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16 possible outcomes when you flip a coin 4 times. 16 possible outcomes. And any one of the possible outcomes would be 1 of 16. So if I wanted to say, so if I were to just say the probability, and I'm just going to not talk about this one head. If I just take a, just maybe this thing that has three heads right here. This exact sequence of events, this is the first flip, second flip, third flip, fourth flip. Getting exactly this, this is exactly 1 out of a possible of 16 events. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
So if I wanted to say, so if I were to just say the probability, and I'm just going to not talk about this one head. If I just take a, just maybe this thing that has three heads right here. This exact sequence of events, this is the first flip, second flip, third flip, fourth flip. Getting exactly this, this is exactly 1 out of a possible of 16 events. Now with that out of the way, let's think about how many possibilities, how many of those 16 possibilities involve getting exactly 1 heads. Well, we could list them. You could get your heads. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
Getting exactly this, this is exactly 1 out of a possible of 16 events. Now with that out of the way, let's think about how many possibilities, how many of those 16 possibilities involve getting exactly 1 heads. Well, we could list them. You could get your heads. So this is equal to the probability of getting the heads in the first flip plus the probability of getting the heads in the second flip plus the probability of getting the heads in the third flip. Remember, exactly 1 heads. We're not saying at least 1, exactly 1 heads. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
You could get your heads. So this is equal to the probability of getting the heads in the first flip plus the probability of getting the heads in the second flip plus the probability of getting the heads in the third flip. Remember, exactly 1 heads. We're not saying at least 1, exactly 1 heads. So probability in the third flip. And then, or the possibility that you get heads in the fourth flip. Tails, heads, and tails. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
We're not saying at least 1, exactly 1 heads. So probability in the third flip. And then, or the possibility that you get heads in the fourth flip. Tails, heads, and tails. And we know already what the probability of each of these things are. There are 16 possible events. And each of these are one of those 16 possible events. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
Tails, heads, and tails. And we know already what the probability of each of these things are. There are 16 possible events. And each of these are one of those 16 possible events. So this is going to be 1 over 16. And so we're really saying the probability of getting exactly 1 heads is the same thing as the probability of getting heads in the first flip or the probability of getting heads in the first flip or heads in the second flip or heads in the third flip or heads in the fourth flip. And we can add the probabilities of these different things because they are mutually exclusive. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And each of these are one of those 16 possible events. So this is going to be 1 over 16. And so we're really saying the probability of getting exactly 1 heads is the same thing as the probability of getting heads in the first flip or the probability of getting heads in the first flip or heads in the second flip or heads in the third flip or heads in the fourth flip. And we can add the probabilities of these different things because they are mutually exclusive. Any two of these things cannot happen at the same time. You have to pick one of these scenarios. And so we can add the probabilities. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And we can add the probabilities of these different things because they are mutually exclusive. Any two of these things cannot happen at the same time. You have to pick one of these scenarios. And so we can add the probabilities. 1 16th plus 1 16th plus 1 16th plus 1 16th. Did I say that four times? Well, assume that I did. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And so we can add the probabilities. 1 16th plus 1 16th plus 1 16th plus 1 16th. Did I say that four times? Well, assume that I did. And so you would get 4 16ths, which is equal to 1 4th. Fair enough. Now let's ask a slightly more interesting question. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
Well, assume that I did. And so you would get 4 16ths, which is equal to 1 4th. Fair enough. Now let's ask a slightly more interesting question. Let's ask ourselves the probability of getting exactly 2 heads. And there's a couple of ways we can think about it. One is just in the traditional way. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
Now let's ask a slightly more interesting question. Let's ask ourselves the probability of getting exactly 2 heads. And there's a couple of ways we can think about it. One is just in the traditional way. And let's just look for the number of possibilities and of those equally likely possibilities. And we can only use this methodology because it's a fair coin. So how many of the total possibilities have 2 heads of the total of equally likely possibilities? | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
One is just in the traditional way. And let's just look for the number of possibilities and of those equally likely possibilities. And we can only use this methodology because it's a fair coin. So how many of the total possibilities have 2 heads of the total of equally likely possibilities? So we know there are 16 equally likely possibilities. How many of those have 2 heads? So I'm actually ahead of time. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
So how many of the total possibilities have 2 heads of the total of equally likely possibilities? So we know there are 16 equally likely possibilities. How many of those have 2 heads? So I'm actually ahead of time. So we save time. I've drawn all of the 16 equally likely possibilities. And how many of these involve 2 heads? | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
