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And so if we want to take the mean of these two numbers, 11 plus 14 is 25. Halfway in between the two is 12.5. So 12.5 is exactly halfway between 11 and 14. And now we've figured out all of the information we need to actually plot or actually create or actually draw our box and whisker plot. So let me draw a number line. So my best attempt at a number line. So that's my number line. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
And now we've figured out all of the information we need to actually plot or actually create or actually draw our box and whisker plot. So let me draw a number line. So my best attempt at a number line. So that's my number line. And let's say that this right over here is 0. I need to make sure I get all the way up to 22 or beyond 22. So let's say that's 0. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
So that's my number line. And let's say that this right over here is 0. I need to make sure I get all the way up to 22 or beyond 22. So let's say that's 0. Let's say this is 5. This is 10. That could be 15. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
So let's say that's 0. Let's say this is 5. This is 10. That could be 15. And that could be 20. This could be 25. We could keep going. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
That could be 15. And that could be 20. This could be 25. We could keep going. 30, maybe 35. So the first thing we might want to think about, there's several ways to draw it. We want to think about the box part of the box and whisker. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
We could keep going. 30, maybe 35. So the first thing we might want to think about, there's several ways to draw it. We want to think about the box part of the box and whisker. It essentially represents the middle half of our data. So it's essentially trying to represent this data right over here. So the data between the medians of the two halves. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
We want to think about the box part of the box and whisker. It essentially represents the middle half of our data. So it's essentially trying to represent this data right over here. So the data between the medians of the two halves. So this is a part that we would attempt to represent with the box. So we would start right over here at this lower, this 2.5. This is essentially separating the first quartile from the second quartile, the first quarter of our numbers from the second quarter of our numbers. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
So the data between the medians of the two halves. So this is a part that we would attempt to represent with the box. So we would start right over here at this lower, this 2.5. This is essentially separating the first quartile from the second quartile, the first quarter of our numbers from the second quarter of our numbers. So let's put it right over here. This is 2.5. 2.5 is halfway between 0 and 5. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
This is essentially separating the first quartile from the second quartile, the first quarter of our numbers from the second quarter of our numbers. So let's put it right over here. This is 2.5. 2.5 is halfway between 0 and 5. So that's 2.5. And then up here we have 12.5. And 12.5 is right over, let's see, this is 10. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
2.5 is halfway between 0 and 5. So that's 2.5. And then up here we have 12.5. And 12.5 is right over, let's see, this is 10. So this right over here would be this halfway between 10 and 15 is 12.5. So let me do this. So this is 12.5 right over here. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
And 12.5 is right over, let's see, this is 10. So this right over here would be this halfway between 10 and 15 is 12.5. So let me do this. So this is 12.5 right over here. 12.5, so that separates the third quartile from the fourth quartile. And then our box is everything in between. So this is literally the middle half of our numbers. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
So this is 12.5 right over here. 12.5, so that separates the third quartile from the fourth quartile. And then our box is everything in between. So this is literally the middle half of our numbers. And we'd want to show where the actual median is. So that was actually one of the things that we wanted to be able to think about in our original, when the owner of the restaurant wanted to think about how far people are traveling from. So the median is 6. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
So this is literally the middle half of our numbers. And we'd want to show where the actual median is. So that was actually one of the things that we wanted to be able to think about in our original, when the owner of the restaurant wanted to think about how far people are traveling from. So the median is 6. So we can plot it right over here. So this right over here looks, this is about 6. So that is, let me do that same pink color. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
So the median is 6. So we can plot it right over here. So this right over here looks, this is about 6. So that is, let me do that same pink color. So this right over here is 6. And then the whiskers of the box and whisker plot essentially show us the range of our data. And so let me do that. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
So that is, let me do that same pink color. So this right over here is 6. And then the whiskers of the box and whisker plot essentially show us the range of our data. And so let me do that. I could do this in a different color that I haven't used yet. I'll do this in orange. So essentially if we want to see, look, the numbers go all the way up to 22. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
And so let me do that. I could do this in a different color that I haven't used yet. I'll do this in orange. So essentially if we want to see, look, the numbers go all the way up to 22. So they go all the way up to, so let's say that this is 22 right over here. Our numbers go all the way up to 22. Our numbers go all the way up to 22. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
So essentially if we want to see, look, the numbers go all the way up to 22. So they go all the way up to, so let's say that this is 22 right over here. Our numbers go all the way up to 22. Our numbers go all the way up to 22. And they go as low as 1. So they go, 1 is right about here. They go as low, let me label that, so that's 1. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
Our numbers go all the way up to 22. And they go as low as 1. So they go, 1 is right about here. They go as low, let me label that, so that's 1. And they go as low as 1. So there you have it. We have our box and whisker plot. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
