For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

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### Mean & Variance for Continuous Distributions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Mean
- Variance
- Standard Deviation
- Example I: Mean & Variance for Continuous Distributions
- Example II: Mean & Variance for Continuous Distributions
- Example III: Mean & Variance for Continuous Distributions
- Example IV: Mean & Variance for Continuous Distributions
- Example V: Mean & Variance for Continuous Distributions

- Intro 0:00
- Mean 0:32
- Mean for a Continuous Random Variable
- Expectation is Linear
- Variance 2:55
- Variance for Continuous random Variable
- Easier to Calculate Via the Mean
- Standard Deviation 5:03
- Standard Deviation
- Example I: Mean & Variance for Continuous Distributions 5:43
- Example II: Mean & Variance for Continuous Distributions 10:09
- Example III: Mean & Variance for Continuous Distributions 16:05
- Example IV: Mean & Variance for Continuous Distributions 26:40
- Example V: Mean & Variance for Continuous Distributions 30:12

### Introduction to Probability Online Course

### Transcription: Mean & Variance for Continuous Distributions

*Hi, welcome back to the probability lectures here on www.educator.com.*0000

*My name is Will Murray, we are learning about continuous probability right now.*0004

*Today, we are going to study the mean and variance for continuous distributions.*0010

*I will teach you how to calculate the mean and variance.*0015

*I also mentioned standard deviation.*0018

*Usually, we do not look at standard deviation so often for the continuous distributions,*0023

*but I will go ahead and show you how to calculate it, if you need to.*0025

*I'm going to spend most of the time though on the mean and variance.*0029

*Let me get started here.*0034

*I will remind you how we calculated the mean for discrete distributions which was as follows.*0036

*For discrete distributions, we calculated the mean, the expected value of our random variable*0044

*was the sum of all possible values of Y of Y × P of Y.*0054

*The mean for continuous distributions is essentially the same thing.*0061

*You still have a Y, instead of P of Y, you have the density function F of Y.*0066

*By the way, remember, there is a density function which we use f to denote the density function.*0071

*There is also a cumulative distribution function for which we use F.*0076

*This is definitely the density function not the cumulative distribution function.*0083

*It is essentially the same formula that we had for discrete distributions.*0090

*The Y is the same, the P of Y is turned into F of Y, and the summation has turned into an integral.*0094

*Really, you do not need to memorize a new formula.*0102

*If you can remember one, it is just kind of translating it into the language of continuous distributions*0106

*to find the expected value for a continuous distribution.*0112

*Remember that, mean and expected value are the same thing, those are completely synonymous.*0116

*If someone asks for the mean, they are asking you for the expected value, and vice versa.*0122

*It is very useful to know that just like with discreet distributions, expectation is linear.*0128

*The expected value of a constant is just a constant.*0137

*If you have two things added together, you can split them up and calculate their expected values separately.*0141

*If you have a constant × a random variable then you can pull a constant outside.*0146

*It is a very useful property of expectation.*0153

*We will be using it again and again, the linearity of expectation.*0156

*By the way, that does not work for variance.*0161

*Variance is not linear, you have to be very careful not to assume that variance is linear.*0164

*You can get yourself in a lot of trouble that way.*0170

*I will go ahead and show you the definition of variance.*0173

*The definition of variance is the same definition that we had for discrete distributions.*0177

*It is the expected value of Y - μ², where that μ was the mean of the distribution,*0183

*same as the expected value of the original variable.*0191

*If you want to calculate that, you would multiply y × μ² × the density function F of Y, and calculate an integral.*0195

*Now, that is not usually how you are going to calculate the variance of a continuous distribution.*0207

*This version of the formula is not usually very useful.*0214

*What is usually easier is to calculate this little formula, the expected value of Y² - the expected value of (Y).*0220

*That is just like we had with the discreet distributions.*0231

*It works the exact same way and it is usually easier to calculate that way.*0235