So I'm actually ahead of time. So we save time. I've drawn all of the 16 equally likely possibilities. And how many of these involve 2 heads? Well, let's see. This one over here has 2 heads. This one over here has 2 heads. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And how many of these involve 2 heads? Well, let's see. This one over here has 2 heads. This one over here has 2 heads. This one over here has 2 heads. Let's see. That's this one over here has 2 heads. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
This one over here has 2 heads. This one over here has 2 heads. Let's see. That's this one over here has 2 heads. And this one over here has 2 heads. And then this one over here has 2 heads. And I believe we are done after that. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
That's this one over here has 2 heads. And this one over here has 2 heads. And then this one over here has 2 heads. And I believe we are done after that. So if we count them, 1, 2, 3, 4, 5, 6 of the possibilities have exactly 2 heads. So 6 of the 16 equally likely possibilities have 2 heads. So we have a 3 8th chance of getting exactly 2 heads. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And I believe we are done after that. So if we count them, 1, 2, 3, 4, 5, 6 of the possibilities have exactly 2 heads. So 6 of the 16 equally likely possibilities have 2 heads. So we have a 3 8th chance of getting exactly 2 heads. Now, that's kind of what we've been doing in the past. But what I want to do is think about a way so we wouldn't have to write out all the possibilities. And the reason why that's useful is we're only dealing with 4 flips now. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
So we have a 3 8th chance of getting exactly 2 heads. Now, that's kind of what we've been doing in the past. But what I want to do is think about a way so we wouldn't have to write out all the possibilities. And the reason why that's useful is we're only dealing with 4 flips now. But if we were dealing with 10 flips, there's no way that we could write out all the possibilities like this. So I really want a different way of thinking about it. And the different way of thinking about it is if we're saying exactly 2 heads, you can imagine we're having the 4 flips. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And the reason why that's useful is we're only dealing with 4 flips now. But if we were dealing with 10 flips, there's no way that we could write out all the possibilities like this. So I really want a different way of thinking about it. And the different way of thinking about it is if we're saying exactly 2 heads, you can imagine we're having the 4 flips. Flip 1, flip 2, flip 3, flip 4. So these are the flips, or you could say the outcome of the flips. And if you're going to have exactly 2 heads, you could say, well, look, I'm going to have 1 head in one of these positions and then 1 head in the other position. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And the different way of thinking about it is if we're saying exactly 2 heads, you can imagine we're having the 4 flips. Flip 1, flip 2, flip 3, flip 4. So these are the flips, or you could say the outcome of the flips. And if you're going to have exactly 2 heads, you could say, well, look, I'm going to have 1 head in one of these positions and then 1 head in the other position. So if I'm picking the first, so you could say, and I have kind of a heads 1 and I have a heads 2. And I don't want you to think that these are somehow the heads in the first flip or the heads in the second flip. What I'm saying is we need 2 heads. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And if you're going to have exactly 2 heads, you could say, well, look, I'm going to have 1 head in one of these positions and then 1 head in the other position. So if I'm picking the first, so you could say, and I have kind of a heads 1 and I have a heads 2. And I don't want you to think that these are somehow the heads in the first flip or the heads in the second flip. What I'm saying is we need 2 heads. We need a total of 2 heads in all of our flips. And I'm just giving one of the heads a name, and I'm giving the other head a name. And what we're going to see in a few seconds is that we actually don't want to double count. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
What I'm saying is we need 2 heads. We need a total of 2 heads in all of our flips. And I'm just giving one of the heads a name, and I'm giving the other head a name. And what we're going to see in a few seconds is that we actually don't want to double count. We don't want to count the situation. We don't want to double count this situation. Heads 1, heads 2, tails, tails, and heads 2, heads 1, tails, tails. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And what we're going to see in a few seconds is that we actually don't want to double count. We don't want to count the situation. We don't want to double count this situation. Heads 1, heads 2, tails, tails, and heads 2, heads 1, tails, tails. For our purposes, these are the exact same outcomes. So we don't want to double count that, and we're going to have to account for that. But if we just think about it generally, how many different spots, how many of different flips can that first head show up in? | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
Heads 1, heads 2, tails, tails, and heads 2, heads 1, tails, tails. For our purposes, these are the exact same outcomes. So we don't want to double count that, and we're going to have to account for that. But if we just think about it generally, how many different spots, how many of different flips can that first head show up in? Well, there's 4 different flips that that first head could show up in. So there's 4 possibilities, 4 flips, or 4 places that it could show up in. Well, if that first head takes up one of these 4 places, let's just say that first head shows up on the third flip, then how many different places can that second head show up in? | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