They go as low, let me label that, so that's 1. And they go as low as 1. So there you have it. We have our box and whisker plot. And you can see, if you have a plot like this, just visually you can immediately see, OK, what is the median? It's the middle of the box, essentially. It shows you the middle half. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
We have our box and whisker plot. And you can see, if you have a plot like this, just visually you can immediately see, OK, what is the median? It's the middle of the box, essentially. It shows you the middle half. So it shows you how far they're spread, or kind of where the meat of the spread is. And then it shows, well, beyond that, we have the range that goes well beyond that, or how far the total spread of our data is. So this gives a pretty good sense of both the median and the spread of our data. | Constructing a box and whisker plot Probability and Statistics Khan Academy.mp3 |
If we only have a few experiments, it's very possible that our experimental probability could be different than our theoretical probability or even very different. But as we have many, many more experiments, thousands, millions, billions of experiments, the probability that the experimental and the theoretical probabilities are very different goes down dramatically. But let's get an intuitive sense for it. This right over here is a simulation created by Macmillan USA. I'll provide the link as an annotation. And what it does is it allows us to simulate many coin flips and figure out the proportion that are heads. So right over here, we can decide if we want our coin to be fair or not. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
This right over here is a simulation created by Macmillan USA. I'll provide the link as an annotation. And what it does is it allows us to simulate many coin flips and figure out the proportion that are heads. So right over here, we can decide if we want our coin to be fair or not. Right now it says that we have a 50% probability of getting heads. We can make it unfair by changing this, but I'll stick with the 50% probability. We want to show that on this graph here. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
So right over here, we can decide if we want our coin to be fair or not. Right now it says that we have a 50% probability of getting heads. We can make it unfair by changing this, but I'll stick with the 50% probability. We want to show that on this graph here. We can plot it. And what this says is at a time, how many tosses do we want to take? So let's say, let's just start with 10 tosses. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
We want to show that on this graph here. We can plot it. And what this says is at a time, how many tosses do we want to take? So let's say, let's just start with 10 tosses. So what this is going to do is take 10 simulated flips of coins with each one having a 50% chance of being heads. And then as we flip, we're gonna see our total proportion that are heads. So let's just talk through this together. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
So let's say, let's just start with 10 tosses. So what this is going to do is take 10 simulated flips of coins with each one having a 50% chance of being heads. And then as we flip, we're gonna see our total proportion that are heads. So let's just talk through this together. So starting to toss. And so what's going on here after 10 flips? So as you see, the first flip actually came out heads. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
So let's just talk through this together. So starting to toss. And so what's going on here after 10 flips? So as you see, the first flip actually came out heads. And if you wanted to say, what's your experimental probability after that one flip? You'd say, well, with only one experiment, I got one head. So it looks like 100% were heads. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
So as you see, the first flip actually came out heads. And if you wanted to say, what's your experimental probability after that one flip? You'd say, well, with only one experiment, I got one head. So it looks like 100% were heads. But in the second flip, it looks like it was a tails because now the proportion that was heads after two flips was 50%. But in the third flip, it looks like it was tails again because now only one out of three, or 33% of the flips have resulted in heads. Now by the fourth flip, we got a heads again, getting us back to 50th percentile. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
So it looks like 100% were heads. But in the second flip, it looks like it was a tails because now the proportion that was heads after two flips was 50%. But in the third flip, it looks like it was tails again because now only one out of three, or 33% of the flips have resulted in heads. Now by the fourth flip, we got a heads again, getting us back to 50th percentile. Now at the fifth flip, it looks like we got another heads. And so now we have three out of five, or 60% being heads. And so the general takeaway here is when you have one, two, three, four, five, or six experiments, it's completely plausible that your experimental proportion, your experimental probability, diverges from the real probability. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
Now by the fourth flip, we got a heads again, getting us back to 50th percentile. Now at the fifth flip, it looks like we got another heads. And so now we have three out of five, or 60% being heads. And so the general takeaway here is when you have one, two, three, four, five, or six experiments, it's completely plausible that your experimental proportion, your experimental probability, diverges from the real probability. And this even continues all the way until we get to our ninth or 10th tosses. But what happens if we do way more tosses? So now I'm gonna do another, well, let's just do another 200 tosses and see what happens. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
And so the general takeaway here is when you have one, two, three, four, five, or six experiments, it's completely plausible that your experimental proportion, your experimental probability, diverges from the real probability. And this even continues all the way until we get to our ninth or 10th tosses. But what happens if we do way more tosses? So now I'm gonna do another, well, let's just do another 200 tosses and see what happens. So I'm just gonna keep tossing here. And you can see, wow, look at this. There was a big run of getting a lot of heads right over here. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
