For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

### Uniform Distribution

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 0:00
- Uniform Distribution 0:15
- Uniform Distribution
- Each Part of the Region is Equally Probable
- Key Properties of the Uniform Distribution 2:45
- Mean
- Variance
- Standard Deviation
- Example I: Newspaper Delivery 5:25
- Example II: Picking a Real Number from a Uniform Distribution 8:21
- Example III: Dinner Date 11:02
- Example IV: Proving that a Variable is Uniformly Distributed 18:50
- Example V: Ice Cream Serving 27:22

### Introduction to Probability Online Course

### Transcription: Uniform Distribution

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

*Today, we are going to talk about the first of several continuous distributions.*0004

*This is probably the easiest one, it is called the uniform distribution.*0010

*We will see very soon why it is called that.*0013

*Let us jump right on in.*0016

*The idea of the uniform distribution is you have a finite range from θ1 to θ2.*0017

*Let me go ahead and draw a graph of this, as I'm talking about this.*0023

*You have 2 constant values, here is θ1 and then you have θ2, somewhere a bit bigger than θ1.*0028

*And then, you just divide your density evenly, you distribute it evenly over that range *0037

*which means you just take a completely horizontal line over that range.*0044

*What that means is, remember the total density always has to be 1, *0051

*that the total area under a density function always has to be 1.*0055

*In order to have that area be 1, the width is θ2 – θ1, the height has to be the constant 1/θ2 – θ1.*0060

*By the way, this triple equal sign, that means always equals 2.*0070

*It means that the density function is constant.*0079

*That is much different from all of the other density functions that we will be studying later.*0082

*That is what makes the uniform distribution a lot easier than some of the later ones.*0087

*It is that the density function is always equal to a constant.*0091

*That constant has to be 111, in order to give your total area 1.*0094

*Each part of the region is equally possible or equally probable.*0101

*It is very easy to calculate probabilities with the uniform distribution, if you have two values A and B here.*0106

*Let me go ahead and draw them in on my graph A and B.*0114

*If you have two values A and B, and we want to find the probability that your random variable *0118

*land somewhere between A and B.*0124

*It is very simple, you just have to calculate the distance between A and B.*0126

*Essentially, you are calculating this area right here.*0131

*And that black area is just going to be B – A/θ2 – tθ1, because it has width B - A and it has height of 111.*0136

*It is very easy to calculate probabilities using uniform distribution.*0149

*You just look at the two ranges that you are interested in, subtract them, *0153

*and then divide that by the appropriate constant which is always θ2 – θ1.*0157

*Let us see how that plays out.*0164

*The key properties in uniform distribution, they mean should be kind of intuitively obvious.*0167

*Let me draw again the graph of the uniform distribution.*0173

*There is θ1 and there is θ2.*0178

*Remember that, we are distributing the density completely evenly between θ1 and θ2.*0181

*You would expect the mean to be halfway between them.*0187

*In fact, that is where it turns out to be.*0191

*The mean is exactly the average of θ1 and θ2.*0193

*You just get θ1 + θ2/2 for the μ.*0198

*That is really intuitively clear, it should not be hard to remember because it should be obvious.*0202

*The variance is less obvious, the variance turns out to be θ2 – θ1²/12.*0207

*I think that is not something that you would guess.*0215

*You probably would guess the mean if you give a little bit of thought.*0218

*The variance is not something that you probably guessed.*0221

*You probably have to calculate it out or just memorize it.*0225

*The standard deviation, remember, it is always the square root of the variance.*0229

*If you take the square root of the variance here, you get θ2 – θ1 ÷ √ 12 which is 2 × √ 3.*0233

*That is the standard deviation of the uniform distribution.*0243

*It should still make a rough intuitive sense because it is really a measure of how spread out the interval is.*0249

*Remember that, variance and standard deviation measure how spread out your dataset is.*0257

*In this case, since we got a uniform distribution, if it is spread out over a wider area *0264