*The way you calculate it, remember the expected value of Y is just the mean,*0242

*which hopefully you already knew or you already calculated before you are worrying about the variance.*0246

*The expected value of Y², what you do is you multiply Y² × the density function, and then you have to integrate that.*0253

*You do have to do an integral but it is usually a simpler one than you would have done,*0261

*if you had calculated this Y - μ² × F of Y.*0266

*This is usually the better way to find the variance of a continuous distribution is, to use this formula down here.*0271

*We will practice that and you will see in the examples.*0281

*We will probably be using this version of the variance formula.*0284

*Usually, it seems to me when the variance problems, with continuous problems,*0291

*you are used to calculating the variance and not often the standard deviation.*0296

*Let me go ahead and give you the formula for standard deviation, because it is very easy.*0300

*It is exactly like the discrete case, if you know the variance, the standard deviation by definition,*0304

*is just the square root of the variance.*0310

*Basically, you just calculate the variance and at the end, if you want standard deviation,*0312

*you calculate the square root of that.*0317

*There is really nothing more and nothing less than that.*0319

*That is possibly why all the problems that you encountered in probability classes just say,*0323

*calculate the variance and I do not even bother to ask you to calculate the standard deviation.*0329

*But if we do ask you to calculate the standard deviation, just calculate the variance first and then take its square root.*0334

*Let us try some examples.*0342

*First example here, it is the same density function that we had in example 3 of the previous video.*0343

*I do not think you really need to understand example 3 of the previous.*0351

*Just in case this looks familiar, this is the density function that you saw before.*0355

*Let Y have the function 1/3 between 0 and 1, 2/3 between 1 and 2.*0361

*That is not a good vertical line at all, I will do a little graph here.*0368

*There is 1, there is 2, and what we have is a density function like that.*0375

*There is our density function and we want to find the expected value of Y.*0384

*We are just going to use the definition of this.*0390

*The expected value of Y is equal to, for any continuous distribution,*0392

*it is equal to the integral from -infinity to infinity of Y × the density function F of Y DY.*0400

*Now, I set the integral from infinity to -infinity but here, there is really no density anywhere outside of the range between 0 and 2.*0408

*I’m just going to integrate from 0 to 2.*0418

*And then, I'm going to break it up from 0 to 1 and 1 to 2 because*0420

*we have two different density functions on those two ranges.*0426

*It is Y × 1/3 DY from 0 to 1 and Y × 2/3 DY from 1 to 2.*0429

*Those are both easy integrals, in fact, you know I'm going to pull out 1/3 from everything because*0440

*that just makes me have to write now less fractions, fewer fractions.*0446

*The integral of Y DY is Y²/2.*0452

*I have to write with that from 0 to 1.*0457

*The integral of 2Y DY is just Y².*0461

*I have to evaluate that from 1 to 2.*0465

*I will keep going here, 1/3 Y²/2, when Y is 1 is 1, Y0 is nothing, + Y² from 1 to 2 is 2² is 4 -1² is 1, 7 + 4 - 1 is 3.*0472

*My mistake, the Y²/2 should have been a ½ not 1, when I plug in Y = 1 before.*0492

*I get 1/3 × 3 1/2 is 7/2.*0504

*I get my expected value of that distribution is 7/6.*0511

*By the way, we can check this, you will get at least an approximate check, if we look at the graph.*0519

*If you look at this graph up here, you will notice that most of the density is concentrated over there between 1 and 2.*0525

*What we found here is that the mean or the expected value is just a little bit bigger than 1, about 1/6.*0535

*It is about right there.*0544

*That is kind of the balancing point of that function.*0545

*It is not surprising that it is a little bit bigger than 1 because there is a little more area to the right of 1 than to the left of 1.*0549

*Our average there or expected value μ = 7/6.*0556

*To recap the steps there, I just used my definition of expected value.*0562

*The integral of Y × the density function F of Y.*0567

*I have to split that up into two different ranges, the range from 0 to 1 and the range from 1 to 2.*0572