So now I'm gonna do another, well, let's just do another 200 tosses and see what happens. So I'm just gonna keep tossing here. And you can see, wow, look at this. There was a big run of getting a lot of heads right over here. And then it looks like there's actually a run of getting a bunch of tails right over here, and then a little run of heads, tails, and then another run of heads. And notice, even after 215 tosses, our experimental probability is still reasonably different than our theoretical probability. So let's do another 200 and see if we can converge these over time. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
There was a big run of getting a lot of heads right over here. And then it looks like there's actually a run of getting a bunch of tails right over here, and then a little run of heads, tails, and then another run of heads. And notice, even after 215 tosses, our experimental probability is still reasonably different than our theoretical probability. So let's do another 200 and see if we can converge these over time. And what we're seeing in real time here should be the law of large numbers. As our number of tosses get larger and larger and larger, the probability that these two are very different goes down and down and down. Yes, you will get moments where you could even get 10 heads in a row or even 20 heads in a row, but over time, those will be balanced by the times where you're getting disproportionate number of tails. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
So let's do another 200 and see if we can converge these over time. And what we're seeing in real time here should be the law of large numbers. As our number of tosses get larger and larger and larger, the probability that these two are very different goes down and down and down. Yes, you will get moments where you could even get 10 heads in a row or even 20 heads in a row, but over time, those will be balanced by the times where you're getting disproportionate number of tails. So I'm just gonna keep going. We're now at almost 800 tosses, and you see now we are converging. Now this is, we're gonna cross 1,000 tosses soon. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
Yes, you will get moments where you could even get 10 heads in a row or even 20 heads in a row, but over time, those will be balanced by the times where you're getting disproportionate number of tails. So I'm just gonna keep going. We're now at almost 800 tosses, and you see now we are converging. Now this is, we're gonna cross 1,000 tosses soon. And you can see that our proportion here is now 51%. It's getting close now. We're at 50.6%, and I could just keep tossing. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
Now this is, we're gonna cross 1,000 tosses soon. And you can see that our proportion here is now 51%. It's getting close now. We're at 50.6%, and I could just keep tossing. This is 1,100. We're gonna approach 1,200 or 1,300 flips right over here. But as you can see, as we get many, many, many more flips, it was actually valuable to see even after 200 flips that there was a difference in the proportion between what we got from the experiment and what you would theoretically expect. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
We're at 50.6%, and I could just keep tossing. This is 1,100. We're gonna approach 1,200 or 1,300 flips right over here. But as you can see, as we get many, many, many more flips, it was actually valuable to see even after 200 flips that there was a difference in the proportion between what we got from the experiment and what you would theoretically expect. But as we get to many, many more flips, now we're at 1,210, we're getting pretty close to 50% of them turning out heads, but we could keep tossing it more and more and more. And what we'll see is as we get larger and larger and larger, it is likely that we're gonna get closer and closer and closer to 50%. It's not to say that it's impossible that we diverge again, but the likelihood of diverging gets lower and lower and lower the more tosses, the more experiments you make. | Experimental versus theoretical probability simulation Probability AP Statistics Khan Academy.mp3 |
So 30 people in a room. They're randomly selected. 30 people. And the question is, what is the probability that at least two people have the same birthday? And this is kind of a fun question because, I don't know, that's the size of a lot of classrooms. What's the probability that at least someone in the classroom shares a birthday with someone else in the classroom? And that's actually another good way to phrase it well. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And the question is, what is the probability that at least two people have the same birthday? And this is kind of a fun question because, I don't know, that's the size of a lot of classrooms. What's the probability that at least someone in the classroom shares a birthday with someone else in the classroom? And that's actually another good way to phrase it well. This is the same thing as saying, what is the probability that someone shares with at least someone else? They could share it with two other people or four other people on the birthday. Someone else. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And that's actually another good way to phrase it well. This is the same thing as saying, what is the probability that someone shares with at least someone else? They could share it with two other people or four other people on the birthday. Someone else. And at first this problem seems really hard because, wow, there's a lot of circumstances that make this true. I could have exactly two people have the same birthday. I could have exactly three people have the same birthday. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Someone else. And at first this problem seems really hard because, wow, there's a lot of circumstances that make this true. I could have exactly two people have the same birthday. I could have exactly three people have the same birthday. I could have exactly 29 people have the same birthday. All of these make this true. So do I add the probability of each of those circumstances and then add them up and then that becomes really hard and then I would have to say, OK, whose birthdays am I comparing and I would have to do combinations. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
I could have exactly three people have the same birthday. I could have exactly 29 people have the same birthday. All of these make this true. So do I add the probability of each of those circumstances and then add them up and then that becomes really hard and then I would have to say, OK, whose birthdays am I comparing and I would have to do combinations. It becomes a really difficult problem. Unless you make kind of one very simplifying, I would say, take on the problem. This is the opposite of, let me draw the probability space. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
