 This is a video looking at some past paper questions on approximations. Now there are several types of approximation that you're expected to be able to do. First of all, there's the Poisson approximation to a binomial distribution. Second, a normal approximation to a binomial distribution. And thirdly, a normal approximation to a Poisson distribution. So I'd like to give them a bit of revision. First of all, suppose that you've got a random variable with a binomial distribution. So x has the binomial distribution with parameters n and p, n being the number of trials and p being the probability of success. There are two possibilities. First of all, sometimes you can make a Poisson approximation. In other words, you can approximate x with a random variable y, which has a Poisson distribution with parameter lambda. And in that case, what you do is you choose lambda to be equal to n times p. And then x and y will have very similar distributions. Now the reason for saying that lambda should be n times p is that n times p is the expected value of x. And remember that lambda by definition is the expected value of y. So by setting lambda equals n times p, we're making sure that x and y have exactly equal means. The other approximation that you can do starting from a binomial distribution is a normal approximation. So in this case, we say that y has a normal distribution with parameters mu and sigma squared. The mean is mu and the variance is sigma squared. And in this situation, we say that mu should be equal to n times p. And sigma squared should be equal to np times 1 minus p. Again, the point of this is that np is the mean of x. So if we say that mu is np, we make sure that x and y have identical means. And also np times 1 minus p is the variance of x. So by setting sigma squared equals np times 1 minus p, we're making sure that x and y have identical variances. Okay, there's another thing to think about, and that's when these approximations are actually valid. So the Poisson approximation to the binomial is valid when n is large and p is very small when p is close to zero. On the other hand, the normal approximation to the binomial is valid when both np is greater than 5 and n times 1 minus p is greater than 5. In this case, we actually want p to be as close as possible to a half so that the normal distribution is symmetrical, not skewed one way or the other. Okay, the other thing that's possible is we're starting from a Poisson distribution. So x has the Poisson distribution with parameter lambda, lambda being expected number of events. And here there's only one possibility which is to use a normal approximation. So y will have a normal distribution with parameters mu and sigma squared. And here we say that mu is lambda, and sigma squared equals lambda. And once again, the point of this is that lambda is the mean of x. So by saying that mu equals lambda, we're making sure that x and y have identical means. And also, lambda is the variance of x. So by setting sigma squared equals lambda, we're making sure that x and y have identical variances. And this approximation is valid when lambda is greater than 10. We want lambda to be as large as possible, and the bigger it is, the better the approximation. Okay, that's almost it with the revision on approximations. We'll get on with the past paper questions very soon. But just a few tips. First of all, it sometimes happens that you have a choice between using a Poisson approximation and a normal approximation to a binomial because it could be that both sets of conditions are satisfied and both approximations appear to be valid. But what you should remember in this case is that you should never use a normal approximation to a binomial distribution if you can use a Poisson approximation instead. There are two reasons for that. First of all, a Poisson approximation is much easier for you to do. And secondly, it's going to be more accurate to use a Poisson approximation rather than a normal approximation. So what this means is that whenever you've got a binomial distribution, you should try and see if you can make a Poisson approximation first. And only if that doesn't work should you try and make a normal approximation instead. The second thing to bear in mind is that if you've got a binomial distribution and you try and make a Poisson approximation to it, you can't then make a normal approximation to your Poisson distribution. You can't go binomial to Poisson to normal because that's bound to be less accurate than going directly from a binomial distribution to a normal approximation. It's very important to remember this because you're going to try and make your Poisson approximation to the binomial first, and sometimes you'll find that that doesn't work. You've got a very high value of lambda and you might be tempted to say, oh well, that doesn't matter because I can now use a normal approximation to that. But that's wrong. Poisson doesn't go