WEBVTT mathematics/probability/murray
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Hi and welcome back to the probability lectures here on www.educator.com, my name is Will Murray.
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Today, we are going to talk about Chebyshev’s inequality.
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This is the second lecture on using inequalities to estimate probability.
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The first one we had was Markov’s inequality.
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If you are looking for Markov’s inequality, there is another video covering that one.
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It is the one video before this one.
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This one is on Chebyshev’s inequality.
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You get some similar kinds of answers which Chebyshev’s inequality.
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The difference is that we use a little more information because now,
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we are going to use the standard deviation of random variable, as well as the expected value or mean.
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In return for using a little more information and doing a little more calculation,
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we get stronger results using Chebyshev’s inequality.
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Let us dive into that.
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Chebyshev’s inequality is a quick way of estimating probabilities.
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It never tells you the exact probability, by the way.
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That gives you an upper and lower bound for the probability but it never tells you it is exactly equal to something.
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It is based on the mean and standard deviation of our random variable.
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We are going to use the Greek letter μ for the mean and we are going to use the greek letter σ, for the standard deviation.
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Let me go ahead and tell you what Chebyshev’s inequality says.
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Suppose y is a random variable and K is a constant.
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K is often a whole number like 2, 3, or 4.
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Chebyshev’s inequality says the probability that the absolute value of 1 - μ is greater than K σ,
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is less than or equal to 1/K².
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That is quite a mouthful, starting with the name Chebyshev’s himself is a little tough to deal with.
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But inequality is a little bit complicated.
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I want to think about the intuition of that first.
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What we are really saying there, first of all, σ is a standard deviation.
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K is a measure of how many standard deviations you are willing to go away you from the mean.
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That is what they are really measuring right here.
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The absolute value of 1 – μ, μ is the mean.
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The absolute value of 1 - μ is the distance, that is the value of A - b is the distance between A and B.
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This is the distance from Y to μ.
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What this is really saying is, how far are you willing to go from your expected value, from your mean, μ?
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The question is, what is the probability of you deviating more than K standard deviations from your mean?
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Maybe, I can try to graph that.
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If we have a certain amount of data here in and it is grouped like that, that is kind of a common distribution of data.
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You have a mean right there in the middle.
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What is the chance the probability of being more than K standard deviations from your mean?
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In other words, how much area could there be out there in the tail of the distribution?
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What Chebyshev’s inequality says, is that there is not that much.
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It is unlikely that the variable will be far from its mean, 1/K² is your balance.
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That is usually a fairly small number, for example if K is 3 then 1/K² would be 1/9.
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The probability would be less than 1/9 that you would be 3 standard deviations from your mean.
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That is what Chebyshev’s inequality is saying.
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It is that, your probability of being many standard deviations from your mean is quite low.
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The more standard deviations you go, the smaller the probability gets.
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There is also a reverse version of Chebyshev’s inequality.
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You just turn around the inequalities there.
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This is the same inequality that I just showed you.
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If you turn that around, if you change this greater than to a less than or equal to, that is the change here.
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We are asking the opposite question, what is the chance that you were close to your mean?
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What is the chance that you will win K standard deviations from your mean?
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Since, the probability of being far away from your mean is very well,
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that means the probability of being close to your mean is very high.
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The answer we get, the probability of being close to your mean is greater than or equal to 1 -1/K², its exact opposite.
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It was 1/K² before, now it is 1 - 1/K² and where we had less than or equal to,
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before, meaning the probability of being very far away from mean is quite low,
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the probability of being very close the mean is very high.
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What we are saying here is that, it is unlikely that the variable will be close to the mean.
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We say close, we mean within a few standard deviations.
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Remember, σ is the standard deviation and K is the number of standard deviations away from your mean that you are willing to accept.
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If your chance of being with many standard deviations of your mean is quite high,
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that is what Chebyshev’s inequality is saying, if you turn it around.
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Let us check that out in the context of some examples and see how it plays out.
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This first example is quite similar to an example we had back when we are studying Markov's inequality.
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You will recognize the numbers.
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The difference is that we are now going to incorporate the standard deviation, before we just incorporated the mean or the expected value.
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Now, we incorporate the standard deviation and we are going to use Chebyshev’s inequality,
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instead of Markov’s inequality, and we get much stronger results this time.
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In this case, we done a survey of students on a campus and we discovered that on average,
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they are carrying about $20.00 in cash with them.