But if we just think about it generally, how many different spots, how many of different flips can that first head show up in? Well, there's 4 different flips that that first head could show up in. So there's 4 possibilities, 4 flips, or 4 places that it could show up in. Well, if that first head takes up one of these 4 places, let's just say that first head shows up on the third flip, then how many different places can that second head show up in? Well, if that first head is in one of the 4 places, then that second head can only be in 3 different places. So that second head can only be in 3 different places. And so it could be in any one of these. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
Well, if that first head takes up one of these 4 places, let's just say that first head shows up on the third flip, then how many different places can that second head show up in? Well, if that first head is in one of the 4 places, then that second head can only be in 3 different places. So that second head can only be in 3 different places. And so it could be in any one of these. It could maybe be right over there, any one of those 3 places. And so when you think about it in terms of the first, and I don't want to say the first head, head 1. Actually, let me call it this way. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And so it could be in any one of these. It could maybe be right over there, any one of those 3 places. And so when you think about it in terms of the first, and I don't want to say the first head, head 1. Actually, let me call it this way. Head A and head B. That way you won't think that I'm talking about the first flip or the second flip. So this is head A, and this right over there is head B. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
Actually, let me call it this way. Head A and head B. That way you won't think that I'm talking about the first flip or the second flip. So this is head A, and this right over there is head B. So if you had a particular, I mean, these heads are identical. These outcomes aren't different. But the way we talk about it right now, it looks like there's 4 places that we could get this head in, and there's 3 places where we could get this head in. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
So this is head A, and this right over there is head B. So if you had a particular, I mean, these heads are identical. These outcomes aren't different. But the way we talk about it right now, it looks like there's 4 places that we could get this head in, and there's 3 places where we could get this head in. And so if you multiply all of the different ways that you could get all of the different scenarios where this is in 4 different places, and then this is in one of the 3 leftover places, you get 12 different scenarios. But there would only be 12 different scenarios if you viewed this as being different than this. And let me rewrite it with our new. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
But the way we talk about it right now, it looks like there's 4 places that we could get this head in, and there's 3 places where we could get this head in. And so if you multiply all of the different ways that you could get all of the different scenarios where this is in 4 different places, and then this is in one of the 3 leftover places, you get 12 different scenarios. But there would only be 12 different scenarios if you viewed this as being different than this. And let me rewrite it with our new. So this is head A, this is head B, this is head B, this is head A. There would only be 12 different scenarios if you viewed these 2 things as fundamentally different. But we don't. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
And let me rewrite it with our new. So this is head A, this is head B, this is head B, this is head A. There would only be 12 different scenarios if you viewed these 2 things as fundamentally different. But we don't. We're actually double counting, because we can always swap these 2 heads and have the exact same outcome. So what you want to do is actually divide it by 2. So you want to divide it by all of the different ways that you can swap 2 different things. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
But we don't. We're actually double counting, because we can always swap these 2 heads and have the exact same outcome. So what you want to do is actually divide it by 2. So you want to divide it by all of the different ways that you can swap 2 different things. If we had 3 heads here, you would think about all of the different ways that you could swap 3 different things. If we had 4 heads here, it would be all of the different ways you could swap 4 different things. So there's 12 different scenarios if you couldn't swap them, but you want to divide it by all of the different ways that you can swap 2 things. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
So you want to divide it by all of the different ways that you can swap 2 different things. If we had 3 heads here, you would think about all of the different ways that you could swap 3 different things. If we had 4 heads here, it would be all of the different ways you could swap 4 different things. So there's 12 different scenarios if you couldn't swap them, but you want to divide it by all of the different ways that you can swap 2 things. So 12 divided by 2 is equal to 6 fundamentally different scenarios, considering that you can swap them. If you assume that head A and head B can be interchangeable, that it's a completely identical outcome for us, because they're really just heads. So there's 6 different scenarios, and we know that there's a total of 16 equally likely scenarios. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
