Hi and welcome back. In this video we're going to continue our study of the Central Limit Theorem. We're going to talk about some of the ramifications, some of the implications, and also how to do some of the actual problems and calculations associated with these types of problems. Before we do that, I want to remind you of a proposition we did last time, we did it right before the central limit theorem and it said, if we have X1 through Xn iid independent identically distributed. And each of them are normally distributed with the same mean and the same variants than the average has a normal distribution with mean mu and then variance sigma squared over n. Now, I want to extend that proposition and I want to say if X1 through Xn are independent with each Xi being normally distributed. But instead of having the same mean, maybe they all have different means and different variances. Then the sum, not the average, but the sum is still normally distributed. The mean of that sum is the sum of the means, and the variance of the sum is the sum of the variances. So what that's saying is that the expected value of the sum, X sub i is the sum i = 1 to n of the expected value of X sub i, and that's equal to the sum i = 1 to n of mu sub i. And the same for the variance. For the variance, it completely depends on the fact that we have an independent sample, 1 to n the variance of X sub i. And that's going to be i = 1 to n the sigma i squared. And right in here this depends on X1 up to Xn being independent. Now, we could extend even further if we wished. And that would be the sum of i = 1 to n of C sub i X sub i. So this would not just be the some of the X sub i but any linear combination of the X sub i's and where the Ci's are constant. And this would be normally distributed with mean 1 through n C sub i, mu sub i and variants i = 1 to n C sub i squared sigma sub i squared. And this again assumes X1 up to Xn are independent. If they're not, then this is not true. Now sometimes we want to add normal random variables. Sometimes we want to average them. So, this proposition gives us a lot of flexibility. Let's do an example. So suppose you have 3 errands that you want to do maybe in three different stores, and let Ti be the time to make the ith purchase for i 1 through 3. And let's assume T4 is the total walking time between stores. I want to assume that the T1, 2, 3 and 4 are all normally distributed and they're all independent. Suppose you're going to leave at 10 in the morning and you want to tell a colleague, I'll be back by time T, what time T should you tell your colleague? So that your chance of being back by that time is 0.99. So there's a lot to unpack in this problem. So let's start by letting T sub 0 be the sum of the four times, so this will be your total time, and then the expected value of T sub 0. That's going to be the some of the four means. So 15 + 5 + 8 + 12 so we get 40. And we could make the units if we want to we could say the units are minutes. And the variance of T sub 0 is the sum of the variances. So that would be 16 + 1 + 4 + 9 and so we get 30. So the previous proposition tells us that T sub 0 the sum has a normal distribution with parameter mean 40 and variance is 30. And we want to find, want to find t so that the probability that our capital T0 is less than or equal to t is 0.99, so T sub 0 is the total time it takes us. And we want to know what time little t will give us the probability of T0 less than or equal to T being 0.99. So let's solve this problem and I'll just rewrite this. T0 is normal, 40 and 30. And as we just said, we want T0 less than t to be 0.99. Now, we've solved problems involving normal random variables before, and this is exactly the same now that we've added them all together and made one normal random variable called T0. So the way we do that is we standardize or normalize, and so we've got T -40 over the standard deviation, that's the square root of 30, and we want that to be 0.99. So that tells us we've got probability. Now, this becomes our Z and it's less than t- 40 over the square root of 30. And that's going to be phi because this Z is normal 01. So we get phi of t- 40 over the square root of 30. And we want that to be 0.99. So we look up in our table and we see that phi of 2.33 is 0.99. So that tells us that t- 40 over the square root of 30 has to equal 2.33. And when we solve for t, we get 52.76 minutes. So, if we leave at 10:00, we will return by 10:52.76 with probability 0.99. And you could imagine lots of things having sums of normal random variables, for example, UPS or Fedex deliveries. Maybe take a certain amount of time for the first delivery than a normally distributed amount of time for the second, perhaps, maybe other kinds of services might be normally distributed. And so the total time spent on some number of services could be a normal random variable. All right, let's go back now to the central limit theorem. So let's let X1 through Xn be a random sample. So remember that's independent and identically distributed. So they have the same mean and the same variance. If n is sufficiently large, and that's always a good question is how large does n have to be, then X bar has approximately a normal distribution with mean mu and variants of X bar is sigma squared over n. Let's apply this to a problem. So suppose you want to verify that 25 kilogram bags of fertilizer are being filled with the appropriate amount of fertilizer, how would you go about doing this? Well, maybe you would select a random sample of 50 bags of fertilizer and you weigh them. So