*Factor the 1/3 out of everything and I dropped in my two different functions for the density functions.*0578

*Factor out a 1/3, did a couple of easy integrals, plug in the values, the limits, and simplify it down to a number of 7/6.*0587

*When I got that number, it really was a surprise looking at the graph because*0596

*it seem to balance a bit bigger than 1 because more of the area is concentrated bigger than 1 there.*0600

*Let us move on, example 2, we got θ1 and θ2 are constants.*0608

*I guess θ2 is the bigger one, I will go ahead and graph this, as we introduce the problem.*0614

*There is θ1 and there is θ2.*0621

*And then, we are going to consider the uniform density function F of Y is just the constant 1/θ2 – θ1.*0625

*It was just constant because there is no Y in there.*0633

*From Y going from θ1 to θ2.*0636

*There is our density function.*0641

*By assumption, that means that it is 0 everywhere else.*0643

*That is our density function, our density is uniformly distributed.*0648

*We will talk more about the uniform distribution, later on.*0654

*We have us some more videos that you will see, if you scroll down about to the different continuous distributions,*0657

*the continuous probability distributions.*0665

*Uniform will be one of them, this is kind of a warm up to that video later on.*0667

*The value here is the constant value 1/θ2 – θ1.*0673

*By the way, it has to be that constant value, in order to make the total area equal to 1.*0679

*Because the width is θ2 – θ1, that means the height must be 1/θ2 – θ1.*0684

*You have to know that, that is the constant value there, if it is going to be constant.*0690

*Let us find E of Y.*0697

*Μ is our expected value, or mean of our distribution.*0701

*E of Y, now I'm going to use again, the generic formula for expected value of -infinity to infinity of Y F of Y DY.*0707

*In this case, the only place where we have positive density is from θ1 to θ2.*0719

*I will fill those in as limits and not worry about everything else, that infinity of Y.*0727

*F of Y is just the constant, it is 1/θ2 – θ1 DY.*0734

*I will pull out our constant out.*0742

*I will just go ahead and do the integral because it is an easy integral.*0745

*Y²/2 × θ2 – θ1.*0749

*I have to evaluate that from θ1 to θ2.*0755

*It is a form the algebra works out with this, stick with me on this.*0760

*I get θ2² – θ1²/2 × θ2 – θ1.*0764

*The numerator factors θ2 + θ1 and θ2 – θ, that is a difference of squares.*0775

*On the bottom, we just get θ2 – θ1.*0787

*That factors cancel and I’m going to write this in a different order.*0791

*I will just write it as θ1 + θ2/2, that is my expected value of the uniform distribution.*0796

*It is nice to work that out arithmetically, but it is also nice to realize that that is a completely intuitive result,*0809

*if you look at the graph.*0817

*Because if you look at the graph, exactly half the area is to the left and exactly half of the area is to the right.*0819

*It is a uniform distribution.*0826

*You are really not surprised at all, to find that the mean is just halfway between θ1 and θ2.*0827

*It is the average of the two values.*0835

*Let me write that a little bigger so you can actually see it.*0839

*Θ1 + θ2/2, it is not surprising at all to get that mean to be halfway between the two endpoints.*0842

*But, because it is uniformly distributed, the density is uniformly distributed, if they were not uniformly distributed,*0852

*we might not expect the mean to fall exactly on the halfway mark.*0857

*But since, it is uniformly distributed, not at all surprising that*0862

*we end up with our mean being just halfway down the middle θ1 + θ2.*0866

*Let me show you how we calculated that quickly.*0873

*Use the definition of the mean or expected value, same thing by the way, of a continuous distribution*0875

*which is the integral from -infinity to infinity of Y × the density function DY.*0882

*Since, our only positive density was between θ1 and θ2, I through away those infinities and just looked at the range from θ1 to θ2.*0889

