the z-score I get is 1.75.Positive 1.75.So to get the answer, I'm looking for the probability
that z is greater than 1.75.The reason I put a greater than signhere is because I'm looking for the shaded region to the right,so greater than that guy.So I'm looking for the shaded region greaterthan that z-score.Now, you cannot look in a table of z-scores and just plugin 1.75 and get the right answer,because the table for the standard normal distribution
of z-scores, it always gives you the shaded region to the left.So because I've got the arrow going the wrong way,I'm going to change it around.I'll put this guy up.I'll flip the arrow around the other way, but to do that,I've got to put a negative sign in front of our z-score.We've done this many, many times before.So now, I can look up this guy, the probability,and I'll look at negative 1.75 in the chart.
Right?When I look at negative 1.75 in the chart, I get 0.0401.So, the answer to this guy is, if I have the characteristicsof this population, and I choose 110 people at random,and I ask them each, how many hours a week do you play games,they're all going to give me 110 different answers.I average the result.
And this is telling me, what's the probabilitythat that average of that sample that I took is greater than 8.So it's over here.What is the probability of that?Turns out the probability is pretty low.It's non-zero.Of course, it's possible that I couldhave 110 people that would end up giving me an average.Maybe I just happen to survey peoplethat play a lot more games.
It's possible.But it's more likely that I'm going to get somethingcloser to 7.5, because that's the mean,and the standard deviation of the sampling distributionis pretty small.But in this case, I'm asking for somethingthat's way far to the right, and so I'm gettinga pretty low probability.So I wanted to put these two concepts close together,these two problems close together,
so that this is what you're usuallylooking for when you're applying the conceptsof the central limit theorem.We've got gone over that many, many times.This previous problem might trick you into thinkingit's a central limit theorem, but it really isn't.A lot of times, they'll throw this problemin on a test when they're testing youon central limit theorem, just to trip up, to see what you do.You need to know what you're looking for.The bottom line is, if you're doing a problem that involves
sampling and a sample distribution,you need to key on those words.There's nothing at all involving a sampling distributionor a sample size or something, thenyou probably don't have anything going on with the central limittheorem.And you may need to look at a sort of simpler type of problemtechnique to solve it.So this is a good way to complete and concludeour introduction to central limit theorem.
I know that it's a little difficultto understand at first, but I'm hopingthat with all of these problems, youget a feel for what's basically going on here.And these different problems, I wantto encourage you to solve them yourself,and grab additional problems from your book.Now in the next section, we're goingto continue working with the central limit theorem,but in all the problems so far, we'vebeen applying it to the sample mean, right,
because it's a very important thing that westudy a lot in statistics.But in the next section, we'll be talking about somethingcalled proportions.Sample proportions.And I think that when you see that,you'll understand that's very, very useful as well/we're still going to be applying the central limit theoremin a slightly different way for a slightly different typeof problem.So follow me on there, and we'll tackle that next.
Jason Gibson walks through two sample problems about the central limit theorem. He also explains when the use of the central limit theorem is appropriate to solve the problem.
Looks like you do not have access to this content.
Jason Gibson walks through two sample problems about the central limit theorem. He also explains when the use of the central limit theorem is appropriate to solve the problem.