 This is a video about using the normal distribution to approximate the binomial distribution. Suppose that x is a random variable with the binomial distribution, and that there are 20 trials and 0.5 is the probability of success. If we draw a bar chart of the probabilities for different possible number of successes, it looks something like this. And what you'll notice straight away is that this has the familiar bell shape that you'll remember from studying the normal distribution. And in fact, we can be more specific. If we work out the mean and the variance of this binomial distribution, we can actually plot a normal distribution curve that will look very similar. Well, the mean of a binomial distribution is given by the number of trials times the probability of success. So in this case, that's 20 times 0.5, which is 10. The variance of a binomial distribution is given by the number of trials times the probability of success times 1 minus the probability of success. So in this case, that's 20 times 0.5 times 0.5 again, which is 5. So now if we draw the normal distribution with mean 10 and variance 5, you'll see that we get something with a very similar shape to the binomial distribution that we started off with. So the binomial distribution with n trials and p as the probability of success can be approximated by the normal distribution with mean mu and variance sigma squared, where mu is n times p and sigma squared is np times 1 minus p. There's something very important that you need to be aware of, though, when using the normal distribution to approximate the binomial distribution. And that's the need to make something called a continuity correction. Suppose that x has the binomial distribution with 20 trials and 0.5 as the probability of success, and y has the normal distribution with mean 10 and variance 5, just as before. X is discrete, and y is continuous. So we can't say that the probability that x is equal to 8 is about the same as the probability that y is exactly equal to 8. To understand why not, think of the graph of the probability density function for y, and remember that we find probabilities by looking at the area under the graph. So to find the probability that y is exactly equal to 8, we'd be looking for the area of this line. And obviously the area of any line is 0. So the probability that y is exactly equal to 8 is 0. And that's why we can't say that the probability that x is 8 is about the same as the probability that y is exactly equal to 8. Instead, if we want to find the probability that x is equal to 8, we need to say that it's about the same as the probability that y is between 7.5 and 8.5. And this technique is known as making a continuity correction. You can see that this is plausible, because if we look at the graph of the probability density function again, and we look at the area under the curve between 7.5 and 8.5, this might be the probability that x is equal to 8. But let's look at this properly. Here's the bar chart that I drew earlier, which shows the probability for each possible number of successes. If we change this into a sort of histogram, where there are no gaps between the bars, then the area of each bar tells us the probability of obtaining each possible number of successes. Let's superimpose on top of that the probability density function for the normal distribution. You can see that the area of each bar is approximately the same as the equivalent area under the curve. For example, if you look at the bar for x is equal to 8, it's almost exactly the same in area as the area under the curve between 7.5 and 8.5. To the left, you give up a little bit of area, but to the right, you gain approximately the same amount of area. So the two red areas there were the same. OK, well, this explains the need to make a continuity correction. You have to remember that if you want to find the probability that x is equal to 8, you should work out the chance that y is between 7.5 and 8.5. Similarly, if you want to find the chance that x is 11, you have to find the probability that y is between 10.5 and 11.5. You always need to make a continuity correction whenever you're approximating a discrete random variable with a continuous one. If you want to find the chance that x is between 8 and 11, you'd have to find the chance that y is between 7.5 and 11.5. And if you want to find the chance that x is between 10 and 15, you'd have to find the chance that y is between 9.5 and 15.5. What do you think you'd have to work out if you wanted to know the probability that x is between 12 and 14? The answer is that you'd need to find the probability that y is between 11.5 and 14.5. OK, there's one other thing that you need to understand in connection with the normal approximation to the binomial distribution. The approximation is only valid when np is greater than 5 and when n times 1 minus p is greater than 5. To understand this, let's look at some examples. Here's the random variable with the binomial distribution with 20 trials and 0.5 is the probability of success. Here, np is 20 times 0.5, which is 10. And n times 1 minus p is 20 times 0.5, which is also 10. So both these numbers are bigger than 5. And you'll see that we can superimpose the probability density function for a normal distribution and there are no problems. So here, it's perfectly OK to use the normal distribution to approximate the binomial distribution. Here's another example. This time we've got 20 trials and 0.4 is the probability of success. And np is 20 times 0.4, which is 8. And at the same time, n times 1 minus p is 20 times 0.6, which is 12. So here again, we should be OK. Both these numbers are greater than 5. And it's clear that we can superimpose the normal distribution and there are no problems. So this is another case where we can use the normal distribution to approximate the binomial distribution. About this situation, though, here there are 20 trials and 0.1 is the probability of success. If you calculate np, this time it's 20 times 0.1, which is only 2. And n times 1 minus p is 20 times 0.9, which is 18. But 2 is the problem value. 