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This would have been a lot harder to do or more time-consuming to do if I had 20 flips. Is there some shortcut here? Is there some other way to think about it? And you couldn't just do it in some simple way. You can't just say, oh, probability of heads times probability of heads, because if you got heads the first time, then now you don't have to get heads anymore. Or you could get heads again, but you don't have to. So it becomes a little bit more complicated. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
And you couldn't just do it in some simple way. You can't just say, oh, probability of heads times probability of heads, because if you got heads the first time, then now you don't have to get heads anymore. Or you could get heads again, but you don't have to. So it becomes a little bit more complicated. But there is an easy way to think about it where you could use this methodology right over here. You'll actually see this on a lot of exams where they make it seem like a harder problem, but if you just think about it the right way, all of a sudden it becomes simpler. One way to think about it is the probability of at least one heads in three flips is the same thing as the probability of not getting all tails. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
So it becomes a little bit more complicated. But there is an easy way to think about it where you could use this methodology right over here. You'll actually see this on a lot of exams where they make it seem like a harder problem, but if you just think about it the right way, all of a sudden it becomes simpler. One way to think about it is the probability of at least one heads in three flips is the same thing as the probability of not getting all tails. If we got all tails, then we don't have at least one head. So these two things are equivalent. The probability of getting at least one head in three flips is the same thing as the probability of not getting all tails in three flips. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
One way to think about it is the probability of at least one heads in three flips is the same thing as the probability of not getting all tails. If we got all tails, then we don't have at least one head. So these two things are equivalent. The probability of getting at least one head in three flips is the same thing as the probability of not getting all tails in three flips. Let me write in three flips. So what's the probability of not getting all tails? Well, that's going to be 1 minus the probability of getting all tails. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
The probability of getting at least one head in three flips is the same thing as the probability of not getting all tails in three flips. Let me write in three flips. So what's the probability of not getting all tails? Well, that's going to be 1 minus the probability of getting all tails. And since it's three flips, it's the probability of tails, tails, and tails. Because any of the other situations are going to have at least one head in them. And that's all of the other possibilities. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
Well, that's going to be 1 minus the probability of getting all tails. And since it's three flips, it's the probability of tails, tails, and tails. Because any of the other situations are going to have at least one head in them. And that's all of the other possibilities. And this is the only other leftover possibility. If you add them together, you're going to get one. Let me write it this way. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
And that's all of the other possibilities. And this is the only other leftover possibility. If you add them together, you're going to get one. Let me write it this way. The probability of not all tails plus the probability of all tails, well, this is essentially exhaustive. This is all of the possible circumstances. So your chances of getting either not all tails or all tails, and these are mutual exclusives, so we can add them, so the probability of not all tails or the probability of all tails is going to be equal to 1. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
Let me write it this way. The probability of not all tails plus the probability of all tails, well, this is essentially exhaustive. This is all of the possible circumstances. So your chances of getting either not all tails or all tails, and these are mutual exclusives, so we can add them, so the probability of not all tails or the probability of all tails is going to be equal to 1. These are mutual exclusives. You're either going to have not all tails, which means a head shows up, or you're going to have all tails, but you can't have both of these things happening. And since they're mutual exclusives, and you're saying the probability of this or this happening, you can add their probabilities. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
So your chances of getting either not all tails or all tails, and these are mutual exclusives, so we can add them, so the probability of not all tails or the probability of all tails is going to be equal to 1. These are mutual exclusives. You're either going to have not all tails, which means a head shows up, or you're going to have all tails, but you can't have both of these things happening. And since they're mutual exclusives, and you're saying the probability of this or this happening, you can add their probabilities. And this is essentially all of the possible events. So this is essentially, if you combine these, this is the probability of any of the events happening, and that's going to be a 1 or 100% chance. So another way to think about it is the probability of not all tails is going to be 1 minus the probability of all tails. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
And since they're mutual exclusives, and you're saying the probability of this or this happening, you can add their probabilities. And this is essentially all of the possible events. So this is essentially, if you combine these, this is the probability of any of the events happening, and that's going to be a 1 or 100% chance. So another way to think about it is the probability of not all tails is going to be 1 minus the probability of all tails. So that's what we did right over here. And the probability of all tails is pretty straightforward. That's the probability of it's going to be 1 half, because you have a 1 half chance of getting a tails on the first flip, times, let me write it here so it becomes a little clearer, so this is going to be 1 minus the probability of getting all tails, well, you have a 1 half chance of getting tails on the first flip, and then you're going to have to get another tails on the second flip, and then you're going to have to get another tails on the third flip. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
So another way to think about it is the probability of not all tails is going to be 1 minus the probability of all tails. So that's what we did right over here. And the probability of all tails is pretty straightforward. That's the probability of it's going to be 1 half, because you have a 1 half chance of getting a tails on the first flip, times, let me write it here so it becomes a little clearer, so this is going to be 1 minus the probability of getting all tails, well, you have a 1 half chance of getting tails on the first flip, and then you're going to have to get another tails on the second flip, and then you're going to have to get another tails on the third flip. And then 1 half times 1 half times 1 half, this is going to be 1 eighth. And then 1 minus 1 eighth, or 8 eighths minus 1 eighth, is going to be equal to 7 eighths. So we can apply that to a problem that is harder to do than writing all of the scenarios like we did in the first problem, we can say, let's say we have 10 flips, the probability of at least 1 head in 10 flips. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