*then you should have a higher variance or a higher standard deviation.*0270

*If it is squished into a smaller area then you should have a smaller variance or a smaller standard deviation.*0274

*In this case, since we have the term of θ2 – θ1, that is the width of the interval there, θ2 – tθ1.*0282

*What we are saying here is that the standard deviation is proportional to the width of the interval.*0291

*If you have a wider interval, if you spread your data out more then you have a larger standard deviation.*0297

*If you compress your interval, then you will have a smaller standard deviation.*0303

*It should be intuitively plausible.*0307

*What is not so obvious I think is, the constant 12 for the variance, 2√ 3 for the standard deviation.*0310

*I think that part would not be obvious unless, you actually calculated them out.*0317

*Let us go ahead and look at some problems involving the uniform distribution.*0322

*They generally tend to be fairly easy to calculate.*0328

*The first example here is, you are sitting on your front doorstep, waiting for your morning newspaper to arrive.*0330

*It always arrives sometime between 7:00 AM and the noon.*0337

*The time at which it arrives follows a uniform distribution.*0342

*We want to find that probability that it will arrive during an odd number hour, *0347

*which means we want it to arrive between 7 and 8, not between 8 and 9 because that would be even number hour.*0352

*Between 9 and 10 we are qualified, 10 and 11 would not qualify, and 11 to 12 that would also qualify as an odd numbered hour.*0360

*It looks like there is 3 hours here that would qualify.*0371

*Our total range here is 7 to noon, 7 to 12, that is θ1 is 7 and θ2 is 12.*0374

*Θ2- θ1 is 12 – 7 is 5, that is our denominator here.*0385

*We want to talk about what range we are interested in.*0401

*We have all these odd number hours, 8 -7 + 10 - 9 + 12 -11, which of course is just 3 separate hours on that 5 hour interval.*0404

*We have a total probability of 3/5.*0423

*If your newspaper is going to arrive sometime between 7:00 AM and 12, and it is uniformly distributed in that interval, *0431

*then there is a 60% chance that it will be an odd numbered hour.*0440

*Let me recap that.*0447

*We got this newspaper arriving in a 5 hour interval, that is where I got that denominator of 5 because it is a 5 hour interval.*0449

*If you want think about it in terms of θ2 – θ1, that is 12 -7, that is where that 5 comes from.*0457

*The 3 comes from the 3 hours that have odd numbers.*0463

*The 7:00 hour, the 9:00 hour, and 11:00 hour, gives you 3 different hours that have odd numbers.*0469

*Our total fraction is 3/5, the probability that it will arrive during an odd numbered hour is exactly 60%.*0477

*Remember, I told you that the uniform distribution is one of the easiest to deal with.*0484

*Problems involving uniform distribution are often very easy computationally.*0489

*This certainly qualifies as an easy one computationally but if you stick around, I got a couple of harder once coming up.*0494

*We will see something a little harder coming up.*0500

*Example 2 is also going to be an easy one, trust me, example 3 will be a little more tricky.*0504

*But example 2, we are going to pick a real number Y from the uniform distribution on the interval from 5 to 12.*0511

*Let me go ahead and graph this out.*0519

*There is 5, there is 6, 7, 8, 9, 10, 11, there is 12.*0521

*We are going to pick a real number Y from somewhere in this interval on the uniform distribution.*0531

*We want to find the probability that that value of Y will turn out to be bigger than 9.*0537

*The probability that Y would be bigger than or equal to 9, is equal to, on the denominator we have θ2 – θ1.*0544

*In the numerator, we have B - A being the interval that we are interested in.*0555

*Let me go ahead and draw that in.*0561

*Θ1 is 5, θ2 is 12, the interval that we are interested in is from 9 to 12 because we want Y to be bigger than 9.*0563

*6, 7, 8, 9, there is 9 right there.*0573

*There is A is 9, and B is the same as θ2, B is 12.*0577

*B - A is 12 – 9, θ2 – θ1 is 12 -5.*0583

*What we have there is 12 – 9 is 3, 12 -5 is 7.*0592

*The probability that our number will be bigger than 9 is exactly 3/7.*0598

*Again, very easy computations for the uniform distribution, it is just a matter of saying how wide is your interval.*0604