Xi will be the weight of the ith bag from i 1 to 50. The expected value is 25 for every single one of those bags. And because they're filled in a mechanical setting, there might be a little bit of over ridge or a little bit of under ridge in each bag, and maybe the variance is 0.5. Let X bar be the average, and for this problem, I want to find the probability that the average is between 24.75 and 25.25. So from the central limit theorem, we know that X bar has a normal distribution, the mean is going to be 25 and the variance is 0.5 divided by the 50. And so that's the same as normal 25 and then 0.01. So to calculate this probability, this is going to be 24.75- 25 over the square root of the variance less than or equal to X bar- 25 over square root of 0.1 less than or equal to 25.25- 25 divided by the square root of 0.1. Solving, we get -2.5 less than or equal to Z, so this is our Z less than or equal to 2.5. This is going to be phi of 2.5- phi of- 2.5 and this is about 0.9876. So, if the bags are truly being filled 225 kilograms, then the probability that the average is within 0.25 of the mean is 0.9876. Now, what happens if the bags are being under filled and maybe they're not under filled by a lot, maybe it's only half a kilogram, so maybe the expected value is 24.5. So then what happens to that probability? So now X bar is normal 24.5 and still 0.01. And the probability that we want to calculate, is this one and we're going to subtract the mean, so that's 24.5 by like that. And we're going to divide by the standard deviation. And this will give us the probability of 2.5 less than or equal to Z less than or equal to 7.5. So that's phi of 7.5- phi of 2.5. This is for all intents and purposes, this is just 1 and then phi of 2.5 is 0.9938. So the probability now is 0.0062. So, if the bags are being systematically underfilled and you average the weight of 50 bags, there's a very good chance the average will not be close to 25, it will be significantly lower than 24.75. And so you'll be able to detect that underfilling. So the previous example dealt with a continuous random variable. So we can think of the weights of the bags as being a continuous random variable. In this example, I want to consider a discreet use of the central limit theorem. In a statistics class of 36 students, past experience indicates that 53 of the students will score at or above 80. For a randomly selected exam, find the probability at least 20 students will score above 80%. So let's let X sub i be a 1 if ith student scores greater than or equal to 80% and a 0 if the score is less than 80%. We're going to let X equaling the sum i = 1 to 36 of X sub i and that's going to be the number of successes, where a success is the student scores above 80%. Now, we know X is binomial with parameters np and np times 1- P. And in this case that's going to be binomial 19.08 and 8.97. And we are assuming to get it to be binomial, we are assuming that the scores on each of those 36 students are independent of each other. And what we want to know is the probability that X is bigger than or equal to 20. Well, that's going to be one minus the probability that X is less than or equal to. And I'm going to do 19 .5 here because this is a discrete random variable, and we're going to overlay a normal random variable on top of it. And so I want to go halfway between 19 and 20 to account for that continuity on the domain. So this is going to be 1- the probability and we will normalize this. So we're using the central limit theorem X -19.08 divided by the square root of the variance. And then we get 19.5 -19.08 divided by the square root of 8.97, and that's going to be1- the probability that Z is less than or equal 2.14, and that's going to be 1- phi of 0.14 and that's approximately = 2.4443. All right, so we have seen this before, this is a normal approximation to the binomial. We did this in one of our previous videos, but it really is an application of the central limit theorem and here it is again. So, if X is binomial counting the number of successes in n independent Bernoulli trials, then we know the expected value and we know the variants. And so the central limit theorem tells us that x- np divided by the square root of the variance is approximately normally distributed. So we've seen several examples of different uses of the central limit theorem, and it also can provide insight into why many random variables have probability distributions that are approximately normal. For example, the measurement error in a scientific experiment can be thought of as the sum of a number of underlying perturbations and errors of small magnitude. And so some of those errors are positive, some of them are negative, and in the aggregate, they give something that resembles or is approximated by a normal random variable. A practical difficulty in applying the central limit theorem is in knowing when and is sufficiently large. And the problem is that the accuracy of the approximation for a particular n depends on the shape of the original underlying distribution being sampled. So if the shape is approximately symmetric about its mean, then you need fewer n. If it's more irregular or more skewed, then you might need a larger n. And the central limit theorem just tells us that it's approximately normal, and really it tells us in the limit it is normal. All right, so we have seen multiple uses of the central limit theorem, and I hope that gives you a good idea of the importance of the central limit theorem. You'll be exploring this further in future courses. Thank you.