*I filled in my F of Y is 1/θ2 – θ1.*0902

*That is a constant, it easily comes through the integral.*0906

*The integral of Y is Y²/2.*0909

*Plug in θ2 as Y² and θ1 as Y.*0912

*We get θ2² – θ1².*0918

*It is a nice algebra here, we factored a difference of squares formula.*0921

*We get θ2 – θ1 canceling, we end up with θ1 + θ2/2 which completely confirms the suspicion that I hope you had when I first graph this,*0925

*which is that the mean would come out to be right in the middle, because the density is uniformly distributed.*0935

*In the next example, we are going to calculate the variance of the uniform distribution.*0942

*Part of that calculation, we will be using the mean.*0947

*I want you to remember this mean.*0951

*We will just plug it right in, when we get to the right point in the formula for the variance on the next problem.*0956

*In example 3 here, it is kind of a continuation of example 2.*0967

*We are still looking at 2 constant values, θ1 and θ2.*0972

*We have the uniform density function between θ1 and θ2.*0976

*We are finding the variance.*0982

*In example 2, we found the expected value E of Y.*0984

*Now, we are finding V of Y.*0988

*We are going to be using the answer from example 2 in examples 3.*0989

*If you did not just watch example 2, if you are just joining us for today for example 3,*0995

*you really want to go back and watch example 2, and make sure you understand the answer to example 2.*1001

*We will be using it at a key step here in examples 3.*1006

*For example 3, we want to find the variance.*1011

*The first step of that is to find the expected value of Y².*1014

*By definition, that means the integral from -infinity to infinity of Y² F of Y DY.*1021

*Just like in example 2, the only range we have here for Y is θ1 to θ2.*1032

*Everywhere else, we can assume that the density function is 0.*1039

*I'm going to cut off those infinities and just integrate from θ1 to θ2.*1043

*My Y², my F of Y is 1/θ2 – θ1.*1054

*Those are all constants, 1/θ2 – θ1 DY.*1060

*That is a pretty easy integral, I get Y³/3 is the integral of Y².*1064

*I still have that constant θ2 – θ1.*1074

*I'm integrating this from Y = θ1 to Y = θ2.*1079

*Let me plug those in.*1086

*Θ2³ – θ1³/3 × θ2 – θ1.*1088

*Just like in example 2, there is a really nice algebra that works out here, if you remember your difference of cubes formula.*1097

*In example 2 was, the difference of squares formula which everybody remembers.*1106

*In example 3, it is the difference of cubes formula which is not quite as well known, it is θ2 – θ1.*1110

*I will remind you A³ – B³.*1118

*A² + AB, not 2AB, + B².*1121

*That is the difference of cubes formula.*1134

*I will write that up here because you might not be so familiar, people who are a little rusty on algebra.*1137

*A - B × A² + AB + B².*1143

*That is what I use right here to factor this.*1150

*That is really nice because in the denominator, we also have θ2 – θ1.*1153

*Those cancel, E of Y² here is equal to θ2² + θ1 θ2 + θ1².*1160

*Let me switch the roles of θ1 and θ2.*1175

*I will write θ1² + θ1 θ2 + θ2².*1178

*That will be a little easier to keep track of in the next step.*1184

*I did say there is a next step, we have not yet found the variance.*1187

*Let me remind you that there is a 2 step procedure for finding the variance.*1191

*The variance of Y which is the same as σ² is always the expected value of Y² - the expected value of (Y)².*1195

*We figure out the expected value of Y² right here, it is θ1² + θ1 θ2 + θ2² ÷ 3.*1215

*The expected value of Y, we figured this out in example 2.*1230

*That is why I said go back and watch example 2.*1234

*Let me remind you of the answer from example 2, I will not work it out again.*1237

*But example 2, we figured out that E of Y is θ1 + θ2/2.*1241

*That is not surprising for the uniform distribution because you just get the average of the 2 endpoints.*1250