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We also know that their standard deviation is $10.00.
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If we need a student at random, we want to estimate the chance that that student is carrying more than $100.
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We also want to estimate the chance that she is carrying less than $80.00.
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Let me start out by writing down Chebyshev’s inequality, just reminding you what the formula was.
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The probability that Y - μ is greater than K σ, it is called the Chebyshev’s inequality, that probability is less than 1/K².
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Let us fill in what we can use here.
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We know that the μ, that is the average value of the variable or the mean or the expected value, which we are given here is 20.
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And we are also given the standard deviation is 10, that is the σ there.
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We want to find the chance that a student is carrying more than $100.
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Let us figure out how many standard deviations away from the mean, we would have to be.
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The mean is 20, if I want to have more than $100 then that $80 more than the mean which is 8 standard deviations away from mean.
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We would have to be 8 standard deviations away from the mean.
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The probability that Y -20 is greater than 8 × 10.
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As you have to be $80.00 away from the mean.
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According to this, that 8 is the K there.
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It is also equal to 1/K² so 1/8² which is 1/64.
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What that is telling us is that, if we meet a student on campus, we say what is the likelihood that that student has more than $100?
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It is very unlikely, the probability is less than 1/64.
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If you interview all your students, most likely every 64 students you interview,
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at most 1 of them will actually be carrying more than $100, according to these numbers here.
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There is a second part to this problem, we also want to estimate the chance that she is carrying less than $80.00.
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That is going to use the reverse incarnation of Chebyshev’s inequality.
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Let me go ahead and write that down.
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The probability of Y - μ is less than or equal to K σ.
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According to Chebyshev’s inequality is greater than or equal to 1 -1/K², that is the reverse version of Chebyshev’s inequality.
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In this case, our μ is still 20, our σ is still 10.
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In this case, we want 80 dollars.
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We want 80 -20 is 60, we want to be $60.00 away from the mean.
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60 is 6 standard deviation, 6 × 10.
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We will use K = 6 there in our own Chebyshev’s inequality.
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We get the probability that Y -20 is less than or equal to 6 × 10, is greater than or equal to,
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that is what Chebyshev’s inequality tells us here, greater than or equal to 1 -1/ the K was 6, 6² there.
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That is 1 - 1/36, and that simplifies down to the probability is greater than 35/36, because that is 1 -1/36
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If we meet a student and we want to say, what is the chance that she is carrying less than $80.00 in cash?
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The chance is at least 35/36.
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Remember, just like the Markov’s inequality, you never want to just give an answer,
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when you are being asked the question using Chebyshev’s inequality
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because it never gives you a specific numerical answer, it always gives you inequality.
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It will never tell you what the probability is, it will just tell you that the probability is less than this or greater than that.
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In all of these cases, it is very important to include the inequality signs in our answers.
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All we are doing is we are giving upper and lower bounds for the probability.
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We are not saying that we know what the probability is.
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Let me show you where I got each of those steps there.
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We start out with the basic version of Chebyshev’s inequality.
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The probability that Y – μ is greater than K σ, is less than 1/K².
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In this case, our standard deviation is 10, that is the σ, the mean is 20.
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We want to have more than $100, that means we really want to be 8 standard deviations away from the mean.
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A hundred is $80.00 away from 20.
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That is 8 standard deviations away from the mean.
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That is where we got the value of K = 8 there.
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Chebyshev’s tells us that the probability is less than 1/K², that is where we got 1/64 for our probability.
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Actually, our probably is less than or equal to 1/64 there.
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For the second part of the problem, we want to estimate the chance that she is carrying less than $80.00.
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I use the less than version of Chebyshev’s inequality and it tells me
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that the probability is greater than or equal to 1/1 – K².
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You are very likely to be within many standard deviations of your mean.
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How many standard deviations are we talking about?
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The mean is 20 and we want to have less than $80.00.
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80 -20 is 60, that is 6 standard deviations.
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I put that value in for K and put that K in, we get 1 -1/6².
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That simplifies down to 35/36, that tells me that the probability that a student will have less than $80.00 is at least 35/36.
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It does not say equal to 35/36 but at least 35/36.
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In example 2 here, we got students on a college campus have completed an average of 50 units and their standard deviation is 15 units.
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We meet a randomly selected student, we want to estimate the chance that this person has completed more than 95 units.