*In this case, our interval is 7 units wide, the interval from 5 to 12, that is where we got that 7 in the denominator.*0612

*How wide is the region that you are interested in, the region that you might call success?*0621

*In this case, success is defined as Y is bigger than 9.*0626

*That region of success will be the interval from 9 to 12.*0631

*The width of that interval from 9 to 12 is just 12 - 9 is 3, that is where we got that 3.*0637

*Our total answer, our total probability there is just 3/7.*0643

*I guess you could convert that into a decimal, I think that would turn out to be about 42%.*0649

*But that does not come out neatly, I’m going to leave it as a fraction as 3/7 there.*0655

*Example 3 is a bit trickier, what is happening here is that you have a dinner date with your best friend.*0663

*You are planning to meet at 6 pm at the restaurant but the problem is that, both of you tend to run a little late.*0670

*In fact, even though you are planning to me at 6:00 pm, you might be a little bit late, your friend might be a little bit late.*0678

*One of you is probably going to end up waiting a little bit for the other one.*0687

*The way it works is you tend to arrive between 0 and 15 minutes late.*0690

*You are never later than 15 minutes and you are never early.*0696

*You are always maybe 7 minutes, 8 minutes, 11 minutes late, somewhere between 0 and 15.*0700

*Your friend is always between 0 and 10 minutes late.*0706

*Your friend is a little bit more prompt than you are.*0709

*The question we are asking here is the probability that you will arrive before your friend.*0713

*In other words, will you be the one who gets there first and has to find the table at dinner*0718

*and will be sitting around waiting for your friend, or will it be the other way around?*0724

*Your friend arrives first and has to deal with the waiter, and your friend would be waiting for you.*0728

*Let me show you how to solve this one.*0733

*We are going to set up two variables here because there are two independent things going on.*0736

*There is you arriving to the restaurant and there is your friend arriving to the restaurant.*0740

*I set up the variable for X which is your arrival time which could be anywhere from 0 to 15 minutes late.*0746

*Y is going to be your friends arrival time which could be anywhere from 0 to 10 minutes late.*0759

*A really useful way to think about this problem is to graph it.*0770

*Let me go ahead and draw a graph of the possibilities here.*0774

*I will put X on the X axis and Y on the Y axis.*0780

*Your arrival time can be anywhere from 0 to, there is you being 5 minutes late, here you are being 10 minutes late,*0785

*and here you are coming in 15 minutes late.*0794

*We know you are not going to be later than that.*0797

*There is your friend arriving 5 minutes late and here is your friend arriving 10 minutes late.*0800

*We know that your friend would not be later than 10 minutes.*0806

*What that means is your combined arrival time, the combination of arrival times*0809

*is going to be somewhere in this rectangle, depending on when you arrive and depending on when your friend arrives.*0817

*Somewhere, you will arrive a certain number of minutes late and your friend will arrive a certain number of minutes late.*0825

*That will give us a point somewhere in this rectangle.*0830

*Then, we will look at that and say did you arrive first or did your friend arrive first.*0834

*The way we want to think about that is we want to calculate the probability that you will arrive before your friend.*0841

*In other words, we want the probability that X is less than or equal to Y, that is you arriving before your friend.*0848

*Let me just turn those variables around because I think it will be easier to graph that way.*0859

*The probability that Y is greater than or equal to X, that is saying the same thing.*0864

*Let me graph the region in which Y is greater than X.*0869

*A little bit of algebra review here.*0873

*Maybe, I will do this in black. *0877

*The line Y equals X is that line right there.*0881

*That is the line Y is equal to X.*0885

*Y greater than X means you go above the lines.*0888

*It is all this triangular region above the line here.*0894

*That is the region where you arrive before your friend.*0900

*Anywhere below the line, that means your friend arrives first and you arrived afterwards.*0907