binomial to Poisson to normal. You've got to go directly from binomial to normal and make a normal approximation to the binomial distribution. And finally, you should remember that whenever you make a normal approximation to either the binomial distribution or the Poisson distribution, you have to make a continuity correction. In fact, you need to make a continuity correction whenever you're approximating a discrete random variable with a continuous one. Okay, so now let's get on with the questions. The first one I want to look at is question number 5 from January 2005. And this is about manufacturing things on a production line. Here you can see some workers at the Ford car plant manufacturing magnetos on a production line. And in the same place, they're making the actual cars. And in a more modern context, here are some planes being manufactured at the Boeing factory again on a sort of production line. So here's the question. And the best thing now will be for you to pause the video and to have a go at this question yourself. So pause the video and work out your answers to these questions before looking at my solutions. Okay, so let's do part A. First of all, the random variable here, the number of articles that are defective, has a binomial distribution with parameters 10 and 0.032. 10 because that's the number of articles, so the number of trials. And 0.032 because that's the probability of success. Success here being that an article is defective. Now we're asked to find the probability that x is equal to 2. And we're not going to be able to do this using the probability tables because the probability is an awkward one, 0.032. So we'll have to use the formula. And the formula says that that probability is 10 choose 2 times 0.032 squared times 0.968 to the power of 8. 0.968 is 1 minus 0.032. And if we work that out, that's going to be 10 times 9 over 2 times 0.032 squared times 0.968 to the power of 8, which is 0.0355 to four decimal places, and that's the answer. Okay, moving on to part B, we now take a sample of 100 articles. So this time the random variable has a binomial distribution with parameters 100 and 0.032. And obviously we're going to use an approximation here. So the first thing we should try is to use a Poisson approximation. And you can see that that's valid because here n, i.e. 100, is large and p, i.e. 0.032, is very small. It's close to zero. So we can say that y has the Poisson distribution with parameter lambda. And then we should fix lambda to be 100 times 0.032 so that the mean's correct. And that's 3.2. So this is going to be fine. We're going to use the Poisson approximation to this binomial distribution. And the question asks us the probability that x is less than 4. Now, again, lambda is an irritating value. It's 3.2. So we're not going to be able to use the tables, which only tell us about values of lambda that are multiples of 0.05. So we're going to have to use the formula. Now, the probability that x is less than 4 is the same as the probability that x is less than or equal to 3. In other words, it's the probability that x is 0, plus the chance that it's 1, plus the chance that it's 2, plus the chance that it's 3. And if we use the formula in each of these cases, we get something like this. Well, we get exactly this. So it's 3.2 to the power of 0 times e to the power of minus 3.2 over nought factorial and so on. Actually, that looks much simpler if we factorize. That's the same as e to the power of minus 3.2 times 3.2 to the power of 0 over 0 factorial, plus 3.2 to the power of 1 over 1 factorial, and so on. You might be able to write it down in this form directly. Okay, so if we look at that, that's e to the power of minus 3.2 times 1, because 3.2 to the power of 0 is 1, and 0 factorial is 1, so that's 1 over 1, plus 3.2 because 3.2 to the power of 1 over 1 factorial is 3.2 over 1, which is 3.2, plus 3.2 squared over 2, plus 3.2 cubed over 6, which is 0.6025 to four decimal places. Okay, now we can do the final part of the question, and this time it tells us that a random sample of a thousand is taken. So now x has the binomial distribution with parameters of thousand and 0.032. So again, we should start by trying to use a Poisson approximation, which is clearly valid because again, n is even larger than before, and p is the same, so n is still large and p is very small. So we could try and say that y has the Poisson distribution with parameter lambda, and then lambda would be a thousand times 0.032 because we have to multiply n and p together, which is 32. But now we've got a problem because we're certainly not going to be able to use the tables for the case where lambda is equal to 32, so we're supposed to be finding the probability that more than 42 are defective, and we don't want to have to use the formula to add up such a large number of different probabilities. So we've tried to use the Poisson approximation, but now we can see that it isn't really