*Let me plug that in, θ1 + θ2/2.*1258

*We want to square that whole thing.*1263

*I accidentally changed my - to +.*1266

*Let me quickly correct that before anybody notices.*1272

*Let me expand this θ1 + θ2² because we are going to do some algebra and combine this.*1277

*This is going to work out really nicely, I practiced this and it worked out just great.*1285

*This is θ1² + 2θ1 θ2 + θ2²/4 because we have to square both top and bottom here.*1289

*We want to combine these two, it looks like it is going to be really messy but it will simplify nicely, I promise.*1304

*I have a common denominator, I got a 3 and 4.*1310

*My common denominator is going to be a 12 and that means I have to multiply the first set of terms by 4.*1312

*I will go ahead and do that.*1320

*4 θ1² + 4 θ1 θ2 + 4 θ2².*1321

*The next sets of terms are all negative and I need to multiply them all by 3.*1330

*-3 θ1² – 3 × 2 is 6 θ1 θ2 – 2 θ2².*1335

*I did not multiply that by 3, not a 2 there but 3 θ2².*1348

*I can simplify this down, I get 4.*1355

*Θ1² – 3 θ1², I got θ1² σ of θ1².*1358

*4 θ1 θ2 – 6 θ1 θ2 – 2 θ1 θ2 + θ2²/12.*1364

*What you will notice is that numerator is exactly the perfect square of either θ1 - θ2 or θ2 – θ1.*1380

*I’m going to say θ2 – θ1.*1391

*It does not matter but I like that, because I have been told that θ1 is less than θ2.*1396

*Θ2- θ1 is positive, let us keep it positive.*1402

*Θ2 – θ1²/12 is my variance, that is quite a nice formula.*1404

*I like the fact that it simplifies that way.*1423

*Unlike the mean, the answer from example 2, I do not think the variance is totally obvious.*1425

*I could not have just look at that and told you off the top of my head that that was going to be the variance.*1431

*It is not totally surprising because it is dependent on the distance from θ1 to θ2.*1438

*Remember, the variance kind of measures how spread out your distribution is.*1447

*If θ1 and θ2 are far apart, then, we get a bigger variance here.*1451

*That makes intuitive sense but I do not think I could have looked at that and just say,*1455

*it is definitely going to be θ2 – θ1²/12.*1460

*That would not have been so obvious.*1463

*Whereas with the expected value, if you are really on top of your game,*1465

*you could probably eyeball that distribution and say, I know it is going to be θ1 + θ2/2.*1469

*Let me show you the steps that we use for that.*1475

*We started out finding E of Y².*1478

*The reason I did that was because I was looking forward to this formula for the variance, E of Y² – E of (Y)².*1482

*I’m going to find me E of (Y)² first.*1491

*By definition, that is Y² × the density function and we will integrate that.*1494

*The range we want to integrate on is θ1 and θ2.*1500

*The density function is just this constant 1/θ2 – θ1.*1504

*That integrates easily Y² integrates to Y³/3, drop in the values of θ, the values of Y at the endpoints.*1507

*Using this nice difference of cubes formula, it expands out and cancels.*1519

*We get θ2 – θ1 canceling from the top and bottom.*1526

*We get a fairly nice little expression for the expected value of Y².*1530

*But that is not the variance yet, you have to subtract off the expected value of (Y)².*1536

*And the expected value of Y is what we figure out in example 2, that is θ1 + θ2/2.*1542

*That is where I’m getting this from.*1549

*Because I want to put everything over a common denominator, I expanded this out.*1553

*It is quite a mess here, especially when we put it over a common denominator of 12,*1559

*and in particular, it simplify down to this very nice formula θ2 – θ1²/12.*1563

*I could have said θ1 – θ2² as well, but I use θ2 – θ1 because that will be a positive number,*1572