So do I add the probability of each of those circumstances and then add them up and then that becomes really hard and then I would have to say, OK, whose birthdays am I comparing and I would have to do combinations. It becomes a really difficult problem. Unless you make kind of one very simplifying, I would say, take on the problem. This is the opposite of, let me draw the probability space. Let's say that this is all of the outcomes. Let me draw it with a thicker line. So let's say that's all of the outcomes in my probability space. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
This is the opposite of, let me draw the probability space. Let's say that this is all of the outcomes. Let me draw it with a thicker line. So let's say that's all of the outcomes in my probability space. So that's 100% of the outcomes. And we want to know, let me draw it in a color that won't be offensive to you. That doesn't look that great, but anyway. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
So let's say that's all of the outcomes in my probability space. So that's 100% of the outcomes. And we want to know, let me draw it in a color that won't be offensive to you. That doesn't look that great, but anyway. Let's say that this is the probability, this area right here, I don't know how big it really is, we'll figure it out, let's say that this is the probability that someone shares a birthday with at least someone else. What's this area over here? What's this green area? | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
That doesn't look that great, but anyway. Let's say that this is the probability, this area right here, I don't know how big it really is, we'll figure it out, let's say that this is the probability that someone shares a birthday with at least someone else. What's this area over here? What's this green area? Well, that means if these are all the cases where someone shares a birthday with someone else, these are all the areas where no one shares a birthday with anyone. Or you could say all 30 people have different birthdays. So we're trying to figure out, this is what we're trying to figure out, the probability, I'll just call it the probability that someone shares. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
What's this green area? Well, that means if these are all the cases where someone shares a birthday with someone else, these are all the areas where no one shares a birthday with anyone. Or you could say all 30 people have different birthdays. So we're trying to figure out, this is what we're trying to figure out, the probability, I'll just call it the probability that someone shares. I'll call it the probability of sharing, probability of s. If this whole area is area 1 or area 100%, this green area right here, this is going to be 1 minus p of s. Or if we said that this is the probability, or another way we could say it, actually this is the best way to think about it. If this is different, so this is the probability of different birthdays. This is the probability that all 30 people have 30 different birthdays, right? | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
So we're trying to figure out, this is what we're trying to figure out, the probability, I'll just call it the probability that someone shares. I'll call it the probability of sharing, probability of s. If this whole area is area 1 or area 100%, this green area right here, this is going to be 1 minus p of s. Or if we said that this is the probability, or another way we could say it, actually this is the best way to think about it. If this is different, so this is the probability of different birthdays. This is the probability that all 30 people have 30 different birthdays, right? No one shares with anyone. The probability that someone shares with someone else plus the probability that no one shares with anyone, they all have distinct birthdays, that's got to be equal to 1. Because we're either going to be in this situation or we're going to be in that situation. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
This is the probability that all 30 people have 30 different birthdays, right? No one shares with anyone. The probability that someone shares with someone else plus the probability that no one shares with anyone, they all have distinct birthdays, that's got to be equal to 1. Because we're either going to be in this situation or we're going to be in that situation. Or you could say they're equal to 100%, either way. 100% and 1 are the same number. It's equal to 100%. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Because we're either going to be in this situation or we're going to be in that situation. Or you could say they're equal to 100%, either way. 100% and 1 are the same number. It's equal to 100%. So if we figure out the probability that everyone has the same birthday, we could subtract it from 100. So let's see, if we could just rewrite this. The probability that someone shares a birthday with someone else, that's equal to 100% minus the probability that everyone has distinct separate birthdays. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
It's equal to 100%. So if we figure out the probability that everyone has the same birthday, we could subtract it from 100. So let's see, if we could just rewrite this. The probability that someone shares a birthday with someone else, that's equal to 100% minus the probability that everyone has distinct separate birthdays. And the reason why I'm doing that is because, as I started off in the video, this is kind of hard to figure out. I could figure out the probability that two people have the same birthday, five people, and it becomes very confusing. But here, if I want to just figure out the probability that everyone has a distinct birthday, it's actually a much easier probability to solve for. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
The probability that someone shares a birthday with someone else, that's equal to 100% minus the probability that everyone has distinct separate birthdays. And the reason why I'm doing that is because, as I started off in the video, this is kind of hard to figure out. I could figure out the probability that two people have the same birthday, five people, and it becomes very confusing. But here, if I want to just figure out the probability that everyone has a distinct birthday, it's actually a much easier probability to solve for. So what's the probability that everyone has a distinct birthday? So let's think about it. Person 1, just for simplicity, let's imagine the case that we only have two people in the room. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