going to work. And the key thing here is that we now need to start again. We're going to end up making a normal approximation, but we don't want to make a normal approximation to this Poisson distribution. We want to go back to the start of the approximation to the binomial. So we'll say that y has the normal distribution with parameters mu and sigma squared, but mu needs to be a thousand times 0.032, so it has the right mean, so 32 as we just worked out, and sigma squared needs to be a thousand times 0.032 times 0.968. Remember, 0.968 was 1 minus 0.032, which is 30.976. So the mean for our normal distribution needs to be 32, and our variance needs to be 30.976. Okay, so we're asked for the probability that x is greater than 42, and that means that x will be 43 or more. And so when we make a normal approximation, we have to remember to make a continuity correction. And we should say that the chance that x is greater than 42 is about the same as the chance that y is greater than 42.5. That's because the cutoff for x is between 42 and 43, 42 we don't want, 43 we do want, and therefore the cutoff for y should be halfway between 42 and 43. Now that's 42.5. Now you find this probability by standardizing. That's the same as the chance that z is greater than 42.5 minus 32 over the square root of 30.976. Remember that z has the standard normal distribution. And if you do that sum, that's about the same as the chance that z is greater than 1.89. So now we need to go to the normal distribution tables and search for 1.89. It's just down the bottom of the page there. So the probability is 0.9706. But, Daly, we're asked for the probability that z is greater than 1.89, and that's going to be a small probability. So we should subtract 0.9706 from 1, giving us the answer 0.0294, which is the final answer to this question. So note that that was a really nice question, which started off with a binomial distribution, and in one part we had to use a calculation to find a binomial probability. In the second part we had to make a Poisson approximation and then do a calculation to find a Poisson probability. And in the final part we had to make a normal approximation and do a calculation to find a normal probability. OK, the next question I'd like to look at is number 3 from January 2003. And this is about a summer meadow with some weeds growing in it. So just as before, the best thing for you to do now is to pause the video and to have a go at this question before you listen to anything that I've got to say about it. OK, so for the first part we just need to say when a Poisson distribution is an appropriate model. And you should remember that there are four conditions. So you have to say that the weeds must grow randomly, independently, one at a time, and at the same rate in every part of the meadow. And that's all we need to say. In fact, you only need to say two of these conditions. I've just listed all four to remind you. OK, moving on, Part B. They've rather given away the fact that our random variable is going to have a Poisson distribution. But you can see why we're talking about a number of events, well, a number of weeds in an interval of space. And they've told us the mean rate of weeds. So our random variable here is going to have a Poisson distribution with some parameter lambda. And here lambda is going to be 4 times 0.7 because it's 0.7 weeds per meter squared and we're talking about 4 meters squared. So the expected number of weeds is going to be 2.8. Now the question is asking us for the probability that x is less than 3, which is the same as the probability that x is less than or equal to 2. But again, we're not going to be able to look up this probability using the tables because they only tell us about cases where lambda is something like 2.5 or 3. And here lambda is an irritating number between those two. It's 2.8. So we're going to have to use the formula again. The chance that x is less than or equal to 2 is the chance that it's 0 plus the chance that it's 1 plus the chance that it's 2. And if you apply the formula to those three probabilities, we get things like 2.8 to the power of 0 times e to the power of minus 2.8 over 0 factorial and so on. Or you might be able to write it straight away in this factorized form. So it's e to the power of minus 2.8 times 2.8 to the power of 0 over 0 factorial and so on. Which is e to the power of minus 2.8 times 1 plus 2.8 plus 2.8 squared over 2 which is 0.4695 to 4 decimal places. So that's the answer to this question. Okay, now we can do the final part of the question. And you can see that this time we need to make an approximation of some kind. But you know that there's only one approximation we can make to a Poisson distribution and that's a normal approximation. So we can see straight away that we're going to have to make a normal approximation to a Poisson distribution. So x has a Poisson distribution with parameter lambda and this time lambda