That's the probability of it's going to be 1 half, because you have a 1 half chance of getting a tails on the first flip, times, let me write it here so it becomes a little clearer, so this is going to be 1 minus the probability of getting all tails, well, you have a 1 half chance of getting tails on the first flip, and then you're going to have to get another tails on the second flip, and then you're going to have to get another tails on the third flip. And then 1 half times 1 half times 1 half, this is going to be 1 eighth. And then 1 minus 1 eighth, or 8 eighths minus 1 eighth, is going to be equal to 7 eighths. So we can apply that to a problem that is harder to do than writing all of the scenarios like we did in the first problem, we can say, let's say we have 10 flips, the probability of at least 1 head in 10 flips. Well, we use the same idea, this is the same thing, this is going to be equal to the probability of not all tails in 10 flips. So we're just saying the probability of not getting all of the flips going to be tails, all of the flips is tails, not all tails in 10 flips. And this is going to be 1 minus the probability of flipping tails 10 times. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
So we can apply that to a problem that is harder to do than writing all of the scenarios like we did in the first problem, we can say, let's say we have 10 flips, the probability of at least 1 head in 10 flips. Well, we use the same idea, this is the same thing, this is going to be equal to the probability of not all tails in 10 flips. So we're just saying the probability of not getting all of the flips going to be tails, all of the flips is tails, not all tails in 10 flips. And this is going to be 1 minus the probability of flipping tails 10 times. So it's 1 minus 10 tails in a row. And so this is going to be equal to, this part right over here, let me write this. So this is going to be this 1, let me just rewrite it, this is equal to 1 minus, and this part is going to be, well, 1 tail, another tail, so it's 1 half times 1 half, and I'm going to do this 10 times. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
And this is going to be 1 minus the probability of flipping tails 10 times. So it's 1 minus 10 tails in a row. And so this is going to be equal to, this part right over here, let me write this. So this is going to be this 1, let me just rewrite it, this is equal to 1 minus, and this part is going to be, well, 1 tail, another tail, so it's 1 half times 1 half, and I'm going to do this 10 times. Let me write this a little neater, because I need a 1 half, so that's 5, 6, 7, 8, 9, and 10. And so we really just have to, the numerator is going to be 1. So this is going to be 1, this is going to be equal to 1, let me do it in that same color of green. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
So this is going to be this 1, let me just rewrite it, this is equal to 1 minus, and this part is going to be, well, 1 tail, another tail, so it's 1 half times 1 half, and I'm going to do this 10 times. Let me write this a little neater, because I need a 1 half, so that's 5, 6, 7, 8, 9, and 10. And so we really just have to, the numerator is going to be 1. So this is going to be 1, this is going to be equal to 1, let me do it in that same color of green. This is going to be equal to 1 minus, our numerator, you just have 1 times itself 10 times, so that's 1. And then on the denominator, you have 2 times 2 is 4, 4 times 2 is 8, 16, 32, 64, 128, 256, 512, 1,024. This is the same exact same thing as 1 is 1024 over 1024 minus 1 over 1024, which is equal to 1023, or 1,023, over 1,024, we have a common denominator here, so 1,000, doing that same blue, over 1,000 and 1,024. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
So this is going to be 1, this is going to be equal to 1, let me do it in that same color of green. This is going to be equal to 1 minus, our numerator, you just have 1 times itself 10 times, so that's 1. And then on the denominator, you have 2 times 2 is 4, 4 times 2 is 8, 16, 32, 64, 128, 256, 512, 1,024. This is the same exact same thing as 1 is 1024 over 1024 minus 1 over 1024, which is equal to 1023, or 1,023, over 1,024, we have a common denominator here, so 1,000, doing that same blue, over 1,000 and 1,024. So if you flip a coin 10 times in a row, a fair coin, you're probability of getting at least one heads in that 10 flips is pretty high. It's 1,023 over 1,024, and you can get a calculator out to figure that out in terms of a percentage. Actually, let me just do that just for fun. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
This is the same exact same thing as 1 is 1024 over 1024 minus 1 over 1024, which is equal to 1023, or 1,023, over 1,024, we have a common denominator here, so 1,000, doing that same blue, over 1,000 and 1,024. So if you flip a coin 10 times in a row, a fair coin, you're probability of getting at least one heads in that 10 flips is pretty high. It's 1,023 over 1,024, and you can get a calculator out to figure that out in terms of a percentage. Actually, let me just do that just for fun. So if we have 1,023 divided by 1,024, that gives us, you have a 99.9% chance that you're going to have at least one heads. So this is, if we round, this is equal to 99.9% chance. And I rounded a little bit. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
Actually, let me just do that just for fun. So if we have 1,023 divided by 1,024, that gives us, you have a 99.9% chance that you're going to have at least one heads. So this is, if we round, this is equal to 99.9% chance. And I rounded a little bit. It's actually even slightly higher than that. And this is a pretty powerful tool, or a pretty powerful way to think about it, because it would have taken you forever to write all of the scenarios down. In fact, there would have been 1,024 scenarios to write down. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
And I rounded a little bit. It's actually even slightly higher than that. And this is a pretty powerful tool, or a pretty powerful way to think about it, because it would have taken you forever to write all of the scenarios down. In fact, there would have been 1,024 scenarios to write down. So doing this exercise for 10 flips would have taken up all of our time. But when you think about it in a slightly different way, when you just say, look, the probability of getting at least one heads in 10 flips is the same thing as the probability of not getting all tails. That's 1 minus the probability of getting all tails. | Coin flipping probability Probability and Statistics Khan Academy.mp3 |
He placed ping pong balls numbered from zero to 32, so I guess that would be, what, 33 ping pong balls, in a drum and mixed them well. Note that the median of the population is 16, right? The median number, of course, yes, and that population is 16. He then took a random sample of five balls and calculated the median of the sample. So we have this population of balls. He takes a, we know the population parameter. We know that the population median is 16, but then he starts taking a sample of five balls, so n equals five, and he calculates a sample median. | Sample statistic bias worked example Sampling distributions AP Statistics Khan Academy.mp3 |