But here, if I want to just figure out the probability that everyone has a distinct birthday, it's actually a much easier probability to solve for. So what's the probability that everyone has a distinct birthday? So let's think about it. Person 1, just for simplicity, let's imagine the case that we only have two people in the room. What's the probability that they have different birthdays? So if I have two, let's see, person 1, their birthday could be 365 days out of 365 days in the year, right? Whatever their birthday is. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Person 1, just for simplicity, let's imagine the case that we only have two people in the room. What's the probability that they have different birthdays? So if I have two, let's see, person 1, their birthday could be 365 days out of 365 days in the year, right? Whatever their birthday is. And then person 2, if we wanted to ensure that they don't have the same birthday, how many days could person 2 be born on? Well, it could be born on any day that person 1 was not born on. So there's 364 possibilities out of 365. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Whatever their birthday is. And then person 2, if we wanted to ensure that they don't have the same birthday, how many days could person 2 be born on? Well, it could be born on any day that person 1 was not born on. So there's 364 possibilities out of 365. So if you had two people, the probability that no one is born on the same birthday, this is just 1, is just going to be equal to 364 over 365, right? Now, what happens if we had three people? So first of all, the first person could be born on any day, then the second person could be born on 364 possible days out of 365, and then the third person, what's the probability that this third person isn't born on either of these people's birthdays? | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
So there's 364 possibilities out of 365. So if you had two people, the probability that no one is born on the same birthday, this is just 1, is just going to be equal to 364 over 365, right? Now, what happens if we had three people? So first of all, the first person could be born on any day, then the second person could be born on 364 possible days out of 365, and then the third person, what's the probability that this third person isn't born on either of these people's birthdays? So two days are taken up, so the probability is 363 over 365. So this is equal to, you multiply them out, you get 365 times 360. Actually, I should rewrite this one. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
So first of all, the first person could be born on any day, then the second person could be born on 364 possible days out of 365, and then the third person, what's the probability that this third person isn't born on either of these people's birthdays? So two days are taken up, so the probability is 363 over 365. So this is equal to, you multiply them out, you get 365 times 360. Actually, I should rewrite this one. Instead of saying this is 1, let me write this as, the numerator is 365 times 364 over 365 squared. Because I want you to see the pattern, right? Here the probability is 365 times 364 times 363 over 365 to the third power. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Actually, I should rewrite this one. Instead of saying this is 1, let me write this as, the numerator is 365 times 364 over 365 squared. Because I want you to see the pattern, right? Here the probability is 365 times 364 times 363 over 365 to the third power. And so in general, if you just kept doing this to 30, if I just kept this process for 30 people, so 30 people, the probability that no one shares the same birthday would be equal to 365 times 364 times 363. We'll have 30 terms up here, right? All the way down to what? | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Here the probability is 365 times 364 times 363 over 365 to the third power. And so in general, if you just kept doing this to 30, if I just kept this process for 30 people, so 30 people, the probability that no one shares the same birthday would be equal to 365 times 364 times 363. We'll have 30 terms up here, right? All the way down to what? 330, all the way down to 336, right? That'll actually be 30 terms divided by 365 to the 30th power, and you could just type this into your calculator right now, it'll take you a little time to type in 30 numbers. And you'll get the probability that no one shares the same birthday with anyone else. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
All the way down to what? 330, all the way down to 336, right? That'll actually be 30 terms divided by 365 to the 30th power, and you could just type this into your calculator right now, it'll take you a little time to type in 30 numbers. And you'll get the probability that no one shares the same birthday with anyone else. But before we do that, let me just show you something that might make it a little bit easier. Is there any way that I can mathematically express this with factorials, or that I could mathematically express this with factorials? Well, let's think about it. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And you'll get the probability that no one shares the same birthday with anyone else. But before we do that, let me just show you something that might make it a little bit easier. Is there any way that I can mathematically express this with factorials, or that I could mathematically express this with factorials? Well, let's think about it. 365 factorial is what? 365 factorial is equal to 365 times 364 times 363 times all the way down to 1, right? You just keep multiplying, it's a huge number. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Well, let's think about it. 365 factorial is what? 365 factorial is equal to 365 times 364 times 363 times all the way down to 1, right? You just keep multiplying, it's a huge number. Now, if I just want the 365 times the 364 in this case, I have to get rid of all of these numbers back here. So one thing I could do is I could divide this thing by all of these numbers. So 363 times 362 all the way down to 1. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
You just keep multiplying, it's a huge number. Now, if I just want the 365 times the 364 in this case, I have to get rid of all of these numbers back here. So one thing I could do is I could divide this thing by all of these numbers. So 363 times 362 all the way down to 1. So that's the same thing as dividing by 363 factorial. 365 factorial divided by 363 factorial is essentially this, because all of these terms cancel out. So this is equal to 365 factorial over 363 factorial over 365 squared. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