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Let me write down Chebyshev’s inequality for that.
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The probability of Y - μ is greater than K σ is less than or equal to 1 /K², that is Chebyshev’s basic inequality.
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We are going use the basic version because we are trying to estimate the chance that something is more than 95.
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Let us try to figure out what the relevant numbers here are.
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The average of 50 units, that is the μ, that is going to be 50.
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The σ is the standard deviation, that is 15 units there.
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We have to figure out what K is here.
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We have to figure out how many standard deviations away from the mean are we expected to be here.
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In this case, we want to have more than 95 units.
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95 -50 is 45 which is 3 standard deviations, three × 15.
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That means that our K value is going to be three here.
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The probability that Y -50 is a greater than three × 15, is according to Chebyshev’s inequality less than 1/3² which is 1/9.
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What we conclude here is that the probability is less than 1/9.
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That means fewer than 1 in 9 students will have more than 95 units.
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That is what we can conclude from that.
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We cannot say it is equal to 1/9, we cannot just give 1/9 as an answer.
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We have to say the probability is less than 1/9.
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To recap how we derived that, we started with the basic version of Chebyshev’s inequality.
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The probability that Y - μ is greater than K σ is less than 1/K².
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And then I filled in what I knew here, the 50 is the average number of units, that is the μ there.
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That is where that 50 comes from, 15 also came from the problem because that is the standard deviation.
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I want to figure out what K should be.
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In order to calculate that, I figured out how many standard deviations away from the mean are we interested in?
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95 -50 is 45 and that is three standard deviations because the standard deviation is 15.
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Our K there is three.
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We pop that right there into Chebyshev’s inequality, we get 1/9.
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1/9 is not our answer by itself, we have to say the probability is less than or equal to 1/9.
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That is how we answer that.
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In example three here, we got the scores on a national exam are symmetrically distributed.
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I see you got a small typo there in the distributed, let me fix that.
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Symmetrically distributed around a mean of 76 with the variance of 64.
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The minimum passing score is 60.
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We want to use Chebyshev’s inequality to estimate the proportion of students that will pass this exam.
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Let me draw a little graph of what is going on here.
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We are given that the scores are symmetrically distributed.
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It will look something like this.
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The mean is 76, let me fill that in.
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The mean here is 76, that is our μ =76, right there in the middle of the data.
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The minimum passing score is 68.
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We want to try to find out how many students are going to be scoring above 60.
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Let me put cutoff down there at 60.
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There is 60, that is the minimum passing score.
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We want to see how many students will be above that cutoff.
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In order to do that, we are going to use Chebyshev’s inequality.
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Let me go ahead and set up Chebyshev’s inequality.
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It says the probability that Y - μ is greater than K σ.
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That probability is less than or equal to 1 /K².
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Let us figure out what we know here.
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We know that μ is 76, we are given that.
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Σ is the standard deviation.
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We were not given the standard deviation.
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The variance is 64, the variance of Y is 64 that is not the same as the standard deviation.
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You have to be careful here.
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The standard deviation is the square root of the variance.
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That standard deviation would just be √ 64, that would be 8.
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If I plug that in, σ is equal to 8.
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We want to figure out what K is going to be, in order to use Chebyshev’s inequality.
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That means I want to figure out how many standard deviations away from the mean do I need to be.
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In this case, we are interested in a cutoff of 60, that is the value we are interested in.
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60-76, absolute of value of that is 16 which is 2 × the standard deviation of 8.
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That tells me that the K value that I’m interested in is 2.
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The probability that Y - μ is greater than 2 × 8, according to Chebyshev’s inequality is less than or equal to 1 or 2², that is just ¼.
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Here is a twist that we have not yet seen before with Chebyshev’s inequality.
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You got to follow me closely on this.
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That is telling me the probability that Y - μ will be bigger than 16 in either direction.
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What that is really doing is, that is giving you a bound on the probability of being less than 60 or bigger then,
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let us see 76 + 16 will be 92.
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That is really going 2 standard deviations down below the mean or 2 standard deviations up above the mean.
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Remember, our σ was 8.
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What we are really found is the probability of being in either one of those together is less than ¼, is less than or equal to ¼.
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We are also given that it was symmetrically distributed.
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That means the probability of each one of those is less than ½ of 1/4 which is 1/8.
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In each one of those, the probability of being in that region is less than ½ of ¼ which is 1/8.