He then took a random sample of five balls and calculated the median of the sample. So we have this population of balls. He takes a, we know the population parameter. We know that the population median is 16, but then he starts taking a sample of five balls, so n equals five, and he calculates a sample median. Sample median. And then he replaced the balls and repeated this process for a total of 50 trials. His results are summarized in the dot plot below, where each dot represents the sample median from a sample of five balls. | Sample statistic bias worked example Sampling distributions AP Statistics Khan Academy.mp3 |
We know that the population median is 16, but then he starts taking a sample of five balls, so n equals five, and he calculates a sample median. Sample median. And then he replaced the balls and repeated this process for a total of 50 trials. His results are summarized in the dot plot below, where each dot represents the sample median from a sample of five balls. So he does this. He takes these five balls, puts them back in, then he does it again, then he does it again, and every time, he calculates the sample median for that sample, and he plots that on the dot plot. So, and he'll do this for 50 samples. | Sample statistic bias worked example Sampling distributions AP Statistics Khan Academy.mp3 |
His results are summarized in the dot plot below, where each dot represents the sample median from a sample of five balls. So he does this. He takes these five balls, puts them back in, then he does it again, then he does it again, and every time, he calculates the sample median for that sample, and he plots that on the dot plot. So, and he'll do this for 50 samples. And each dot here represents that sample statistic, so it shows that four times we got a sample median. In four of those 50 samples, we got a sample median of 20. In five of those sample medians, we got a sample median of 10. | Sample statistic bias worked example Sampling distributions AP Statistics Khan Academy.mp3 |
So, and he'll do this for 50 samples. And each dot here represents that sample statistic, so it shows that four times we got a sample median. In four of those 50 samples, we got a sample median of 20. In five of those sample medians, we got a sample median of 10. And so what he ends up creating with these dots is really an approximation of the sampling distribution of the sample medians. Now, to judge whether it is a biased or unbiased estimator for the population median, well, actually, pause the video. See if you can figure that out. | Sample statistic bias worked example Sampling distributions AP Statistics Khan Academy.mp3 |
In five of those sample medians, we got a sample median of 10. And so what he ends up creating with these dots is really an approximation of the sampling distribution of the sample medians. Now, to judge whether it is a biased or unbiased estimator for the population median, well, actually, pause the video. See if you can figure that out. All right, now let's do this together. Now, to judge it, let's think about where the true population parameter is, the population median. It's 16, we know that. | Sample statistic bias worked example Sampling distributions AP Statistics Khan Academy.mp3 |
See if you can figure that out. All right, now let's do this together. Now, to judge it, let's think about where the true population parameter is, the population median. It's 16, we know that. And so that is right over here, the true population parameter. So if we were dealing with a biased, a biased estimator for the population parameter, then as we get our approximation of the sampling distribution, we would expect it to be somewhat skewed. So for example, if the sampling, if this approximation of the sampling distribution looks something like that, then we say, okay, that looks like a biased estimator. | Sample statistic bias worked example Sampling distributions AP Statistics Khan Academy.mp3 |
It's 16, we know that. And so that is right over here, the true population parameter. So if we were dealing with a biased, a biased estimator for the population parameter, then as we get our approximation of the sampling distribution, we would expect it to be somewhat skewed. So for example, if the sampling, if this approximation of the sampling distribution looks something like that, then we say, okay, that looks like a biased estimator. Or if it was looking something like that, we'd say, okay, that looks like a biased estimator. But if this approximation for our sampling distribution that Alejandro is constructing, where we see that roughly the same proportion of the sample statistics came out below as came out above the true parameter, and it's not, it doesn't have to be exact, but it seems roughly the case, this seems pretty unbiased. And so to answer the question, based on these results, it does appear that the sample median is an unbiased estimator of the population median. | Sample statistic bias worked example Sampling distributions AP Statistics Khan Academy.mp3 |
And so it has various outcomes of those two free throws and then the corresponding probability, missing both free throws, 0.2, making exactly one free throw, 0.5, and making both free throws, 0.1. Is this a valid probability model? Pause this video and see if you can make a conclusion there. So let's think about what makes a valid probability model. One, the sum of the probabilities of all the scenarios need to add up to 100%, so we would definitely want to check that. And also, they would all have to be positive values, or I guess I should say, none of them can be negative values. You could have a scenario that has a 0% probability. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
So let's think about what makes a valid probability model. One, the sum of the probabilities of all the scenarios need to add up to 100%, so we would definitely want to check that. And also, they would all have to be positive values, or I guess I should say, none of them can be negative values. You could have a scenario that has a 0% probability. And so all of these look like positive probabilities, so we meet that second test, that all the probabilities are non-negative, but do they add up to 100%? So if I had 0.2 to 0.5, that is 0.7 plus 0.1, they add up to 0.8, or they add up to 80%. So this is not a valid probability model. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
You could have a scenario that has a 0% probability. And so all of these look like positive probabilities, so we meet that second test, that all the probabilities are non-negative, but do they add up to 100%? So if I had 0.2 to 0.5, that is 0.7 plus 0.1, they add up to 0.8, or they add up to 80%. So this is not a valid probability model. In order for it to be valid, they would all, all the various scenarios need to add up exactly to 100%. In this case, we only add up to 80%. If we added up to 1.1 or 110%, then we would also have a problem, but we can just write no. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
So this is not a valid probability model. In order for it to be valid, they would all, all the various scenarios need to add up exactly to 100%. In this case, we only add up to 80%. If we added up to 1.1 or 110%, then we would also have a problem, but we can just write no. Let's do another example. So here, we are told, you are a space alien. You visit planet Earth and abduct 97 chickens, 47 cows, and 77 humans. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