So 363 times 362 all the way down to 1. So that's the same thing as dividing by 363 factorial. 365 factorial divided by 363 factorial is essentially this, because all of these terms cancel out. So this is equal to 365 factorial over 363 factorial over 365 squared. And of course, for this case, it's almost silly to worry about the factorials, but it becomes useful once we have something larger than two terms up here. So by the same logic, this right here is going to be equal to 365 factorial over 362 factorial over 365 squared. And actually, just another interesting point. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
So this is equal to 365 factorial over 363 factorial over 365 squared. And of course, for this case, it's almost silly to worry about the factorials, but it becomes useful once we have something larger than two terms up here. So by the same logic, this right here is going to be equal to 365 factorial over 362 factorial over 365 squared. And actually, just another interesting point. How did we get this 365? Sorry, how did we get this 363 factorial? Well, 365 minus 2 is 363, right? | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And actually, just another interesting point. How did we get this 365? Sorry, how did we get this 363 factorial? Well, 365 minus 2 is 363, right? And that makes sense, because we only wanted two terms up here. So we wanted to divide by a factorial that's 2 less. And so we'd only get the highest two terms left. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Well, 365 minus 2 is 363, right? And that makes sense, because we only wanted two terms up here. So we wanted to divide by a factorial that's 2 less. And so we'd only get the highest two terms left. So this is also equal to, you could write this as 365 factorial divided by 365 minus 2 factorial. 365 minus 2 is 363 factorial. And then you just end up with those two terms, and that's that there. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And so we'd only get the highest two terms left. So this is also equal to, you could write this as 365 factorial divided by 365 minus 2 factorial. 365 minus 2 is 363 factorial. And then you just end up with those two terms, and that's that there. And then likewise, this right here, this numerator, you could rewrite as 365 factorial divided by 365 minus 3. And we had 3 people. Factorial. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And then you just end up with those two terms, and that's that there. And then likewise, this right here, this numerator, you could rewrite as 365 factorial divided by 365 minus 3. And we had 3 people. Factorial. And that should hopefully make sense, right? This is the same thing as 365 factorial. Well, 365 divided by 3 is 362 factorial. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Factorial. And that should hopefully make sense, right? This is the same thing as 365 factorial. Well, 365 divided by 3 is 362 factorial. And so that's equal to 365 times 364 times 363, all the way down, divided by 362 times all the way down. And that'll cancel out with everything else, and you'd be just left with that. And that's that right there. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Well, 365 divided by 3 is 362 factorial. And so that's equal to 365 times 364 times 363, all the way down, divided by 362 times all the way down. And that'll cancel out with everything else, and you'd be just left with that. And that's that right there. So by that same logic, this top part here can be written as 365 factorial over what? 365 minus 30 factorial. And I did all of that just so I could show you kind of the pattern. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And that's that right there. So by that same logic, this top part here can be written as 365 factorial over what? 365 minus 30 factorial. And I did all of that just so I could show you kind of the pattern. And because this is, frankly, easier to type into a calculator if you know where the factorial button is. So let's figure out what this entire probability is. So turning on the calculator, we want, so let's do the numerator, 365 factorial divided by, what was 365 minus 30, that's 335, right? | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And I did all of that just so I could show you kind of the pattern. And because this is, frankly, easier to type into a calculator if you know where the factorial button is. So let's figure out what this entire probability is. So turning on the calculator, we want, so let's do the numerator, 365 factorial divided by, what was 365 minus 30, that's 335, right? Divided by 335 factorial. And that's that whole numerator. And now we want to divide the numerator, divided by 365 to the 30th power. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
So turning on the calculator, we want, so let's do the numerator, 365 factorial divided by, what was 365 minus 30, that's 335, right? Divided by 335 factorial. And that's that whole numerator. And now we want to divide the numerator, divided by 365 to the 30th power. Let the calculator think, and we get 0.2936. Equals 0.2936, it keeps, actually, 37 if you round it. Which is equal to 29.37%. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And now we want to divide the numerator, divided by 365 to the 30th power. Let the calculator think, and we get 0.2936. Equals 0.2936, it keeps, actually, 37 if you round it. Which is equal to 29.37%. Now, just so you remember what we were doing all along, this was the probability that no one shares a birthday with anyone. This was the probability of everyone having distinct different birthdays from everyone else. And we've said, well, the probability that someone shares a birthday with someone else, or maybe more than one person, is equal to all of the possibilities, kind of the 100%, the probability space, minus the probability that no one shares a birthday with anybody. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Which is equal to 29.37%. Now, just so you remember what we were doing all along, this was the probability that no one shares a birthday with anyone. This was the probability of everyone having distinct different birthdays from everyone else. And we've said, well, the probability that someone shares a birthday with someone else, or maybe more than one person, is equal to all of the possibilities, kind of the 100%, the probability space, minus the probability that no one shares a birthday with anybody. So that's equal to 100% minus 29.37%. Or another way you could write it is, that's 1 minus 0.2937. Which is equal to, so if I want to subtract that from 1, 1 minus, that just means the answer, that means 1 minus 0.29. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