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The probability that Y is less than 60 is less than 1/8.
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What we are interested in, is the probability that Y is bigger than 60 because those are the students that are passing the exam.
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In that case, you turn it around and you do 1 -1/8.
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Notice, I'm not worrying about all those scores bigger than 92.
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I’m not worrying about the 1/8 of the students that score bigger than 92, because all the students passed anyway.
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I’m just interested in these 4 students below 60, those students fail the exams.
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That is less than 1/8 of the population, that means more then 7/8 population will pass the exam,
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will get higher than a 60 score on the exam.
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Let me write this in words.
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At least 7/8 of the students will pass the exam.
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That is what Chebyshev’s inequality let us conclude.
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It does not tell us that exactly 7/8 will pass.
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It tells us that at least 7/8 students will pass.
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It could be higher but Chebyshev’s inequality would not give us a specific value.
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To recap what happened here, we start out with Chebyshev’s inequality.
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The probability of Y - μ being greater than K σ is less than 1 /K².
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In this case, our μ was 76, our σ was 8.
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We got that from the variance of 64.
00:23:20.100 --> 00:23:23.100
But remember, standard deviation is the square root of variance.
00:23:23.100 --> 00:23:27.700
The standard aviation is √ 64 which is just 8.
00:23:27.700 --> 00:23:34.100
I’m wondering, how many standard deviations away from the mean are we interested in going here?
00:23:34.100 --> 00:23:39.600
We are interested in a cutoff of 60, because that is the passing score for the exam.
00:23:39.600 --> 00:23:41.900
How far is that from 76?
00:23:41.900 --> 00:23:46.800
That is 16 units away, that is 8 × the standard deviation of 8.
00:23:46.800 --> 00:23:51.100
That means we use K = 2 in Chebyshev’s inequality.
00:23:51.100 --> 00:23:59.900
The probability that you are more than 2 standard deviations away from the mean is less than 1/2² or ¼.
00:23:59.900 --> 00:24:07.900
Here is the subtlety, that ¼ of the population could be 2 standard deviations below the mean
00:24:07.900 --> 00:24:11.300
or that could be 2 standard deviations above the mean.
00:24:11.300 --> 00:24:19.600
Since, we are given that the scores are symmetric, we know that half of them are below the mean and half of them are above the mean.
00:24:19.600 --> 00:24:26.500
We divide that 1/4 into two parts and we find that the probability of being less than 60,
00:24:26.500 --> 00:24:33.400
being 2 standard deviations on the low side is less than 1/8.
00:24:33.400 --> 00:24:38.500
That really came from doing 1/4/2,that is how we got that 1/8.
00:24:38.500 --> 00:24:41.700
The probability of being bigger than 60, in other words,
00:24:41.700 --> 00:24:50.400
the probability of scoring higher than 60 passing the exam is at least 1 -1/8 which is 7/8.
00:24:50.400 --> 00:24:55.000
In the end, we know that at least 7/8 of the students will pass.
00:24:55.000 --> 00:25:01.200
It would not be accurate, if you are given this problem and you just gave your answer and you gave 7/8.
00:25:01.200 --> 00:25:03.100
My probability students sometimes do that.
00:25:03.100 --> 00:25:05.600
I will give a complicated problem and they will just say 7/8.
00:25:05.600 --> 00:25:09.500
That does not really tell us exactly what is going on.
00:25:09.500 --> 00:25:17.800
You really have to say, it is at least 7/8, the proportion of students that will pass is greater than or equal to 7/8.
00:25:17.800 --> 00:25:22.800
That is what Chebyshev’s inequality tells you.
00:25:22.800 --> 00:25:28.800
In example 4, and this is one that is very similar to one we had back in the lecture on Markov’s inequality
00:25:28.800 --> 00:25:31.100
but there is a little more information than it now.
00:25:31.100 --> 00:25:32.700
We are going to get a different answer.
00:25:32.700 --> 00:25:36.200
Be very careful that you do not get this two problems mixed up.
00:25:36.200 --> 00:25:42.800
In example 4, we have seismic data telling us that California has a major earthquake on average, once every 10 years.
00:25:42.800 --> 00:25:49.200
Up to there, it is the exact same as a problem we had back in the video on Markov’s inequality.
00:25:49.200 --> 00:25:51.100
Now, here is the new stuff.