So there's 12 different scenarios if you couldn't swap them, but you want to divide it by all of the different ways that you can swap 2 things. So 12 divided by 2 is equal to 6 fundamentally different scenarios, considering that you can swap them. If you assume that head A and head B can be interchangeable, that it's a completely identical outcome for us, because they're really just heads. So there's 6 different scenarios, and we know that there's a total of 16 equally likely scenarios. So we could say that the probability of getting exactly 2 heads is 6 times 6 scenarios. And there's a couple of ways. You could say there's 6 scenarios that give us 2 heads of a possible 16, or you could say there are 6 possible scenarios, and the probability of each of those scenarios is 1 16th. | Getting exactly two heads (combinatorics) Probability and Statistics Khan Academy.mp3 |
The results are shown in the two plots below. And so the first statement that we have to complete is the mean number of fruits is greater for, and actually let me go to the actual screen, is greater for, we have to pick between freshmen and seniors and then they say the mean is a good measure for the center of distribution of, and we pick either freshmen or seniors. So let me go back to my scratch pad here and let's think about this. So let's first think about the first part. So let's just calculate the mean for each of these distributions. And I encourage you to pause the video and try to calculate it out on your own. So let's first think about the mean number of fruit for freshmen. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
So let's first think about the first part. So let's just calculate the mean for each of these distributions. And I encourage you to pause the video and try to calculate it out on your own. So let's first think about the mean number of fruit for freshmen. So essentially we're just gonna take each of these data points, add them all together, and then divide by the number of data points that we have. So we have one data point at zero. So we have one data point at zero, so I'll write zero. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
So let's first think about the mean number of fruit for freshmen. So essentially we're just gonna take each of these data points, add them all together, and then divide by the number of data points that we have. So we have one data point at zero. So we have one data point at zero, so I'll write zero. And then we have two data points at one. So we could say plus two times one. And then we have two data points at two. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
So we have one data point at zero, so I'll write zero. And then we have two data points at one. So we could say plus two times one. And then we have two data points at two. So we write plus two times two. And then let's see, we have a bunch of data. We have four data points at three. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
And then we have two data points at two. So we write plus two times two. And then let's see, we have a bunch of data. We have four data points at three. So we could say we have four threes. So let me circle that. So we have four threes plus four times three. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
We have four data points at three. So we could say we have four threes. So let me circle that. So we have four threes plus four times three. And then we have three fours, so plus three times four. And then we have a five, so plus five. And then we have a six. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
So we have four threes plus four times three. And then we have three fours, so plus three times four. And then we have a five, so plus five. And then we have a six. Let me do this in a color that you can see. And then we have a six right over here, plus six. And how many total points did we have? | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
And then we have a six. Let me do this in a color that you can see. And then we have a six right over here, plus six. And how many total points did we have? Well, we had one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14. Oh, actually, be careful, we had 15 points. And I didn't put that one in there. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
And how many total points did we have? Well, we had one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14. Oh, actually, be careful, we had 15 points. And I didn't put that one in there. So actually, let me just. So we have 15 points. And I can't forget this one over here. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
And I didn't put that one in there. So actually, let me just. So we have 15 points. And I can't forget this one over here. So plus, my pen is acting a little funny right now, but we'll power through that. Plus 19. Plus 19. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
And I can't forget this one over here. So plus, my pen is acting a little funny right now, but we'll power through that. Plus 19. Plus 19. So what is this going to be? So this is just going to be zero. This is going to be two. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
Plus 19. So what is this going to be? So this is just going to be zero. This is going to be two. This is going to be four. This is going to be 12. My pen is really acting up. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
This is going to be two. This is going to be four. This is going to be 12. My pen is really acting up. It's almost like it's running out of digital ink or something. And this is going to be another 12. And then we have five, six, and 19. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
My pen is really acting up. It's almost like it's running out of digital ink or something. And this is going to be another 12. And then we have five, six, and 19. So what is this going to be? Two plus four is six. Plus 24 is 30. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
And then we have five, six, and 19. So what is this going to be? Two plus four is six. Plus 24 is 30. Plus 11 is 41. Plus 19 gets us to 60. 60 divided by 15 is four. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
Plus 24 is 30. Plus 11 is 41. Plus 19 gets us to 60. 60 divided by 15 is four. So the mean number of fruit per day for the freshman is four pieces of fruit per day. So this right over here, that right over there, is our mean for the, let me put that in a color that you can actually see. Now let's do the same calculation for the seniors. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