*Let us try to calculate that the probability of being in that black shaded region.*0915

*It is the total shaded area ÷ the total area in the rectangle.*0921

*Let us try to figure out what those areas are.*0932

*The shaded area, I see I have a triangle with 10 units on a side here.*0934

*That is base × height/2, that is 10 × 10/2, 100/2 is 50.*0942

*50 units in your shaded triangle.*0950

*The total area is a rectangle, it is 10 by 15 on the side, that is 10 × 15 is 150.*0955

*Here is very nice, it simplifies lovely to 1/3.*0963

*The probability that you will arrive before your friend is exactly 1/3.*0970

*If you make lots and lots of dates with the same friend, and you guys both follow the same habits over the years, *0977

*what will happen is 1/3 of the time you will be sitting around waiting for your friend.*0985

*2/3 of the time your friend will be sitting around waiting for you.*0990

*That is really the result of the fact that your friend is a little bit more prompt than you.*0995

*Most of the time your friend will end up waiting for you, at some of the time, 1/3 of the time, *1003

*you will wait for your friend.*1007

*Let me recap that.*1009

*The way we approach this is, we noticed first that we really had two independent uniform distributions.*1011

*There is one for your arrival time and there is one for your friend’s arrival time.*1017

*That is the first thing I did was to set up variables to indicate your arrival time and your friend’s arrival time.*1023

*Your arrival time, I put on the X axis that goes from 0 to 15 because you can be anywhere from 0 to 15 minutes late.*1030

*Your friend is anywhere from 0 to 10 minutes late.*1040

*I got those from the stem of the problem here.*1043

*I graph those here and I got this nice rectangle of possible combinations of arrival ×.*1048

*Once I have this rectangle, I know that while you arrive at a particular time, your friend will arrive at a particular time,*1057

*that means essentially we are choosing a point at random in this rectangle.*1063

*And then, we have to ask whether you will arrive before your friend?*1069

*You arriving before your friend means your arrival time is less than or equal to your friend’s arrival time,*1074

*which can be rewritten as Y greater than X.*1081

*We graph the shaded region, represents the region where Y is greater than or equal to X.*1084

*And then, it was just a matter of calculating the areas which turned into *1091

*a little old fashion geometry of calculating the area of a triangle.*1094

*The triangle was ½ base × height, ½ × 10 × 10.*1099

*The rectangle was base × height, that is 15 × 10.*1105

*We get 50/150 simplifies down to a probability of 1/3.*1113

*That represents the chance that you have arrived at the restaurant before your friend.*1119

*You will be the one who has to sit around and wait.*1125

*In example 4, we have a problem that is a great interest to computer programmers.*1132

*The reason is that most computer languages have a random number function.*1139

*It uses something rand or random, or something like that.*1146

*If you type rand into a computer program and the appropriate language, it will give you a number between 0 and 1.*1151

*We usually try to arrange it so that the random numbers are uniformly distributed between 0 and 1, *1162

*which is very good if you need a number between 0 and 1.*1168

*In lots of cases, when you need a random number in a computer program, *1172

*you need a random number on another range which might be from θ1 to θ2.*1177

*What this problem really does is, it shows you how to convert a uniform distribution on 0-1*1184

*into a uniform distribution onto θ1 and θ2.*1193

*That is the point of this problem, it is very useful for computer programmers*1197

*but that is not actually what we are doing here.*1203

*What we are doing is we are making a little transformation and we are starting with Y*1205

*being uniformly distributed on the interval from 0 to 1.*1210

*We are looking at another variable X which is defined to be, by the way that colon means define to be.*1215

*X is defined to be θ2 – tθ1 Y + θ1.*1230

*We want to show that that is a uniform distribution on the interval from θ1 to θ2.*1236

*To show that, let me first find the range of values for X.*1243

*Notice that, the range for Y is from 0 to 1, I have Y =0 to Y =1, if I plug those values into X, *1250