If we added up to 1.1 or 110%, then we would also have a problem, but we can just write no. Let's do another example. So here, we are told, you are a space alien. You visit planet Earth and abduct 97 chickens, 47 cows, and 77 humans. Then, you randomly select one Earth creature from your sample to experiment on. Each creature has an equal probability of getting selected. Create a probability model to show how likely you are to select each type of Earth creature. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
You visit planet Earth and abduct 97 chickens, 47 cows, and 77 humans. Then, you randomly select one Earth creature from your sample to experiment on. Each creature has an equal probability of getting selected. Create a probability model to show how likely you are to select each type of Earth creature. Input your answers as fractions or as decimals rounded to the nearest hundredth. So in the last example, we wanted to see whether a probability model was valid, whether it was legitimate. Here, we wanna construct a legitimate probability model. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
Create a probability model to show how likely you are to select each type of Earth creature. Input your answers as fractions or as decimals rounded to the nearest hundredth. So in the last example, we wanted to see whether a probability model was valid, whether it was legitimate. Here, we wanna construct a legitimate probability model. Well, how would we do that? Well, the estimated probability of getting a chicken is going to be the fraction that you're sampling from that is, are chickens, because any one of the animals are equally likely to be selected. 97 of the 97 plus 47 plus 77 animals are chickens. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
Here, we wanna construct a legitimate probability model. Well, how would we do that? Well, the estimated probability of getting a chicken is going to be the fraction that you're sampling from that is, are chickens, because any one of the animals are equally likely to be selected. 97 of the 97 plus 47 plus 77 animals are chickens. And so, what is this going to be? It's going to be 97 over 97, 47, and 77. You add them up. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
97 of the 97 plus 47 plus 77 animals are chickens. And so, what is this going to be? It's going to be 97 over 97, 47, and 77. You add them up. Three sevens is 21. And then, let's see, two plus nine is 11, plus four is 15, plus seven is 22, so 221. So 97 of the 221 animals are chickens. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
You add them up. Three sevens is 21. And then, let's see, two plus nine is 11, plus four is 15, plus seven is 22, so 221. So 97 of the 221 animals are chickens. And so, I'll just write 97, two 21s. They say that we can answer as fractions, so the problem's gonna go that way. What about cows? | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
So 97 of the 221 animals are chickens. And so, I'll just write 97, two 21s. They say that we can answer as fractions, so the problem's gonna go that way. What about cows? Well, 47 of the 221 are cows. So there's a 47, two 21st probability of getting a cow. And then, last but not least, you have 77 of the 221s are human. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
What about cows? Well, 47 of the 221 are cows. So there's a 47, two 21st probability of getting a cow. And then, last but not least, you have 77 of the 221s are human. Is this a legitimate probability distribution? Well, add these up. If you add these three fractions up, the denominator's gonna be 221, and we already know that 97 plus 47 plus 77 is 221, so it definitely adds up to one, and none of these are negative. | Valid discrete probability distribution examples Random variables AP Statistics Khan Academy.mp3 |
And in that presidential election, there are two candidates. There's candidate A and candidate B. And there's some reality. Let's say I live in a very decisive country, and everyone is going to vote for either, and everyone participates in the election, and everyone is going to vote for either candidate A or candidate B. And so there's some percentage. There's some reality there that P, let me write it over here, maybe 1 minus P percent. Let me do the P first. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Let's say I live in a very decisive country, and everyone is going to vote for either, and everyone participates in the election, and everyone is going to vote for either candidate A or candidate B. And so there's some percentage. There's some reality there that P, let me write it over here, maybe 1 minus P percent. Let me do the P first. There's some reality that maybe P percent will vote for B. And I could switch them around if I wanted. So P percent are going to vote for B, and the rest of the people are going to vote for A. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Let me do the P first. There's some reality that maybe P percent will vote for B. And I could switch them around if I wanted. So P percent are going to vote for B, and the rest of the people are going to vote for A. So maybe 1 minus P percent are going to vote for A. 1 minus P. And you might already recognize that this is a Bernoulli distribution. There's one of two values for a sample I can get. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So P percent are going to vote for B, and the rest of the people are going to vote for A. So maybe 1 minus P percent are going to vote for A. 1 minus P. And you might already recognize that this is a Bernoulli distribution. There's one of two values for a sample I can get. And right here, the values I said, you're either voting for candidate A or you're voting for candidate B. It's very hard to deal with those values. You can't calculate a mean between A and B and all of that. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
There's one of two values for a sample I can get. And right here, the values I said, you're either voting for candidate A or you're voting for candidate B. It's very hard to deal with those values. You can't calculate a mean between A and B and all of that. Those are letters. They're not numbers. So to make it manipulatable mathematically, we're going to say sampling someone who's going to vote for A is equivalent to sampling a 0. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
You can't calculate a mean between A and B and all of that. Those are letters. They're not numbers. So to make it manipulatable mathematically, we're going to say sampling someone who's going to vote for A is equivalent to sampling a 0. And sampling someone who's going to vote for B is equivalent to sampling a 1. And if you do that with a Bernoulli distribution, we learned in the video on Bernoulli distributions that the mean of this distribution right here is going to be equal to P. And it's a pretty straightforward proof for how we got that. So the mean of this distribution, which will actually be not a value that this distribution can take on, is going to be someplace over here. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So to make it manipulatable mathematically, we're going to say sampling someone who's going to vote for A is equivalent to sampling a 0. And sampling someone who's going to vote for B is equivalent to sampling a 1. And if you do that with a Bernoulli distribution, we learned in the video on Bernoulli distributions that the mean of this distribution right here is going to be equal to P. And it's a pretty straightforward proof for how we got that. So the mean of this distribution, which will actually be not a value that this distribution can take on, is going to be someplace over here. And it is going to be equal to P. Now, my country has 100 million people. It is practically or is definitely impossible for me to be able to go and ask all 100 million people who are they going to vote for. So I won't be able to exactly figure out what these parameters are going to be, what my mean is, what P is going to be. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So the mean of this distribution, which will actually be not a value that this distribution can take on, is going to be someplace over here. And it is going to be equal to P. Now, my country has 100 million people. It is practically or is definitely impossible for me to be able to go and ask all 100 million people who are they going to vote for. So I won't be able to exactly figure out what these parameters are going to be, what my mean is, what P is going to be. But instead of doing that, what I'm going to do is do a random survey. I'm going to sample this population, look at that data, and then get an estimate of what P really is. Because this is what I really care about. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So I won't be able to exactly figure out what these parameters are going to be, what my mean is, what P is going to be. But instead of doing that, what I'm going to do is do a random survey. I'm going to sample this population, look at that data, and then get an estimate of what P really is. Because this is what I really care about. I really care about P. So I'm going to try to estimate P with a sample. And then we're also going to think about how good of an estimate that is. So let's say I am going to randomly survey 100 people. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Because this is what I really care about. I really care about P. So I'm going to try to estimate P with a sample. And then we're also going to think about how good of an estimate that is. So let's say I am going to randomly survey 100 people. And let's say I got the following results. Let's say that 57 people say that they were going to vote for person A. Let me write it this way. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So let's say I am going to randomly survey 100 people. And let's say I got the following results. Let's say that 57 people say that they were going to vote for person A. Let me write it this way. So 57 people say they're going to vote for A. Or that's equivalent to getting 57 samples of 0. And then the rest of the people, once again, very decisive population. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Let me write it this way. So 57 people say they're going to vote for A. Or that's equivalent to getting 57 samples of 0. And then the rest of the people, once again, very decisive population. No one is undecided. The rest of the people, so 43 people, say they're going to vote for B. Or that's the equivalent of sampling 43 ones. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
And then the rest of the people, once again, very decisive population. No one is undecided. The rest of the people, so 43 people, say they're going to vote for B. Or that's the equivalent of sampling 43 ones. Now, given this sample here, what is my sample mean and my sample variance? My sample mean right here, well, that's just going to be the average of these 0's and 1's. So I got 57 0's. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Or that's the equivalent of sampling 43 ones. Now, given this sample here, what is my sample mean and my sample variance? My sample mean right here, well, that's just going to be the average of these 0's and 1's. So I got 57 0's. So it's going to be 57 times 0 plus my 43 1's. So the sum of all of my samples. So it's 43 1's plus 43 times 1 over the total number of samples I took, over 100. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So I got 57 0's. So it's going to be 57 times 0 plus my 43 1's. So the sum of all of my samples. So it's 43 1's plus 43 times 1 over the total number of samples I took, over 100. So what does this get me? So 57 times 0 is 0. 43 times 1 divided by 100 is 0.43. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So it's 43 1's plus 43 times 1 over the total number of samples I took, over 100. So what does this get me? So 57 times 0 is 0. 43 times 1 divided by 100 is 0.43. That is my sample mean, the mean of just the 100 data points that I actually got. Now, what is my sample variance? Is going to be equal to the sum of my squared distances to the mean divided by my samples minus 1. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
43 times 1 divided by 100 is 0.43. That is my sample mean, the mean of just the 100 data points that I actually got. Now, what is my sample variance? Is going to be equal to the sum of my squared distances to the mean divided by my samples minus 1. Remember, this is a sample variance. And we want to get the best estimator of the real variance of this distribution. And to do that, you don't divide by 100. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Is going to be equal to the sum of my squared distances to the mean divided by my samples minus 1. Remember, this is a sample variance. And we want to get the best estimator of the real variance of this distribution. And to do that, you don't divide by 100. You're going to divide by 100 minus 1. We learned that many, many videos ago. So I have 57 samples of 0. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
And to do that, you don't divide by 100. You're going to divide by 100 minus 1. We learned that many, many videos ago. So I have 57 samples of 0. So I have 57 samples of 0. And so each of those samples are 0 minus 0.43 away from the mean. Each of those samples are 0. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So I have 57 samples of 0. So I have 57 samples of 0. And so each of those samples are 0 minus 0.43 away from the mean. Each of those samples are 0. You subtract 0.43. This is the difference between 0 and 0.43. And if I want the squared distance, I square it. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Each of those samples are 0. You subtract 0.43. This is the difference between 0 and 0.43. And if I want the squared distance, I square it. That's how we calculate variance. There's 57 of those. And then there's 43 times that I sampled a 1 in my sample population. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
And if I want the squared distance, I square it. That's how we calculate variance. There's 57 of those. And then there's 43 times that I sampled a 1 in my sample population. 43 times I sampled a 1. And the 1 is 1 minus 0.43 away from the mean, because that is the mean. And I want to square that distance. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
And then there's 43 times that I sampled a 1 in my sample population. 43 times I sampled a 1. And the 1 is 1 minus 0.43 away from the mean, because that is the mean. And I want to square that distance. And then I don't want to just divide it by n. I don't want to just divide it by 100. Remember, I'm trying to estimate the true population mean in order for this to be the best estimator of that. And I gave you an intuition of why many, many videos ago. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