*that is because θ2 is the bigger one.*1579

*That is the variance of the uniform distribution.*1582

*We are going to do a whole video on the uniform distribution later on,*1585

*if you just scroll down you should be able to see that.*1588

*You will see that we will be using this mean and variance, that we worked out here in examples 2 and 3.*1591

*Let us look at example 4 here.*1602

*We have Y has density function ½ × 2 - Y on the range 0 to 2, and 0 elsewhere.*1604

*We want to find the expected value of Y.*1613

*In example 5, we are going to keep going with the same random variable.*1615

*We will find the variance of Y.*1621

*You want to make sure that you watch these two examples in tandem because*1624

*we are going to be using the results from one in the next one.*1629

*Let us find the expected value of Y.*1632

*The expected value of Y by definition, remember, it is always the integral from -infinity to infinity of Y ×*1635

*the density function f of Y DY.*1644

*In this case, I do not need to integrate from infinity to infinity because the only place where the density is positive is from 0 to 2.*1647

*I’m just going to integrate from 0 to 2 of Y.*1656

*I will fill in the density function ½ × 2 – Y DY.*1660

*I think I will distribute that to make it a little easier to integrate.*1666

*Y × ½ × 2 is just Y – Y × ½ × Y.*1670

*That is ½ Y², just distributing those.*1680

*I cannot put off the calculus any longer, integrate those, I get Y²/2 - the integral of Y² is Y³/3 × ½ is Y³/6.*1685

*I need to evaluate that from Y = 0 to Y = 2.*1702

*I will plug those values in, I get 2²/2 is 4/2 is 2 – Y³/6 is 8/6.*1708

*That is 2 - 4/3 and that is 2 and 6/3 - 4/3 is 2/3.*1719

*That is my answer for the expected value.*1740

*That was fairly straightforward, it came straight out of the definition here.*1742

*The expected value of a random variable is, you integrate Y × the density function from -infinity to infinity.*1746

*But in practice, you end up integrating your density function is defined to be positive.*1758

*That is on this range right here 0 to 2 because everywhere else, it is 0.*1764

*Let me integrate Y ×, I just plug in the densely function from here.*1769

*I do not need to worry about the region on which it is 0.*1774

*Plug that in, distribute the terms to make it easier to integrate, do a little integral,*1779

*drop in the numbers and I get my expected value of 2/3.*1786

*You want to hang onto this value for the next example.*1790

*In example 5, we are going to come back to the same density function and we are going to calculate the variance.*1794

*Remember, a key step in calculating the variance is using the expected value.*1801

*Remember this value of 2/3, we are going to use it again in the next example.*1806

*In example 5, we are looking at the same density function that we had in example 4,*1813

*F of Y is ½ × 2 – Y on the range from 0 to 2, and at 0 everywhere else.*1818

*We want to find the variance.*1825

*Let me remind you of the useful way to calculate the variance.*1827

*The variance is E of Y² – E of (Y)².*1832

*What we will do first is we will calculate E of (Y)², and then we will come back and + it into this formula.*1842

*E of Y², by itself here, I’m not calculating the whole variance yet.*1850

*I’m just finding E of Y².*1855

*By definition, that is the integral from -infinity to infinity of Y² × whatever density function we have for this variable.*1857

*In this case, our density function is 0 everywhere outside the range is 0 to 2.*1868

*I just need to integrate from 0 to 2 Y².*1874

*We look at that density function is ½ × 2 – Y DY, that is the integral from 0 to 2.*1879

*If I distribute that Y² and ½, Y² × ½ × 2 is just Y².*1888

*And then, Y² × ½ × Y is ½ Y³.*1896

*I have to integrate all that with respect to Y.*1903

*My integral is, Y² integrates to Y³/3.*1907

*Y³ integrates to Y⁴/4, but I also got a ½ here.*1913

*I have to make that Y⁴/8.*1919

*I want to integrate all of that from 0 to 2, Y = 2.*1923

*I get to Y³/3 that is 8/3 – y⁴/8 that is 16/8.*1934

*If I plug in 0 nothing happens there, I do not need to worry about that.*1944

*That is 8/3 -2, 2 is 6/3, the expected value Y² is just 2/3.*1948

*That is not the answer to the problem yet, we are supposed to find the variance.*1960