2 is too small. And it reflects the fact that the histogram is butting up against 0. The distribution is no longer symmetrical. In fact, it's positively skewed. You can see that this time, if we try to superimpose a normal distribution, it doesn't really work. The area under the graph at x equals 0 and x equals 1 is much smaller than the relevant bars. And the area under the graph at x equals 3 is much bigger than the bar at 3. So this time the normal approximation doesn't work. The probabilities it gives are wrong. We can also get into trouble at the other end of the spectrum. This time we've got 20 trials and 0.95 is the probability of success. Here np is 20 times 0.95, which is 19. But n times 1 minus p is 20 times 0.05, which is only 1. And this is a problem value. This is too small. And it reflects the fact that we're too shoved up against the right-hand side of the graph. And again, it's not symmetrical. This time it's negatively skewed. If we try to superimpose a normal distribution, you'll see that the area under the curve at 20 is much smaller than the bar at 20. And the area under the curve at 18 is much too big. It's much greater than the area of the bar at 18. So this is another situation where we can't use the normal distribution to approximate the binomial distribution. OK, one more example. This is just to show that what matters is not only the value of p, but also the number of trials. Here, p is small, but it turns out not to matter because the number of trials is very big. Here np is 100 times 0.1, which is 10. And n times 1 minus p is 100 times 0.9, which is 90. So 10 and 90 are both OK. They're both bigger than 5. And so it should be OK to approximate this with the normal distribution. And you can see that by superimposing the normal distribution curve, it is. So you need to remember that we can only use the normal distribution to approximate the binomial distribution when both np and n times 1 minus p are greater than 5. If np is too small, then you've got a distribution which is bunching up at the left-hand end and has positive skew. And if n times 1 minus p is less than 5, you'll have a distribution which is bunching up at the right-hand end and has a negative skew. And in both those cases, you won't be able to use the normal distribution to approximate the binomial distribution. Let's look at an example of the normal approximation in action. Suppose that x has the binomial distribution with 200 trials and 0.75 as the probability of success. Let's find the approximate value of the probability that x is greater than or equal to 140 and less than 155. Well, the first thing to do is to find the mean, which is equal to np. And that's 200 times 0.75, which is 150. Next, we must find the variance, which is np times 1 minus p. So that's 200 times 0.75 times 0.25, which is 37.5. So that means we can use the normal distribution with the mean 150 and the variance 37.5 to approximate this binomial distribution. Now for the next step, we have to be really careful. We're asked for the probability that x is greater than or equal to 140 and less than 155. So that means that the minimum possible value of x is 140 and the maximum possible value of x is 154. OK, so here's a circle which indicates the possible values of x. Now we have to make a continuity correction. So what we say is that the probability that x is greater than or equal to 140 and less than 155 is about the same as the probability that y is greater than or equal to 139.5 and less than or equal to 154.5. Now it's easy to remember that it should be 139.5 and 154.5 because you can see in the picture that the cutoff points are between 139.140 and between 154 and 155. So it's really useful to draw a little picture like this. OK, so now we know that we want the probability that y is between 139.5 and 154.5. Well, the next step is to standardize. So we say that this is the same as the probability that z is between 139.5 minus 150 over the square root of 37.5 because you remember, you standardize by subtracting the mean and dividing by the standard deviation. Well, this is about the same as the probability that z is between minus 1.71 and plus 0.73. So we need to do the probability that z is less than or equal to 0.73 and subtract the probability that z is less than or equal to minus 1.71. Well, to know these probabilities, we better look at the tables. First of all, the table tells us that the probability that z is less than 0.73 is 0.7673. And of course, we can't look up directly the probability that z is less than minus 1.71. But we can look up the probability that it's less than positive 1.71 and you'll see that that's 0.9564 and then we'll need to subtract that from 1. So we need to do 0.7673 take away 1 minus 0.9564 which is 0.7673 take away 0.0436 which is 0.7237 and that's the answer. Okay, that's nearly the end of my video about the normal approximation to the binomial distribution. I just want to remind you about a couple of things. First of all, you've already learned about the Poisson approximation to the binomial. You should remember that that works when n is large and p is close to 0. And when you use the Poisson approximation to the binomial, you say that the binomial distribution with parameters n and p is approximately the same as the Poisson distribution with parameter lambda where lambda is equal to n times p. Here we've just been looking at the normal approximation to the binomial. That works when np is greater than 5 and n times 1 minus p is greater than 5. In that case, the binomial distribution with parameters n and p can be approximated by the normal distribution with parameters mu and sigma squared where mu is n times p and sigma squared is np times 1 minus p. Okay, I hope you found this video useful. Thank you very much for watching.