00:25:51.100 --> 00:25:55.500
We have a standard deviation of 10 years, what can we say about the probability that
00:25:55.500 --> 00:25:58.200
there will be an earthquake in the next 30 years?
00:25:58.200 --> 00:26:06.500
This is all exactly the same as the problem we have for Markov’s inequality, except for this one key phrase here,
00:26:06.500 --> 00:26:11.300
with a standard deviation of 10 years, that is the new information.
00:26:11.300 --> 00:26:18.200
That is going to let us use Chebyshev’s inequality, instead of Markov’s inequality.
00:26:18.200 --> 00:26:23.100
Because Chebyshev’s inequality depends on the standard deviation and the mean,
00:26:23.100 --> 00:26:26.500
Markov’s inequality do not use the standard deviation.
00:26:26.500 --> 00:26:29.000
Let me set up Chebyshev’s inequality.
00:26:29.000 --> 00:26:41.700
The probability that Y - μ is greater than or equal to K σ.
00:26:41.700 --> 00:26:50.200
I think when we originally get Chebyshev’s equality is greater than in that place.
00:26:50.200 --> 00:26:57.500
This says, the probability of being that far away from your mean is less than ¼.
00:26:57.500 --> 00:27:01.000
We have to be careful what Y is here.
00:27:01.000 --> 00:27:07.400
Here, what I mean by Y is, Y is the waiting time until the next major earthquake.
00:27:07.400 --> 00:27:23.300
Starting today, how long do we expect to wait in years before there is a next major earthquake in California?
00:27:23.300 --> 00:27:28.800
What we are given here is that, the average waiting time is 10 years.
00:27:28.800 --> 00:27:31.600
That means that my μ is 10 years.
00:27:31.600 --> 00:27:37.300
That also says that the standard deviation is 10 years too, σ is also 10.
00:27:37.300 --> 00:27:40.100
We want to figure out what K is.
00:27:40.100 --> 00:27:49.800
I wrote down 1/4 there, that was kind of looking ahead to starting to solve the problem.
00:27:49.800 --> 00:27:54.400
Chebyshev’s inequality just says that it is less than or equal to 1 / K².
00:27:54.400 --> 00:28:03.800
We need to figure out what K is and I spoiled the ending here but we will go ahead and figure it.
00:28:03.800 --> 00:28:13.500
The question here is how many standard deviations away from the mean are we expected to be here?
00:28:13.500 --> 00:28:21.700
We want to talk about the probability that there will be an earthquake in the next 30 years.
00:28:21.700 --> 00:28:26.600
That means we want to find the probability that Y will be less than 30.
00:28:26.600 --> 00:28:36.900
30 -10 is 20 which is 2 × our σ of 10 here.
00:28:36.900 --> 00:28:43.100
That means that our K is 2, as I inadvertently let slip earlier, K is 2 here.
00:28:43.100 --> 00:28:45.400
I will plug that value of 2 in.
00:28:45.400 --> 00:28:57.400
The probability that Y - 10 is greater than 2 × 10, according to Chebyshev’s inequality
00:28:57.400 --> 00:29:02.400
is less than or equal to 1/2² which of course is ¼.
00:29:02.400 --> 00:29:05.900
That is why I had written down before.
00:29:05.900 --> 00:29:11.200
That is the probability that Y -10 is greater than 20.
00:29:11.200 --> 00:29:22.000
What we really found there is, the probability that Y is greater than 30 is less than or equal to ¼.
00:29:22.000 --> 00:29:27.900
That is not exactly what the problem is asking for, because this is saying, if Y is greater than 30 that means
00:29:27.900 --> 00:29:35.000
we are waiting longer than 30 years for an earthquake, which means we do not have an earthquake in the next 30 years.
00:29:35.000 --> 00:29:41.900
The problem is asking, what is the probability that there will be an earthquake in the next 30 years?
00:29:41.900 --> 00:29:50.900
That means that are waiting time will be less than 30, which is the opposite of it being greater than 30.
00:29:50.900 --> 00:29:59.400
We are going to switch this around, it is greater than 1 -1/4 which is ¾.
00:29:59.400 --> 00:30:35.700
Our conclusion here is, there is a greater than or equal to 3/4 chance that there will be an earthquake in the next 30 years.
00:30:35.700 --> 00:30:40.300
Notice that, I'm being very careful to include the inequality in there.