60 divided by 15 is four. So the mean number of fruit per day for the freshman is four pieces of fruit per day. So this right over here, that right over there, is our mean for the, let me put that in a color that you can actually see. Now let's do the same calculation for the seniors. So we have one data point where they didn't eat any fruit at all each day, not too healthy. Then you have one one. So I'll just write that as, we could write that as one times one, but I'll just write that as one. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
Now let's do the same calculation for the seniors. So we have one data point where they didn't eat any fruit at all each day, not too healthy. Then you have one one. So I'll just write that as, we could write that as one times one, but I'll just write that as one. Then we have two twos. So plus two times two. Then we have one, two, three, four, five threes. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
So I'll just write that as, we could write that as one times one, but I'll just write that as one. Then we have two twos. So plus two times two. Then we have one, two, three, four, five threes. Five threes. So plus five times three. And then we have, we have three three fours. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
Then we have one, two, three, four, five threes. Five threes. So plus five times three. And then we have, we have three three fours. So plus three times four. And then we have two fives. Plus two times five. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
And then we have, we have three three fours. So plus three times four. And then we have two fives. Plus two times five. And then we have a six. We have a six plus six. And we have a seven. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
Plus two times five. And then we have a six. We have a six plus six. And we have a seven. Someone eats seven pieces of fruit each day. A lot of fiber. Plus seven. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
And we have a seven. Someone eats seven pieces of fruit each day. A lot of fiber. Plus seven. And now, how many data points did we have? Well we have one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16 data points. So we're gonna divide this by 16. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
Plus seven. And now, how many data points did we have? Well we have one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16 data points. So we're gonna divide this by 16. So what is this going to be? This is just zero. Let's see, this is just right over, that's zero. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
So we're gonna divide this by 16. So what is this going to be? This is just zero. Let's see, this is just right over, that's zero. This is four. This is 15. This is 12. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
Let's see, this is just right over, that's zero. This is four. This is 15. This is 12. This is 10. So we have one plus four is five. Plus 15 is 20. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
This is 12. This is 10. So we have one plus four is five. Plus 15 is 20. Plus 12 is 32. Plus 10 is 42. 42 plus six is 48. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
Plus 15 is 20. Plus 12 is 32. Plus 10 is 42. 42 plus six is 48. 48. Am I doing, 42 plus six is 48. Plus seven, 48 plus seven is 55. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
42 plus six is 48. 48. Am I doing, 42 plus six is 48. Plus seven, 48 plus seven is 55. Did I do that right? Let me do that one more time. One plus four is five. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
Plus seven, 48 plus seven is 55. Did I do that right? Let me do that one more time. One plus four is five. Plus 15 is 20. 32, 42. 42 plus 13 is 55. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
One plus four is five. Plus 15 is 20. 32, 42. 42 plus 13 is 55. So this is equal to 55 over 16, which is the same thing as, let's see, that's the same thing as three. And three times 16 is 48. So 3 7 16ths. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
42 plus 13 is 55. So this is equal to 55 over 16, which is the same thing as, let's see, that's the same thing as three. And three times 16 is 48. So 3 7 16ths. So the mean for the seniors, the mean for the seniors, 3 7 16ths, is right around, let's see, this is three, that's four. So 7 16ths, it's a little less than a half. It's right around there. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
So 3 7 16ths. So the mean for the seniors, the mean for the seniors, 3 7 16ths, is right around, let's see, this is three, that's four. So 7 16ths, it's a little less than a half. It's right around there. So the mean number of fruits is definitely greater for the freshmen. They have four, their mean number of fruit eaten per day is four versus three and 7 16ths. The mean is a good measure for the center of the distribution of. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
It's right around there. So the mean number of fruits is definitely greater for the freshmen. They have four, their mean number of fruit eaten per day is four versus three and 7 16ths. The mean is a good measure for the center of the distribution of. So when we think about whether it's freshmen or seniors, so the mean is fairly sensitive to when you have outliers here. For example, someone here was eating 19 pieces of fruit per day. That's an enormous amount of fruit. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
The mean is a good measure for the center of the distribution of. So when we think about whether it's freshmen or seniors, so the mean is fairly sensitive to when you have outliers here. For example, someone here was eating 19 pieces of fruit per day. That's an enormous amount of fruit. They must be only eating fruit. You can imagine if it was even a bigger outlier, someone was eating 20 or 30 pieces of fruit, just that one data point will skew the entire mean upwards. That wouldn't be the effect on the mode because the mode is the middle number. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