And I want to square that distance. And then I don't want to just divide it by n. I don't want to just divide it by 100. Remember, I'm trying to estimate the true population mean in order for this to be the best estimator of that. And I gave you an intuition of why many, many videos ago. We divide by 100 minus 1. We divide by 100 minus 1, or 99. Let's get the calculator out to actually figure out our sample variance. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
And I gave you an intuition of why many, many videos ago. We divide by 100 minus 1. We divide by 100 minus 1, or 99. Let's get the calculator out to actually figure out our sample variance. Let me get the calculator out. And we have, so I'll do the numerator first. I have 57 times 0 minus 0.43 squared plus 43 times 1 minus 0.43 squared. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Let's get the calculator out to actually figure out our sample variance. Let me get the calculator out. And we have, so I'll do the numerator first. I have 57 times 0 minus 0.43 squared plus 43 times 1 minus 0.43 squared. And then all of that divided by 100 minus 1, or 99, is equal to 0.2475. So this is equal to, so my sample variance is equal to 0.2475. And if I want to figure out my sample standard deviation, I just take the square root of that. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
I have 57 times 0 minus 0.43 squared plus 43 times 1 minus 0.43 squared. And then all of that divided by 100 minus 1, or 99, is equal to 0.2475. So this is equal to, so my sample variance is equal to 0.2475. And if I want to figure out my sample standard deviation, I just take the square root of that. My sample standard deviation is just going to be the square root of my sample variance. So I take the square root of that value that I just had, which is 0.497. So actually, let me just round that up as 0.50. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
And if I want to figure out my sample standard deviation, I just take the square root of that. My sample standard deviation is just going to be the square root of my sample variance. So I take the square root of that value that I just had, which is 0.497. So actually, let me just round that up as 0.50. So my sample standard deviation is 0.50. Now, if you just look at this, you say, OK, well your best estimate of the percentage of people voting for A or B is really what you just saw here. Your best estimate of the mean is that 43% of people are going to vote for B and everyone else is going to vote for A. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So actually, let me just round that up as 0.50. So my sample standard deviation is 0.50. Now, if you just look at this, you say, OK, well your best estimate of the percentage of people voting for A or B is really what you just saw here. Your best estimate of the mean is that 43% of people are going to vote for B and everyone else is going to vote for A. But an interesting question is, how good of a sample is that? And let's take it to the next level. Let's try to think of an interval around 43% for which we are 95% we're reasonably confident, or roughly 95% sure, that the real mean is in that interval. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Your best estimate of the mean is that 43% of people are going to vote for B and everyone else is going to vote for A. But an interesting question is, how good of a sample is that? And let's take it to the next level. Let's try to think of an interval around 43% for which we are 95% we're reasonably confident, or roughly 95% sure, that the real mean is in that interval. And let me make it very clear. Let me draw. So when we get our sample mean, we are sampling from the sampling distribution of the sampling mean. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Let's try to think of an interval around 43% for which we are 95% we're reasonably confident, or roughly 95% sure, that the real mean is in that interval. And let me make it very clear. Let me draw. So when we get our sample mean, we are sampling from the sampling distribution of the sampling mean. Let's let me draw that. The sampling distribution of the sample mean. So since we're sampling from a discrete distribution, it's actually going to be a discrete distribution, but it's going to have 100 possible values. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So when we get our sample mean, we are sampling from the sampling distribution of the sampling mean. Let's let me draw that. The sampling distribution of the sample mean. So since we're sampling from a discrete distribution, it's actually going to be a discrete distribution, but it's going to have 100 possible values. This can take on 100 different values here, really anything between 0 and 1. But I'll draw it kind of continuous, because it would be hard for me to draw 100 different bars. If I did, you'd have a bar there, you'd have a bar there. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So since we're sampling from a discrete distribution, it's actually going to be a discrete distribution, but it's going to have 100 possible values. This can take on 100 different values here, really anything between 0 and 1. But I'll draw it kind of continuous, because it would be hard for me to draw 100 different bars. If I did, you'd have a bar there, you'd have a bar there. The odds that your sample mean would be 1 would be very low probability, and then you would have one more bar, a bar like that, a bar like that. But that takes forever to draw, so I'm just going to approximate it with this normal curve right over there. And so the sampling distribution of the sample mean, let me write it over here. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
If I did, you'd have a bar there, you'd have a bar there. The odds that your sample mean would be 1 would be very low probability, and then you would have one more bar, a bar like that, a bar like that. But that takes forever to draw, so I'm just going to approximate it with this normal curve right over there. And so the sampling distribution of the sample mean, let me write it over here. So this is the sampling distribution of the sample mean, it has some mean here. And I can denote it with the mu sub x bar. This tells us this is the mean of the sampling of the sample distribution. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
And so the sampling distribution of the sample mean, let me write it over here. So this is the sampling distribution of the sample mean, it has some mean here. And I can denote it with the mu sub x bar. This tells us this is the mean of the sampling of the sample distribution. But we know from many, many videos that this is going to be the same thing as the population mean that we are sampling from, that each sample comes from, each of these 100 samples come from. So this is going to be equal to mu, which is going to be equal to p. So this is going to be equal to mu, which is equal to p. Now, this variance over here, the variance of this distribution, let me draw it like this. Or even better, let's say the standard deviation of this distribution. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