*if I plug Y =0 in to X, I get X = θ2 – θ1 × 0, that drops out.*1260

*I just get X = θ1.*1268

*If I plug Y = 1 into X, I get X = θ2- θ1 × 1 + θ1.*1270

*That simplifies down to θ2.*1282

*That tells me the range for X, X goes from θ1 to θ2.*1287

*That is hopeful, at least I know that X is distributed somehow on the range from θ1 to θ2.*1294

*But I want to really show that X is a uniform distribution.*1302

*I want to calculate the probability of the line between any 2 values.*1305

*Let me find that probability here.*1313

*The probability that is X is between any two values A and B, I can calculate that as the probability that,*1317

*X, just by definition is θ2 – θ1 × Y + θ1, that should be between A and B.*1330

*What I want to do is to solve this into a set of probabilities for Y.*1343

*First, subtract off θ1 from all 3 sides there.*1350

*The probability of A – θ1 being less than or equal to θ2 -θ1 × Y, less than or equal to B – θ1.*1354

*I’m trying to solve for Y, I'm trying to get Y by itself.*1368

*Next, I’m going to divide by θ2 – θ1.*1372

*This is the probability of A – θ1/θ2- θ1.*1377

*Less than or equal to just Y by itself now, less than or equal to B –θ1/θ2 – θ1.*1385

*I'm remembering that Y itself is a uniformly distributed random variable.*1397

*The probability that Y would be between any two bounds is just the difference between those two bounds.*1404

*You just subtract those two bounds.*1411

*I will just do B – θ1/θ2- θ1- A – θ1/θ2 – θ1.*1413

*I see now that I have a common denominator, θ2- θ1.*1425

*I got –θ1 in the first term and - and -, + θ1 in the second term.*1434

*Those θ1 cancel with each other and I just get down to B – A.*1441

*If you look at that, what I did was I started out with the probability that X is going to be between A and B.*1448

*What I came up with is, the probability is equal to exactly B - A ÷ θ2 – θ1.*1458

*That is exactly the formula for a uniform distribution.*1468

*X has a uniform distribution, X is uniformly distributed, distribution on the interval from θ1 to θ2.*1475

*I should start with θ1 and go to θ2.*1501

*X has a uniform distribution because the probability of X falling in any interval is exactly equal to the width of that interval B – A.*1507

*To recap what I did there, first, I looked at the range for Y, Y goes from 0 to 1.*1519

*Based on that, I calculated the range for X.*1526

*I plugged in those values of Y 0 and 1 into the formula for X here, and calculated the bounds for X being θ1 to θ2.*1529

*I know that X takes on the right range of values.*1541

*And then, I found the probability of any particular sub interval from A to B by converting X into terms of Y.*1545

*Solving out to isolate the variable Y, and then I use the fact that Y is uniformly distributed.*1556

*The probability of Y being between any two limits is just the width of those limits, the difference between those two limits.*1563

*B – θ1/θ2 –θ1- A – θ1/θ2 – θ1.*1572

*That simplify down to B –A /θ2 – θ1, which is exactly the formula for probability with a uniform distribution.*1580

*That tells me that X has a uniform distribution on the interval from θ1 to θ2.*1594

*That is very useful if you are computer programmer because that means *1601

*you can use the random number generator given by most computer programming languages.*1605

*And then, you can use this formula to convert it into a uniform distribution whatever range you want.*1613

*If you want for example, a random number between 80 and 100, and then you just plug in θ1 = 80 and θ2 =100.*1620

*You can use this formula to generate a random number between 80 and 100,*1631

*that will be uniformly distributed between 80 and 100.*1638

*In example 5, we are going leave the world of the random numbers behind.*1644

*We are going to look at the rough and tumble world of ice cream dispensary.*1649

*We have an ice cream machine which gives you servings of ice cream.*1653

*The servings vary a little bit, if you are unlucky the machine will be stingy with you, and it will give you just 206 ml of ice cream.*1659

*If it is a good day for you, if the machine is feeling generous, it will give you up to 238 ml.*1668