That's an enormous amount of fruit. They must be only eating fruit. You can imagine if it was even a bigger outlier, someone was eating 20 or 30 pieces of fruit, just that one data point will skew the entire mean upwards. That wouldn't be the effect on the mode because the mode is the middle number. Even if you change this one point all the way out here, it's not going to change what the middle number is. So the mean is more sensitive to these outliers, to these points that are really, really high, really, really low. So and because the seniors don't seem to have any outliers like that, I would say that the mean is a good measure for the center of distribution for the seniors or a better measure for the center of distribution for the seniors. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
That wouldn't be the effect on the mode because the mode is the middle number. Even if you change this one point all the way out here, it's not going to change what the middle number is. So the mean is more sensitive to these outliers, to these points that are really, really high, really, really low. So and because the seniors don't seem to have any outliers like that, I would say that the mean is a good measure for the center of distribution for the seniors or a better measure for the center of distribution for the seniors. So let's fill both of those out. So the mean number of fruit is greater for the freshmen and the mean is a good measure for the center of distribution for the seniors. And you actually even see it here. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
So and because the seniors don't seem to have any outliers like that, I would say that the mean is a good measure for the center of distribution for the seniors or a better measure for the center of distribution for the seniors. So let's fill both of those out. So the mean number of fruit is greater for the freshmen and the mean is a good measure for the center of distribution for the seniors. And you actually even see it here. We saw that the mean number for freshmen was at four. But if you just ignored this person right over here and just you kind of thought about the bulk of this distribution right over here, four really doesn't look like the center of it. The center of it looks closer to three here. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
And you actually even see it here. We saw that the mean number for freshmen was at four. But if you just ignored this person right over here and just you kind of thought about the bulk of this distribution right over here, four really doesn't look like the center of it. The center of it looks closer to three here. And what happened is this one person eating 19 pieces of fruit per day skewed the mean upwards. While here, that three and 716s really did look closer to the actual distribution, closer to the, actually I shouldn't say, I mean, in both times, we actually did calculate the mean of the actual distribution. But here, since there's no outliers, it does seem the mean seemed much closer to, I guess you could say the middle of this pile right over here. | Comparing means of distributions Probability and Statistics Khan Academy.mp3 |
This is the scores on midterm and final exams. So this axis, the vertical axis, is the scores. And then it's by student. And the blue bar is the midterm. And the yellow bar is the final. And the question they ask us is, by how many points did Nadia's score improve from the midterm to the final exam? So let's look at Nadia. | Reading bar charts comparing two sets of data Pre-Algebra Khan Academy.mp3 |
And the blue bar is the midterm. And the yellow bar is the final. And the question they ask us is, by how many points did Nadia's score improve from the midterm to the final exam? So let's look at Nadia. So this is who we're talking about, Nadia. And we care about how many points did she improve from the midterm to the final. Midterm is blue, final is yellow. | Reading bar charts comparing two sets of data Pre-Algebra Khan Academy.mp3 |
So let's look at Nadia. So this is who we're talking about, Nadia. And we care about how many points did she improve from the midterm to the final. Midterm is blue, final is yellow. So in the midterm, it looks like she scored, and if I were to eyeball it, it looks like 75 points. And on the final, it looks like she scored 80. Looks like she scored 85 points. | Reading bar charts comparing two sets of data Pre-Algebra Khan Academy.mp3 |
Midterm is blue, final is yellow. So in the midterm, it looks like she scored, and if I were to eyeball it, it looks like 75 points. And on the final, it looks like she scored 80. Looks like she scored 85 points. So it looks like her score improved by 10 points. 10 points. Let's try one more. | Reading bar charts comparing two sets of data Pre-Algebra Khan Academy.mp3 |
Looks like she scored 85 points. So it looks like her score improved by 10 points. 10 points. Let's try one more. How many students improve their scores from the midterm to the final exam? So to improve from the midterm to the final, that means that the yellow bar for a given student, which is the final, is going to be higher than the midterm bar. That's the only way you can improve from the midterm to the final. | Reading bar charts comparing two sets of data Pre-Algebra Khan Academy.mp3 |
Let's try one more. How many students improve their scores from the midterm to the final exam? So to improve from the midterm to the final, that means that the yellow bar for a given student, which is the final, is going to be higher than the midterm bar. That's the only way you can improve from the midterm to the final. So Brandon improved from the midterm to the final. Vanessa improved from the midterm to the final. Daniel improved from the midterm to the final. | Reading bar charts comparing two sets of data Pre-Algebra Khan Academy.mp3 |
That's the only way you can improve from the midterm to the final. So Brandon improved from the midterm to the final. Vanessa improved from the midterm to the final. Daniel improved from the midterm to the final. Kevin improved from the midterm to the final. William got a lower score on the final than the midterm. So he did not improve. | Reading bar charts comparing two sets of data Pre-Algebra Khan Academy.mp3 |