This tells us this is the mean of the sampling of the sample distribution. But we know from many, many videos that this is going to be the same thing as the population mean that we are sampling from, that each sample comes from, each of these 100 samples come from. So this is going to be equal to mu, which is going to be equal to p. So this is going to be equal to mu, which is equal to p. Now, this variance over here, the variance of this distribution, let me draw it like this. Or even better, let's say the standard deviation of this distribution. That distance right over here. The standard deviation of the sampling distribution of the sample mean, we've seen it multiple times already, it's going to be this standard deviation. It's going to be the standard deviation of our population distribution. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Or even better, let's say the standard deviation of this distribution. That distance right over here. The standard deviation of the sampling distribution of the sample mean, we've seen it multiple times already, it's going to be this standard deviation. It's going to be the standard deviation of our population distribution. So that standard deviation is going to be that distance over there. So there's some standard deviation associated with this distribution. It's going to be that standard deviation divided by the square root of our sample size. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
It's going to be the standard deviation of our population distribution. So that standard deviation is going to be that distance over there. So there's some standard deviation associated with this distribution. It's going to be that standard deviation divided by the square root of our sample size. And we saw many videos ago why that at least experimentally makes sense, or why it intuitively makes sense. So it's going to be the square root of 100. So it's going to be this guy divided by 10. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
It's going to be that standard deviation divided by the square root of our sample size. And we saw many videos ago why that at least experimentally makes sense, or why it intuitively makes sense. So it's going to be the square root of 100. So it's going to be this guy divided by 10. Now, we do not know what this guy is. The only way to figure out what that guy is is to actually survey 100 million people, which would have been impossible. So to estimate the standard deviation of this, we will use our sampling standard deviation as our best estimate for the population standard deviation. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So it's going to be this guy divided by 10. Now, we do not know what this guy is. The only way to figure out what that guy is is to actually survey 100 million people, which would have been impossible. So to estimate the standard deviation of this, we will use our sampling standard deviation as our best estimate for the population standard deviation. So we could say, now remember, this is an estimate. We cannot come up with the exact number for this just from a sample, but we can estimate it. Because this is our best estimator for this standard deviation, and if we divide it by 10, we will have our best estimator for the standard deviation of the sampling distribution of the sampling mean. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So to estimate the standard deviation of this, we will use our sampling standard deviation as our best estimate for the population standard deviation. So we could say, now remember, this is an estimate. We cannot come up with the exact number for this just from a sample, but we can estimate it. Because this is our best estimator for this standard deviation, and if we divide it by 10, we will have our best estimator for the standard deviation of the sampling distribution of the sampling mean. So remember, this is just an estimate. So you kind of have to take everything after this point with a little bit of a grain of salt. So it's going to be roughly equal to, or an estimate of it is going to be 0.5. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
Because this is our best estimator for this standard deviation, and if we divide it by 10, we will have our best estimator for the standard deviation of the sampling distribution of the sampling mean. So remember, this is just an estimate. So you kind of have to take everything after this point with a little bit of a grain of salt. So it's going to be roughly equal to, or an estimate of it is going to be 0.5. And remember, every time we take a different sample from here, this number is going to change. So this isn't like something in stone. This is dependent on our sample. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So it's going to be roughly equal to, or an estimate of it is going to be 0.5. And remember, every time we take a different sample from here, this number is going to change. So this isn't like something in stone. This is dependent on our sample. So it's going to wiggle around a little bit, depending on what numbers we actually get in our sample. But it's going to be 0.50. This is the s right here. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
This is dependent on our sample. So it's going to wiggle around a little bit, depending on what numbers we actually get in our sample. But it's going to be 0.50. This is the s right here. This is the s right over here. This 0.50 divided by 10, which is equal to 0.05. So our best estimate of this standard deviation is 0.05, or you could even view it as 5%. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
This is the s right here. This is the s right over here. This 0.50 divided by 10, which is equal to 0.05. So our best estimate of this standard deviation is 0.05, or you could even view it as 5%. Now, what I want to do is come up with an interval around this sample mean where I'm reasonably confident, using all of my estimates and all of that, where I'm reasonably confident that there's a 95% chance that the true mean is within two standard deviations. Or let me put it this way. There's a 95% chance that the true mean is in that interval. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
So our best estimate of this standard deviation is 0.05, or you could even view it as 5%. Now, what I want to do is come up with an interval around this sample mean where I'm reasonably confident, using all of my estimates and all of that, where I'm reasonably confident that there's a 95% chance that the true mean is within two standard deviations. Or let me put it this way. There's a 95% chance that the true mean is in that interval. So let me write this down. I want to find an interval such that I am reasonably confident, and I'm putting this kind of touchy-feely language over here, because it's all around the fact that I don't know for a fact that this standard deviation is 0.05. I'm just estimating. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
There's a 95% chance that the true mean is in that interval. So let me write this down. I want to find an interval such that I am reasonably confident, and I'm putting this kind of touchy-feely language over here, because it's all around the fact that I don't know for a fact that this standard deviation is 0.05. I'm just estimating. But I'm reasonably confident that there is a 95% chance that the true mean of the population, which is the same thing as the proportion of the population who are going to vote for person B, or the proportion of the population that are going to be a 1. So this is also, we just have to remember, that mu is equal to p, that there's a 95% chance that the true p is in that interval. And actually, since I've already gone 14 minutes into this video, I'm going to pause this video. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