And we've said, well, the probability that someone shares a birthday with someone else, or maybe more than one person, is equal to all of the possibilities, kind of the 100%, the probability space, minus the probability that no one shares a birthday with anybody. So that's equal to 100% minus 29.37%. Or another way you could write it is, that's 1 minus 0.2937. Which is equal to, so if I want to subtract that from 1, 1 minus, that just means the answer, that means 1 minus 0.29. You get 0.7063. So the probability that someone shares a birthday with someone else is 0.7063, it keeps going, which is approximately equal to 70.6%. Which is kind of a neat result. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Which is equal to, so if I want to subtract that from 1, 1 minus, that just means the answer, that means 1 minus 0.29. You get 0.7063. So the probability that someone shares a birthday with someone else is 0.7063, it keeps going, which is approximately equal to 70.6%. Which is kind of a neat result. Because if you have 30 people in a room, you might say, oh wow, what are the odds that someone has the same birthday as someone else? It's actually pretty high. Most, 70% of the time, if you have a group of 30 people, at least one person shares a birthday with at least one other person in the room. | Birthday probability problem Probability and Statistics Khan Academy.mp3 |
Based on the data below, which student score improved the most between the midterm and final exams? So they give us this data in terms of a bar graph for each student. We actually have two bars that show the midterm in blue and then the final exam in red. And they tell us that here, midterm in blue and final exam. Sometimes this is called a two-column bar graph because for each student here you have two columns of data. So if you were to actually look at the data itself, you have the midterm data and then you have the final exam data. Now, they're asking us which student score improved the most between the midterms and the final exams. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
And they tell us that here, midterm in blue and final exam. Sometimes this is called a two-column bar graph because for each student here you have two columns of data. So if you were to actually look at the data itself, you have the midterm data and then you have the final exam data. Now, they're asking us which student score improved the most between the midterms and the final exams. So if we look at Jasmine right over here, might as well start with her since she's the most to the left. It looks like she definitely did improve from the midterm to the final. It looks like in the midterm, if I had to guess, this looks like about a, I don't know, it looks like she got maybe about a 72 or 73 on the midterm. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
Now, they're asking us which student score improved the most between the midterms and the final exams. So if we look at Jasmine right over here, might as well start with her since she's the most to the left. It looks like she definitely did improve from the midterm to the final. It looks like in the midterm, if I had to guess, this looks like about a, I don't know, it looks like she got maybe about a 72 or 73 on the midterm. I'm just guessing because I don't know the exact number. And it looks like on the final she got, I don't know, that looks like maybe a 77 or 78 approximately. So she improved a little bit, about five points from the midterm to the final. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
It looks like in the midterm, if I had to guess, this looks like about a, I don't know, it looks like she got maybe about a 72 or 73 on the midterm. I'm just guessing because I don't know the exact number. And it looks like on the final she got, I don't know, that looks like maybe a 77 or 78 approximately. So she improved a little bit, about five points from the midterm to the final. The way that they've given us this information, we don't know the exact numbers because it's not super precise in terms of marking off the bars. But hopefully it will become obvious when we look through everyone's scores. Let's look at Jeff. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
So she improved a little bit, about five points from the midterm to the final. The way that they've given us this information, we don't know the exact numbers because it's not super precise in terms of marking off the bars. But hopefully it will become obvious when we look through everyone's scores. Let's look at Jeff. So Jeff actually did worse from the midterm to the final exam. He got, looks like over an 85 on the midterm and then on the final he got about an 84 or 85. So he actually went down. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
Let's look at Jeff. So Jeff actually did worse from the midterm to the final exam. He got, looks like over an 85 on the midterm and then on the final he got about an 84 or 85. So he actually went down. So it's definitely not Jeff. He definitely did not improve the most. He actually went down. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
So he actually went down. So it's definitely not Jeff. He definitely did not improve the most. He actually went down. Next, if you think about Nevin, it looks like Nevin actually improved about the same amount as Jasmine. On both scores, I don't know if Nevin is a boy or a girl's name, this person on both tests, this person did better than Jasmine. But it looks like the actual improvement is about the same. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
He actually went down. Next, if you think about Nevin, it looks like Nevin actually improved about the same amount as Jasmine. On both scores, I don't know if Nevin is a boy or a girl's name, this person on both tests, this person did better than Jasmine. But it looks like the actual improvement is about the same. It looks like they went from about an 83 to, I don't know, about an 88. I'm just estimating it, just trying to look at this axis right over there and estimating what those scores are. So Nevin and Jasmine right now, based on the three we've seen, are tied for the lead. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
But it looks like the actual improvement is about the same. It looks like they went from about an 83 to, I don't know, about an 88. I'm just estimating it, just trying to look at this axis right over there and estimating what those scores are. So Nevin and Jasmine right now, based on the three we've seen, are tied for the lead. Now let's look at Alejandra. Now Alejandra, this is, okay, so this jumps out. She definitely improved dramatically from the midterm to the final exam. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