Paul has the option between a high deductible or a low deductible health insurance plan. And when we talk about the deductible and health insurance, if someone says that they have a plan with a $1,000 deductible, that means that the insurance company only pays the medical costs after the first $1,000. So if you have a $1,000 deductible and you say incur medical costs of $800, you're going to pay that $800. The insurance company won't pay anything. If you have a deductible of $1,000 and your total medical costs are $1,200, you're going to pay the first $1,000 and then the insurance company will kick in after that. So with that out of the way, let's think about his two plans. If Paul chooses the low deductible plan, he will have to pay the first $1,000 of any, let me do that in purple, the first $1,000 of any medical costs. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
The insurance company won't pay anything. If you have a deductible of $1,000 and your total medical costs are $1,200, you're going to pay the first $1,000 and then the insurance company will kick in after that. So with that out of the way, let's think about his two plans. If Paul chooses the low deductible plan, he will have to pay the first $1,000 of any, let me do that in purple, the first $1,000 of any medical costs. The low deductible plan costs $8,000 for a year. So in this situation, he's going to pay $8,000 to get the insurance. If he has $900 of medical expenses, the insurance company still pays nothing. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
If Paul chooses the low deductible plan, he will have to pay the first $1,000 of any, let me do that in purple, the first $1,000 of any medical costs. The low deductible plan costs $8,000 for a year. So in this situation, he's going to pay $8,000 to get the insurance. If he has $900 of medical expenses, the insurance company still pays nothing. If he has $2,000 of medical expenses, then he pays the first $1,000 and then the insurance company would pay the next $1,000. If he has $10,000 in medical expenses, he would pay the first $1,000 and then the insurance company would pay the next $9,000. If Paul chooses the high deductible plan, he will have to pay the first $2,500 of any medical costs. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
If he has $900 of medical expenses, the insurance company still pays nothing. If he has $2,000 of medical expenses, then he pays the first $1,000 and then the insurance company would pay the next $1,000. If he has $10,000 in medical expenses, he would pay the first $1,000 and then the insurance company would pay the next $9,000. If Paul chooses the high deductible plan, he will have to pay the first $2,500 of any medical costs. The high deductible plan costs $7,500 a year. And it makes sense that the high deductible plan costs less than the low deductible plan because here the insurance doesn't kick in until he has over $2,500 of medical expenses while here it was only 1,000. To help himself choose a plan, Paul found some statistics about common health problems for people similar to him. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
If Paul chooses the high deductible plan, he will have to pay the first $2,500 of any medical costs. The high deductible plan costs $7,500 a year. And it makes sense that the high deductible plan costs less than the low deductible plan because here the insurance doesn't kick in until he has over $2,500 of medical expenses while here it was only 1,000. To help himself choose a plan, Paul found some statistics about common health problems for people similar to him. Assume that the table below, and I put it up here on the right, correctly shows the probabilities and costs of total medical incidents within the next year. So this right over here, it's saying what's the probability he has zero in medical costs? What's the probability, he has a 25% probability of $1,000, 20% probability $4,000. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
To help himself choose a plan, Paul found some statistics about common health problems for people similar to him. Assume that the table below, and I put it up here on the right, correctly shows the probabilities and costs of total medical incidents within the next year. So this right over here, it's saying what's the probability he has zero in medical costs? What's the probability, he has a 25% probability of $1,000, 20% probability $4,000. And this is a simplification, a pretty dramatic simplification from the real world. In the real world, the way this makes it sound is there's only five possible medical costs that someone might have, zero, 1,000, 4,000, 7,000, 15,000. In the real world, you could have $20 in medical costs, you could have 20,000, you could have $999. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
What's the probability, he has a 25% probability of $1,000, 20% probability $4,000. And this is a simplification, a pretty dramatic simplification from the real world. In the real world, the way this makes it sound is there's only five possible medical costs that someone might have, zero, 1,000, 4,000, 7,000, 15,000. In the real world, you could have $20 in medical costs, you could have 20,000, you could have $999. So in the real world, there's many, many more situations here that you would have to kind of redistribute the probabilities accordingly. But with that said, this isn't a bad approximation. It's just saying, okay, roughly, if you wanted to construct this so it's easier to do the math, which the problem writers have done, say, look, okay, 30%, zero, 25%, $1,000. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
In the real world, you could have $20 in medical costs, you could have 20,000, you could have $999. So in the real world, there's many, many more situations here that you would have to kind of redistribute the probabilities accordingly. But with that said, this isn't a bad approximation. It's just saying, okay, roughly, if you wanted to construct this so it's easier to do the math, which the problem writers have done, say, look, okay, 30%, zero, 25%, $1,000. This is pretty indicative if you had to kind of group all of the possible costs into some major bucket. So it's probably at least a pretty good method for figuring out which insurance policy someone should use. So they say, including the cost of insurance, what are Paul's expected total medical costs with the low deductible plan? | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