00:30:40.300 --> 00:30:43.900
I'm not saying that there is exactly ¾ chance, I do not know that.
00:30:43.900 --> 00:30:46.900
I cannot figure that out from the information that we are given.
00:30:46.900 --> 00:30:55.000
What Chebyshev’s inequality does is, it gives me a bound, it allows me to say that there is at least a ¾ chance,
00:30:55.000 --> 00:31:01.400
at least a 75% chance that there will be an earthquake in the next 30 years.
00:31:01.400 --> 00:31:06.200
To recap that, I started out with the generic form of Chebyshev’s inequality.
00:31:06.200 --> 00:31:14.500
The probability of being K standard deviations away from your mean is less than 1/K².
00:31:14.500 --> 00:31:22.500
I filled in the mean is 10, that is this 10 right here is where we get the μ from.
00:31:22.500 --> 00:31:25.800
The standard deviation is 10 as well.
00:31:25.800 --> 00:31:30.400
That is where we get the σ from there.
00:31:30.400 --> 00:31:34.000
I just had to figure out what the value of K was.
00:31:34.000 --> 00:31:39.800
To do that, I remember I was interested in the probability of Y being 30.
00:31:39.800 --> 00:31:43.900
That is why I got that 30 from the problem here.
00:31:43.900 --> 00:31:50.700
Filled in 30 for Y, 10 from μ, and that came out to be 20.
00:31:50.700 --> 00:31:56.200
20 is 2 × 10, that 10 is the σ, that 10 was the μ.
00:31:56.200 --> 00:31:59.900
It is a little confusing because there is a lot of 10 going around here.
00:31:59.900 --> 00:32:12.100
That 20 is 2 × σ, that is where we get our K being 2 which tells us that the probability is less than or equal to 1/2² which is ¼.
00:32:12.100 --> 00:32:17.600
Our probability of Y being greater than 30 is less than ¼.
00:32:17.600 --> 00:32:19.800
That is not what the problem asks, the problem asks what is the probability that
00:32:19.800 --> 00:32:26.400
we will have an earthquake in the next 30 years, which means our waiting time is less than 30 years.
00:32:26.400 --> 00:32:29.000
We will see an earthquake within 30 years.
00:32:29.000 --> 00:32:32.200
It will be less than 30 years until we see an earthquake.
00:32:32.200 --> 00:32:41.400
We have to turn that around, probability that Y is less than 30 is greater than or equal to 1 -1/4 which is ¾.
00:32:41.400 --> 00:32:45.300
I do not just say there is a ¾ chance, I do not just give ¾ as an answer.
00:32:45.300 --> 00:32:51.000
I’m giving a complete sentence which is that there is a greater than or equal to ¾ chance,
00:32:51.000 --> 00:32:59.300
at least a 75% chance that we will see a quake in the next 30 years.
00:32:59.300 --> 00:33:05.300
In example 5, we are looking at housing prices in a small town USA.
00:33:05.300 --> 00:33:13.100
Apparently, they are symmetrically distributed with a mean of $50,000 and a standard deviation of $20,000.
00:33:13.100 --> 00:33:21.600
We are going to use Chebyshev’s inequality to estimate, what proportion of the houses cost less than $90,000?
00:33:21.600 --> 00:33:25.500
Let me do a little graph here because again, we are dealing with a symmetric distribution.
00:33:25.500 --> 00:33:30.100
We will actually find out that this problem is quite similar to an earlier one that we did.
00:33:30.100 --> 00:33:33.400
I think it was examples 3 in this lecture.
00:33:33.400 --> 00:33:40.000
If you remember how to do that one, you might want to try doing this yourself before you watch me give the answers here.
00:33:40.000 --> 00:33:43.000
It works out pretty similarly.
00:33:43.000 --> 00:33:47.800
Let me make a little graph of the housing prices in small town.
00:33:47.800 --> 00:33:53.600
They are symmetrically distributed, I'm going to draw something nice and symmetric here.
00:33:53.600 --> 00:33:56.100
Something that looks like a bell curve.
00:33:56.100 --> 00:34:03.700
We will learn later that this is actually the normal distribution but we have not gotten to that point in the videos yet.
00:34:03.700 --> 00:34:09.900
They are distributed around a mean of $50,000.
00:34:09.900 --> 00:34:15.200
Μ here is 50, I would not bother with the thousands.