So Nevin and Jasmine right now, based on the three we've seen, are tied for the lead. Now let's look at Alejandra. Now Alejandra, this is, okay, so this jumps out. She definitely improved dramatically from the midterm to the final exam. It looks like on the midterm she maybe got an 81 or an 82. So maybe this was an 82 she got on the midterm and then on the final it looks like she got about a 95. A 95 on the final. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
She definitely improved dramatically from the midterm to the final exam. It looks like on the midterm she maybe got an 81 or an 82. So maybe this was an 82 she got on the midterm and then on the final it looks like she got about a 95. A 95 on the final. So it was a dramatic improvement. So right now Alejandra is the leading contender for most improved from the midterm to the final. And finally Marta right over here, it looks like she actually got worse from the midterm to the final. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
A 95 on the final. So it was a dramatic improvement. So right now Alejandra is the leading contender for most improved from the midterm to the final. And finally Marta right over here, it looks like she actually got worse from the midterm to the final. She scored in the mid-90s in the midterm and then low 90s in the final. So she's definitely not the most improved. So the winner here is Alejandra. | Reading bar graphs Applying mathematical reasoning Pre-Algebra Khan Academy.mp3 |
Amelia only likes to use the drive-through for restaurants where the average wait time is in the bottom 10% for that town. What is the maximum average wait time for restaurants where Amelia likes to use the drive-through? Round to the nearest whole second. Like always, if you feel like you can tackle this, pause this video and try to do so. I'm assuming you paused it. Now let's work through this together. So let's think about what's going on. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
Like always, if you feel like you can tackle this, pause this video and try to do so. I'm assuming you paused it. Now let's work through this together. So let's think about what's going on. They're telling us that the distribution of average wait times is approximately normal. So let's get a visualization of a normal distribution. And they tell us several things about this normal distribution. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
So let's think about what's going on. They're telling us that the distribution of average wait times is approximately normal. So let's get a visualization of a normal distribution. And they tell us several things about this normal distribution. They tell us that the mean is 185 seconds. So that's 185 there. The standard deviation is 11 seconds. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
And they tell us several things about this normal distribution. They tell us that the mean is 185 seconds. So that's 185 there. The standard deviation is 11 seconds. So for example, this is going to be 11 more than the mean. So this would be 196 seconds. This would be another 11. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
The standard deviation is 11 seconds. So for example, this is going to be 11 more than the mean. So this would be 196 seconds. This would be another 11. Each of these dotted lines are one standard deviation more. So this would be 207. This would be 11 seconds less than the mean. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
This would be another 11. Each of these dotted lines are one standard deviation more. So this would be 207. This would be 11 seconds less than the mean. So this would be 174, and so on and so forth. And we wanna find the maximum average wait time for restaurants where Amelia likes to use the drive-through. Well, what are those restaurants? | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
This would be 11 seconds less than the mean. So this would be 174, and so on and so forth. And we wanna find the maximum average wait time for restaurants where Amelia likes to use the drive-through. Well, what are those restaurants? That's where the average wait time is in the bottom 10% for that town. So how do we think about it? Well, there's going to be some value. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
Well, what are those restaurants? That's where the average wait time is in the bottom 10% for that town. So how do we think about it? Well, there's going to be some value. Let me mark it off right over here in this red color. So we're gonna have some threshold value right over here where this is anything that level or lower is going to be in the bottom 10%. Well, another way to think about it is this is the largest wait time for which you are still in the bottom 10%. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
Well, there's going to be some value. Let me mark it off right over here in this red color. So we're gonna have some threshold value right over here where this is anything that level or lower is going to be in the bottom 10%. Well, another way to think about it is this is the largest wait time for which you are still in the bottom 10%. And so this area right over here is going to be 10% of the total, or it's going to be 0.10. So the way we can tackle this is we can get up a z-table and figure out what z-score gives us a proportion of only 0.10 being less than that z-score. And then using that z-score, we can figure out this value, the actual wait time. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
Well, another way to think about it is this is the largest wait time for which you are still in the bottom 10%. And so this area right over here is going to be 10% of the total, or it's going to be 0.10. So the way we can tackle this is we can get up a z-table and figure out what z-score gives us a proportion of only 0.10 being less than that z-score. And then using that z-score, we can figure out this value, the actual wait time. So let's get our z-table out. And since we know that this is below the mean, the mean would be the 50th percentile, we know we're gonna have a negative z-score. So I'm gonna take out the part of the table that has the negative z-scores on it. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
And then using that z-score, we can figure out this value, the actual wait time. So let's get our z-table out. And since we know that this is below the mean, the mean would be the 50th percentile, we know we're gonna have a negative z-score. So I'm gonna take out the part of the table that has the negative z-scores on it. And remember, we're looking for 10%, but we don't wanna go beyond 10%. We wanna be sure that that value is within the 10th percentile, that any higher will be out of the 10th percentile. So let's see, when we have these really negative z's, so far it only doesn't even get to the first percentile yet. | Threshold for low percentile Modeling data distributions AP Statistics Khan Academy.mp3 |
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