is going to be 100 times 0.7 because we've got a plot of 100 square meters and it's not 0.7 weeds per square meter. So lambda is 70, the expected number of weeds is 70. So when we make our normal approximation we say that y has the normal distribution with parameters mu and sigma squared but mu needs to be 70 so that they have the same mean and sigma squared also needs to be 70 so that they have the same variance. So we're dealing with a normal distribution with mean 70 and variance 70. Now the question asks us the probability that x is greater than 66 and again we need to make a continuity correction. So when we say that the probability that x is greater than 66 that's about the same as the probability that y is greater than 66.5. So the cutoff for x is between 66 and 67. 66 isn't included in what we want but 67 is and so for y, if the cutoff is between 66 and 67 we should be looking at the probability that y is greater than 66.5. We find that probability by standardizing is the chance that z is greater than 66.5 minus 70 over the square root of 70 which if you work it out is the chance that z is greater than minus 0.42. So again we look at the tables and we search for 0.42 which is down the bottom and the probability for 0.42 is 0.6628. So that's the chance of getting less than positive 0.42, 0.6628. Clearly the probability of getting greater than minus 0.42 will be the same as the probability of getting less than positive 0.42 and so that's actually the answer 0.6628. Okay let's move on to one final question and this is question 4 from June 2008. And this one is about genes in a cell. Here you can see a wonderful photograph of some onion cells with the cells dividing and the chromosomes inside the cells in various states of replication. Okay well this is the last question but again the best thing that you could do now is to pause the video and have a go at it yourself before hearing the answers and the methods. Okay well Part A is asking us to say how the number of damaged genes in the cell will be distributed. And the answer is that it will have a binomial distribution with parameters 11,000 and 0.0005. 11,000 because there are 11,000 genes so that's the number of trials and 0.0005 because that's the probability of a gene being damaged so that's the probability of success. I know it's a bit odd to think about damage being a success but there we go. Okay so moving on to the next part it asks us to find the mean and variance of the number of damaged genes in the cell and remember that X has the binomial distribution with 11,000 and 0.0005 as its parameters so the expected value of X will be 11,000 times 0.0005 which is 5.5 and the variance will be 11,000 times 0.0005 times 1 minus 0.0005 which is 0.9995 and if you do that sum you'll get 5.49725 so that's the variance. Okay we can now get on with the approximation part of the question and remember that because we've got a random variable with the binomial distribution we should start first by trying to make a Poisson approximation and that's clearly going to be valid because here n was extremely large it was 11,000 and p was extremely small it was 0.0005 Okay so for the Poisson approximation we need to choose lambda so that it's equal to the mean value of X so lambda is going to have to be 5.5 because that's the mean number of damaged genes Okay so this is going to work we're going to be able to make a Poisson approximation there's no problem here lambda equals 5.5 so we can use the tables to find probabilities we're asked to find the probability that X is less than or equal to 2 and that's going to be about the same as the probability that Y is less than or equal to 2 and we can look that up straight away we search in the table for where lambda equals 5.5 we find where X is 2 and look along and we see that the probability is 0.0884 so the probability that Y is less than or equal to 2 is 0.0884 and that's the answer a word of caution in this question it looks as though a normal approximation is valid in fact it is valid because np is 5.5 so that's more than 5 and n times 1 minus p would be huge so that's certainly more than 5 and so you might think well I could make a normal approximation here but what you must remember is that you should never make a normal approximation to a binomial distribution if a Poisson approximation would also be valid because the Poisson approximation is first of all much easier to do and secondly gives a more accurate answer so you should do exactly what I just did you should try to make a Poisson approximation and if it works carry on and it's all fine you should only make a normal approximation to a binomial distribution if you've tried to make a Poisson approximation it doesn't work because lambda ends up being too big to use the probability tables okay that's the end of my video about approximations and pass paper questions I hope you found it useful in your revision thank you very much for watching