00:34:15.200 --> 00:34:21.600
We want to estimate the proportion of houses that cost less than $90,000.
00:34:21.600 --> 00:34:28.700
Let me put in the 90,000 here, at somewhere out beyond 50, that is 90.
00:34:28.700 --> 00:34:34.000
We are told that we have a standard deviation of 20,000.
00:34:34.000 --> 00:34:40.000
I guess that means the distance from the mean, to the cutoff we are interested in is 40.
00:34:40.000 --> 00:34:45.300
That is 2 σ, 2 standard deviations because that is 2 × 20.
00:34:45.300 --> 00:34:50.100
Because we are going to be using that, let me go ahead and put in 2 σ in the other direction.
00:34:50.100 --> 00:34:59.100
2 σ in the other direction which I guess will get you down to 50 -40 is 10 on the low side there.
00:34:59.100 --> 00:35:03.200
That is just setting up a picture, we still need to bring Chebyshev’s inequality.
00:35:03.200 --> 00:35:17.000
Chebyshev’s says the probability that Y will be more than K standard deviations away from its mean,
00:35:17.000 --> 00:35:21.500
is less than or equal to 1/K².
00:35:21.500 --> 00:35:25.700
In this case, we are interested in being 2 standard deviations away from the mean.
00:35:25.700 --> 00:35:37.600
Where I got that was 90 -50 is 40 which is 2 × 20, that is 2 standard deviations there.
00:35:37.600 --> 00:35:49.100
The probability that Y -50 is the mean here, is greater than or equal to K was 2,
00:35:49.100 --> 00:35:53.800
σ is the standard deviation, 2 × 20, that will be 40.
00:35:53.800 --> 00:36:01.800
According to Chebyshev’s inequality that is less than 1/K², 1/2² which is ¼.
00:36:01.800 --> 00:36:06.900
The probability of being that far away from the mean is less than ¼.
00:36:06.900 --> 00:36:09.900
Let me go ahead and fill that in here.
00:36:09.900 --> 00:36:18.600
That is this probability but it is also the probability on the lower end because
00:36:18.600 --> 00:36:23.700
we are told that we have a symmetric distribution here.
00:36:23.700 --> 00:36:30.600
What we know is that, all that shaded region there has combined probability.
00:36:30.600 --> 00:36:35.900
The probability of the shaded region is less than or equal to ¼.
00:36:35.900 --> 00:36:43.900
Since, we know it is symmetric, we know that each one of those tails must be less than 1/8.
00:36:43.900 --> 00:36:51.000
1/8 being ½ of 1/4 there, the probability is less than or equal to 1/8.
00:36:51.000 --> 00:36:53.900
Let me go ahead and fill that in.
00:36:53.900 --> 00:37:04.500
In particular, that Y is bigger than 90, the probability that Y is bigger than 90, according to Chebyshev’s inequality,
00:37:04.500 --> 00:37:11.200
since, we are allowed to split up between the top and the low end, we said that the distribution was symmetrical.
00:37:11.200 --> 00:37:19.700
It is less than or equal to ½ × 1/4 of which is 1/8.
00:37:19.700 --> 00:37:23.800
What was the probability that the house was greater than 90?
00:37:23.800 --> 00:37:28.400
We want to estimate the proportion of houses that cost less than 90,000.
00:37:28.400 --> 00:37:34.200
The probability that Y is less than 90, let me turn that around.
00:37:34.200 --> 00:37:41.700
It is greater than or equal to 1 – 1/8, that is 7/8.
00:37:41.700 --> 00:37:48.400
If we convert that into a percentage, that is halfway between ¾ and 1.
00:37:48.400 --> 00:37:59.700
It is halfway between 75 and 100, that is 87.5%.
00:37:59.700 --> 00:38:09.300
The proportion or the probability, the proportion of houses that cost less than $90,000 in this town is,
00:38:09.300 --> 00:38:16.200
I cannot say it is the equal to 87% but it is at least 87.5%.
00:38:16.200 --> 00:38:23.600
I can say that at least 87% of the houses in this town must cost less than $90,000,
00:38:23.600 --> 00:38:26.800
that is the interpretation that I can put on that.
00:38:26.800 --> 00:38:32.600
At least 87% of these houses cost less than $90,000.
00:38:32.600 --> 00:38:35.100
Let me recap where that is coming from.
00:38:35.100 --> 00:38:41.800
The probability that Y - μ is greater that K σ is less than 1/K².