I'm just estimating. But I'm reasonably confident that there is a 95% chance that the true mean of the population, which is the same thing as the proportion of the population who are going to vote for person B, or the proportion of the population that are going to be a 1. So this is also, we just have to remember, that mu is equal to p, that there's a 95% chance that the true p is in that interval. And actually, since I've already gone 14 minutes into this video, I'm going to pause this video. I'm going to stop this video here. And maybe I'll even let you think about it, just based on what everything we've done so far. We figured out the sample mean right over here. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
And actually, since I've already gone 14 minutes into this video, I'm going to pause this video. I'm going to stop this video here. And maybe I'll even let you think about it, just based on what everything we've done so far. We figured out the sample mean right over here. We've figured out an estimate for the, and remember, this is just a sampling mean. We don't know the true, this is the mean of our sample. We don't know the true mean of the sampling distribution. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
We figured out the sample mean right over here. We've figured out an estimate for the, and remember, this is just a sampling mean. We don't know the true, this is the mean of our sample. We don't know the true mean of the sampling distribution. And we also don't know the true standard deviation of the sampling distribution. But we were able to estimate it with the sample standard deviation. Now, everything that we have so far, and based on what we've seen before on confidence intervals and all of that, how can we find an interval such that roughly, and I'm saying roughly because we had to estimate the standard deviation, that there's a 95% chance that the true mean of our population, or that p, the proportion of the population saying 1, is in that interval. | Margin of error 1 Inferential statistics Probability and Statistics Khan Academy.mp3 |
I'm using it essentially to get some practice on some statistics problems. So here, number two. The grades on a statistics midterm for a high school are normally distributed with a mean of 81 and a standard deviation of 6.3. All right. Calculate the z-scores for each of the following exam grades, draw and label a sketch for each example. We could probably do it all on the same example, but the first thing we'd have to do is just remember what is a z-score? What is a z-score? | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
All right. Calculate the z-scores for each of the following exam grades, draw and label a sketch for each example. We could probably do it all on the same example, but the first thing we'd have to do is just remember what is a z-score? What is a z-score? Z-score is literally just measuring how many standard deviations away from the mean. Just like that. So we literally just have to calculate how many standard deviations each of these guys are from the mean, and that's their z-scores. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
What is a z-score? Z-score is literally just measuring how many standard deviations away from the mean. Just like that. So we literally just have to calculate how many standard deviations each of these guys are from the mean, and that's their z-scores. So let me do number or part a. So we have 65. So first we can just figure out how far is 65 from the mean, let me just draw one chart here that we could use the entire time. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
So we literally just have to calculate how many standard deviations each of these guys are from the mean, and that's their z-scores. So let me do number or part a. So we have 65. So first we can just figure out how far is 65 from the mean, let me just draw one chart here that we could use the entire time. So if this is our distribution, let's see, we have a mean of 81. So we have a mean of 81, that's our mean. And then a standard deviation of 6.3. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
So first we can just figure out how far is 65 from the mean, let me just draw one chart here that we could use the entire time. So if this is our distribution, let's see, we have a mean of 81. So we have a mean of 81, that's our mean. And then a standard deviation of 6.3. So our distribution, they're telling us that it's normally distributed, so I can draw a nice bell curve here. They're saying that it's normally distributed, so that's as good of a bell curve as I'm capable of drawing. This is the mean right there at 81. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
And then a standard deviation of 6.3. So our distribution, they're telling us that it's normally distributed, so I can draw a nice bell curve here. They're saying that it's normally distributed, so that's as good of a bell curve as I'm capable of drawing. This is the mean right there at 81. And the standard deviation is 6.3. So one standard deviation above and below is going to be 6.3 away from that mean. So if we go 6.3 in the positive direction, that value right there is going to be 87.3. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
This is the mean right there at 81. And the standard deviation is 6.3. So one standard deviation above and below is going to be 6.3 away from that mean. So if we go 6.3 in the positive direction, that value right there is going to be 87.3. If we go 6.3 in the negative direction, where does that get us, that is close to what, 74.7. If we add 6, it'll get us to 80.7, and then 0.3 will get us to 81. So that's one standard deviation below and above the mean, and then you would add another 6.3 to go to two standard deviations. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
So if we go 6.3 in the positive direction, that value right there is going to be 87.3. If we go 6.3 in the negative direction, where does that get us, that is close to what, 74.7. If we add 6, it'll get us to 80.7, and then 0.3 will get us to 81. So that's one standard deviation below and above the mean, and then you would add another 6.3 to go to two standard deviations. So on and so forth. So that's at least a drawing of the distribution itself. So let's figure out the z-scores for each of these scores. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
So that's one standard deviation below and above the mean, and then you would add another 6.3 to go to two standard deviations. So on and so forth. So that's at least a drawing of the distribution itself. So let's figure out the z-scores for each of these scores. So 65, or each of these grades. 65 is how far? 65 is, you know, it's maybe going to be here someplace. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
So let's figure out the z-scores for each of these scores. So 65, or each of these grades. 65 is how far? 65 is, you know, it's maybe going to be here someplace. So we first want to say, well, how far is it just from our mean, from our mean? So the distance is, we just want a positive number here. Well, actually, you want a negative number, because you want your z-score to be positive or negative. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
65 is, you know, it's maybe going to be here someplace. So we first want to say, well, how far is it just from our mean, from our mean? So the distance is, we just want a positive number here. Well, actually, you want a negative number, because you want your z-score to be positive or negative. Negative would mean to the left of the mean, and positive would mean to the right of the mean. So we say 65 minus 81. So that's literally how far away we are, but we want that in terms of standard deviations. | ck12.org normal distribution problems z-score Probability and Statistics Khan Academy.mp3 |
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