*Essentially, it picks a random amount in between 206 and 230, and they are uniformly distributed.*1676

*The question we are trying to answer is, if you go up there with your bowl and*1684

*you want to predict how much ice cream you will get, you want to describe what the expected amount of ice cream is in a serving,*1691

*and also the standard deviation in that quantity.*1698

*This is really asking, the expected value and mean are the same thing.*1705

*We are trying to calculate the expected value or the mean of this uniform distribution, and also the standard deviation.*1710

*I gave you formulas for those as few slides back, in a slide called key properties of the uniform distribution.*1717

*You can go back and look those up, I will remind you what they are here.*1728

*The mean which is always the same as the expected value, by definition those are the same thing.*1732

*For the uniform distribution is θ1 + tθ2 ÷ 2.*1738

*The θ1 and θ2 are the ranges of the endpoints of the interval.*1746

*In this case that is 206 + 230 ÷ 2, that is 436 ÷ 2 which is 218.*1752

*Our units here are ml, the average amount you expect to get when you fill up your bowl *1767

*at this ice cream machine is 218 ml of ice cream.*1775

*Of course that is not at all surprising, if you are going to get a random amount between 206 and 230, *1780

*it is not surprising that in the long run, you will get about halfway between 206 and 230 which is 218.*1789

*That is really not surprising at all.*1798

*The standard deviation, I also gave you on that slide, several slides ago.*1801

*It is θ2- θ1 ÷ 2 √s 3.*1805

*In this case, θ2, the big one is 230, Θ1 is 206, we want to divide that by 2 √3.*1812

*230 -206 is 24, the 200 part cancels, ÷ 2 and 3, that simplifies down to 12/3.*1823

*Since, 12 is 4 × 3, this just gives us 4 × √ 3 which I put that into a calculator,*1837

*it works out to be just a little bit less than 7 ml.*1849

*It is about 6.9 ml.*1854

*If you fill up your bowl at this ice cream machine, you expect on average to get about 218 ml.*1859

*The standard deviation on that will be 6.9 ml, about 7 ml + or - from 200 and 18.*1867

*To recap where these numbers came from.*1876

*The formulas for the mean and standard deviation, I give this to you on early slide in this talk.*1879

*It was called key properties of the uniform distribution.*1885

*They are fairly straightforward formulas.*1888

*In particular, the mean is what you would guess.*1889

*It is just the average of the upper and lower bounds, 206 + 230/2 gives you an average of 280 ml of ice cream per serving.*1893

*The standard deviation is probably not something you would guess but if you have a formula handy, it is θ2 – θ1/203.*1905

*I will just plug in the θ2 and θ1 into those values there.*1914

*And I simplified it down to 4 √3 and I have this decimal approximation that is about 6.9 or about 7 ml.*1920

*That is your standard deviation in an ice cream serving.*1928

*That is the last example and that wraps up this lecture on the uniform distribution.*1932

*The uniform distribution is just the first, it is the easiest of several continuous distributions.*1938

*We will be moving on from here and looking at the famous normal distribution, not the same as the uniform distribution, *1944

*and also the gamma distribution which includes the exponential distribution and chi square distribution.*1951

*Those are all coming up in the next few lectures here in the probabilities series on www.educator.com.*1958

*You are watching Will Murray with www.educator.com, and thank you very much for joining us, bye.*1965

1 answer

Last reply by: Dr. William Murray

Mon Mar 9, 2015 9:42 PM

Post by Ash Niazi on March 7, 2015

Love your lectures - they're really helping me understand the material.

Question, for Ex 3: I did it at first a bit differently:

My Arrival Time: P = 10 - 0 / 15 - 0 = 10 / 15 = 2/3.

Friend Arrival Time: P = 10-0 / 10- 0 = 1.

P[Friends Time] - P[My Time] = 1 - 2/3 = 1/3.

Is that acceptable?

1 answer

Last reply by: Dr. William Murray

Thu Mar 5, 2015 5:47 PM

Post by Nick Nick on March 4, 2015

Thanks