*The expected value of Y² is just one step in finding the variance.*1965

*Here is the rest of it.*1968

*E of Y² is 2/3, we just calculated that.*1970

*The expected value of Y, I would have to do another whole calculation to find that.*1975

*But I did that calculation in example 4, I hope you did it with me.*1981

*If you have not watched example 4, now is the time when we are using the answer.*1985

*It came out to be the same thing, it was 2/3 by example 4.*1993

*That is really coincidence, the fact that it came out to be the same as E of Y².*2001

*I would not put too much stock in the fact that those numbers came out to be the same.*2006

*It is not going to happen usually, there is no sort of property that made those the same.*2012

*It is just the way that the integrals worked out.*2018

* This is to 2/3 - E of Y², -2/3².*2020

*This is 2/3 – 4/9, 2/3 is 6/9.*2029

*Multiply top and bottom by 3 - 4/9 and I finally get my answer here for the variance is 2/9,*2037

*the variance of this random variable.*2047

*In case you are a little fuzzy on any of the steps, let us check those out again.*2055

*My generic formula for the variance is always E of Y² – E of (Y)².*2061

*The E of Y by itself, I figure it out in example 4.*2069

*You can go back and see the video in example 4, where that came from.*2073

*We figure out that it was 2/3, that is where I dropped that in for E of Y.*2076

*This E of Y² takes more work.*2081

*E of Y², we actually have to calculate that using an integral of Y² × the density function.*2084

*I put in Y², I put in the density function that came from right here.*2090

*These bounds from 0 to 2 that came from these limits right here, because everywhere else the function is 0.*2096

*I really only need to do my integral from 0 to 2.*2105

*Once I got in that form, it is relatively easy algebra to distribute the terms to do the calculus there.*2110

*It is just the power rule, drop in the limits 0 and 2, and simplify it down to 2/3.*2118

*That 2/3 gives me that 2/3 for E of Y².*2125

*The other 2/3 came from E of Y, that was from example 4.*2131

*It is really a coincidence that those numbers were both 2/3.*2135

*That is not any property manifesting itself here, do not read too much into that.*2138

*Those could easily have been different numbers.*2144

*Now, I just simplify the fractions 2/3 - 4/9 simplifies down to 2/9.*2147

*That is my variance for that random variable.*2150

*That wraps up this lecture on mean and variance for continuous distributions.*2156

*This is part of the chapter on continuous probability which is part of our larger lecture series on probability.*2162

*You are watching the probability lectures here on www.educator.com, with your host Will Murray.*2171

*Thank you very much for joining me today, bye.*2177

1 answer

Last reply by: Dr. William Murray

Sun Jan 4, 2015 7:21 PM

Post by Joseph Szmulewicz on January 1, 2015

sorry. I just figured out what happened to the 2 in the second integral. the integral of y is y squared over 2 and the the 2 from 2/3 gets cancelled by the 2 in the denominator. Silly of me to miss this.

1 answer

Last reply by: Dr. William Murray

Sun Jan 4, 2015 7:21 PM

Post by Joseph Szmulewicz on January 1, 2015

also on example 1, it says above that the variance is also calculated, but only the mean got calculated. misprint?

1 answer

Last reply by: Dr. William Murray

Sun Jan 4, 2015 7:21 PM

Post by Joseph Szmulewicz on January 1, 2015

I found a mistake on example 1, I think. Correct me if I am wrong, but you pulled out the 1/3 from the first integral which is right, but the second integral has a 2/3 constant, not 1/3. When you integrate the second part, shouldn't it be 2 times y squared, not just y squared?