It's just saying, okay, roughly, if you wanted to construct this so it's easier to do the math, which the problem writers have done, say, look, okay, 30%, zero, 25%, $1,000. This is pretty indicative if you had to kind of group all of the possible costs into some major bucket. So it's probably at least a pretty good method for figuring out which insurance policy someone should use. So they say, including the cost of insurance, what are Paul's expected total medical costs with the low deductible plan? Round your answer to the nearest cent. So actually, I'll get the calculator out for this. So at the low, whoops, with the low deductible plan here, low deductible plan, he's going to have to spend, his total cost, he's gonna spend 8,000 no matter what. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
So they say, including the cost of insurance, what are Paul's expected total medical costs with the low deductible plan? Round your answer to the nearest cent. So actually, I'll get the calculator out for this. So at the low, whoops, with the low deductible plan here, low deductible plan, he's going to have to spend, his total cost, he's gonna spend 8,000 no matter what. Eight, whoops, what happened to my calculator? He's gonna spend 8,000, my God, I'm having issues. He's gonna spend 8,000 no matter what. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
So at the low, whoops, with the low deductible plan here, low deductible plan, he's going to have to spend, his total cost, he's gonna spend 8,000 no matter what. Eight, whoops, what happened to my calculator? He's gonna spend 8,000, my God, I'm having issues. He's gonna spend 8,000 no matter what. So that is $8,000. And then let's see. There's a 30% probability he spends nothing. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
He's gonna spend 8,000 no matter what. So that is $8,000. And then let's see. There's a 30% probability he spends nothing. I could just write that as plus.3 times zero, and I will write it just so that you see I'm taking that into account. Then there's a 25% chance that he has $1,000 in medical costs. And in the low deductible plan, he still has to pay that $1,000. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
There's a 30% probability he spends nothing. I could just write that as plus.3 times zero, and I will write it just so that you see I'm taking that into account. Then there's a 25% chance that he has $1,000 in medical costs. And in the low deductible plan, he still has to pay that $1,000. So plus.25% chance that he pays $1,000. $1,000. And then you might say, okay, plus.2 times 4,000, but remember, if his medical costs are 4,000, he's not going to pay that 4,000. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
And in the low deductible plan, he still has to pay that $1,000. So plus.25% chance that he pays $1,000. $1,000. And then you might say, okay, plus.2 times 4,000, but remember, if his medical costs are 4,000, he's not going to pay that 4,000. He's only going to pay the first 1,000. So it's really plus.2. In this situation, his out-of-pocket costs are only $1,000. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
And then you might say, okay, plus.2 times 4,000, but remember, if his medical costs are 4,000, he's not going to pay that 4,000. He's only going to pay the first 1,000. So it's really plus.2. In this situation, his out-of-pocket costs are only $1,000. The insurance company will pay the next 3,000. So.2 times 1,000. And then plus.2, 20% chance, even if he has 7,000 in medical costs, he's only going to have to pay the first 1,000. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
In this situation, his out-of-pocket costs are only $1,000. The insurance company will pay the next 3,000. So.2 times 1,000. And then plus.2, 20% chance, even if he has 7,000 in medical costs, he's only going to have to pay the first 1,000. So.2 times 1,000 again. And then plus.05 times, once again, 1,000. If he has 15,000 in expenses, he's only going to have to pay the first 1,000. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
And then plus.2, 20% chance, even if he has 7,000 in medical costs, he's only going to have to pay the first 1,000. So.2 times 1,000 again. And then plus.05 times, once again, 1,000. If he has 15,000 in expenses, he's only going to have to pay the first 1,000. Times 1,000. And we get $8,700. And one way you could have thought about it is, okay, he's going to have to pay $8,000 no matter what. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
If he has 15,000 in expenses, he's only going to have to pay the first 1,000. Times 1,000. And we get $8,700. And one way you could have thought about it is, okay, he's going to have to pay $8,000 no matter what. And all of the situations where he ends up paying, that's these four situations right over here, there's a 70% probability of falling into one of these four situations. And in any one of these, he only has to pay $1,000. The insurance company pays everything after that. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
And one way you could have thought about it is, okay, he's going to have to pay $8,000 no matter what. And all of the situations where he ends up paying, that's these four situations right over here, there's a 70% probability of falling into one of these four situations. And in any one of these, he only has to pay $1,000. The insurance company pays everything after that. So you could say 8,000 plus, there's a 70% chance that he's going to pay 1,000. And once again, this table is a pretty big simplification from the real world. There's probably a lot of scenarios where you'd have to pay $500 or $600 or whatever it might be. | Comparing insurance with expected value Probability and Statistics Khan Academy.mp3 |
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