00:38:41.800 --> 00:38:45.400
That is just the original version of Chebyshev’s inequality.
00:38:45.400 --> 00:38:53.000
In this case, my μ was 50, that came from the mean housing price their.
00:38:53.000 --> 00:39:00.800
The standard deviation, the σ is 20, that was also given to us in the problem.
00:39:00.800 --> 00:39:02.900
I have not figure out what K would be.
00:39:02.900 --> 00:39:07.300
In order to figure out what K would be, I want to know what I was being asked about.
00:39:07.300 --> 00:39:11.900
I was being asked about houses costing less than $90,000.
00:39:11.900 --> 00:39:18.100
We are going to use 90 as our cut off and 90 -50.
00:39:18.100 --> 00:39:20.600
50 is the mean, 90 is what we are interested in.
00:39:20.600 --> 00:39:24.700
The difference there is 40 which is 2 standard deviations, 2 × 20.
00:39:24.700 --> 00:39:30.700
That is where I get my K is equal to 2 there, K =2.
00:39:30.700 --> 00:39:37.600
I plug that in to Chebyshev’s inequality and I get that the probability is less than ¼.
00:39:37.600 --> 00:39:44.300
That is the probability of being 2 standard deviations away from mean, in either direction.
00:39:44.300 --> 00:39:49.800
That includes both of these regions here, both the high region and the low region.
00:39:49.800 --> 00:39:57.900
But I'm really only interest in, how many of the houses are costing too much on the high side?
00:39:57.900 --> 00:40:02.700
Let me cut that region in half and get a probability of less than 1/8.
00:40:02.700 --> 00:40:08.700
That is why that is less than 1/8/.
00:40:08.700 --> 00:40:17.200
That was describing the proportion of houses that cost more than $90,000.
00:40:17.200 --> 00:40:21.100
I want to find the proportion of houses that cost less than $90,000.
00:40:21.100 --> 00:40:22.100
Let us switch that around.
00:40:22.100 --> 00:40:26.200
Instead of talking about 1/8, I will talk about 1 -1/8.
00:40:26.200 --> 00:40:30.100
Instead of talking about less than or equal to, I have a greater than or equal to.
00:40:30.100 --> 00:40:35.000
1 -1/8 simplifies down to 87.5%.
00:40:35.000 --> 00:40:39.600
My answer here is, I do not just say 87.5% is my answer.
00:40:39.600 --> 00:40:48.100
My answer is that, at least 87.5% of the houses in this town cost more than $90,000.
00:40:48.100 --> 00:40:54.000
The at least part is very important part of the answer there.
00:40:54.000 --> 00:40:58.600
That wraps up our lecture on Chebyshev’s inequality.
00:40:58.600 --> 00:41:02.300
You can figure this as companion to the lecture on Markov’s inequality which
00:41:02.300 --> 00:41:07.800
we have on the previous videos is on Markov’s inequality, they go hand in hand.
00:41:07.800 --> 00:41:10.600
Markov’s inequality, you just need to know the mean.
00:41:10.600 --> 00:41:16.600
Chebyshev’s inequality, you need to know the mean and the standard deviation because of that σ there.
00:41:16.600 --> 00:41:19.100
You need to know both for Chebyshev’s inequality.
00:41:19.100 --> 00:41:27.800
You do a little more computation for Chebyshev’s inequality but the trail off that is that you usually get stronger results.
00:41:27.800 --> 00:41:35.000
You usually get some more information about the probabilities for Chebyshev’s than you do for Markov.
00:41:35.000 --> 00:41:40.700
Remember, for either one of these inequalities, you have to give your answer as an inequality.
00:41:40.700 --> 00:41:45.800
You never give numerical value because those numerical values are just lower or upper bounds.
00:41:45.800 --> 00:41:52.400
Your answer will always be that the probability is less than this or greater than that.
00:41:52.400 --> 00:41:57.700
That does it for Chebyshev’s inequality, kind of wrapping up a chapter here.
00:41:57.700 --> 00:42:01.400
We will jump in later on with the binomial distribution.
00:42:01.400 --> 00:42:03.400
I hope you will stick around for that.
00:42:03.400 --> 00:42:08.300
You are enjoying the probability lecture series here on www.educator.com.
00:42:08.300 --> 00:42:11.000
My name is Will Murray, thank you so much for watching, bye.