 Hi everyone it's MJ and in this video we're going to be looking at question 2 from the September 2017 paper and this is quite a tricky question so we're going to go through it slowly and hopefully you're going to be following along. What the question says it says assume that n independent random variables x1 to xn are observed which all have the discrete distribution given by the following. So x can either be 1, 2 or 3 and it has the following probabilities of which it can take that on. We then get told that the mean is this and I think they're just doing this they're trying to be nice they're giving us a bit of a clue because this is a difficult question a lot of people do struggle with it. Because then they come they say determine the exact distribution of the sample mean this is quite difficult because we're used to doing what question 2 says the approximate distribution you know we like oh yeah approximate distribution central limit theorem we can do this so question 2 is quite easy to do and just as well because you know that is for the bulk of the marks worth 5 but these first three marks can be a little bit tricky because there's no direct formula what you actually have to do here is think a bit and where your first bit of thinking comes in is you need to think of you know what are the possible values of our sample mean okay what are the possible values and I mean the possible values could be a whole range of things fortunately for us and this is the big clue is we have it for n equals to 2 okay if n was equal to 1 okay the possible values for x bar would be equal to either 1 2 or 3 because it will be one of those but now the fact that n is equal to 2 okay what this is going to do okay our possible values for when n is equal to 2 it means we could have 1 okay that's when both values any is equal to 1 and 1 we're going to have one but if we have 1 and 2 we could get the answer of 1.5 if both our values were 2 and 2 we get the value of 2 if we have 2 and 3 we're going to get 2.5 and if we have 3 and 3 we're going to get 3 and the nice thing is that if we have 3 and 1 we are going to get 2 again so these are the possible values that we can get and now this is the tricky well this is the cool part just by saying this by saying the possible values you're going to be getting half a mark just for saying that and that is the hardest part of the question if you figured that out if you figured out what the possible values were now to do the actual distribution it's going to be a piece of cake because what you're going to be doing now is let's just write these values out here probability that we get 1 is 0.6 2 is 0.3 and 3 is 0.1 because now what we can do is say the following the probability that x bar is equal to 1 this is going to be the situation where the probability x 1 is equal to 1 and x 2 is equal to 1 and that is going to be equal to 0.6 times 0.6 I'm getting that from these values over here and that is going to be equal to 0.36 the probability that x bar is equal to 1.5 okay this is going to equal the probability that x 1 equals 1 and x 2 equals 2 union with x 2 equal to union x whoops x 1 equals 2 and x 2 is equal to 1 okay you can see where probability is coming in to to help us those two things are the same so we can just say 2 times and then we have 0.6 times 0.3 0.6 times 0.3 and our answer is then going to be 0 comma 36 again look at that the probability that x bar is equal to 2 well this is going to be the situation where we have 2 2 and the situation where we have 3 1 and 1 3 so we can be having these three situations so we first have the one situation which is 0.3 times 0.3 okay that's when both are equal to 2 plus the two situations we have 1 and 3 and 3 and 1 which is 0.6 times 0.1 okay and what that's going to give us is 0.21 okay now the probability that x bar is equal to 2.5 okay this is now going to be equal so we looking at this this is in the situation where we have 2 and 3 and 3 and 2 which means we're going to have 0.3 times 0.1 times that by 2 hopefully you guys are starting to see a bit of the pattern we have going here and then the probability that x bar is equal to 3 it's only in the situation where both of our values are equal to 3 and that's 0.1 times 0.1 which is equal to 0.01 what we must always do is then add these values together to just double check that they do equal 1 and this is going to be the probability distribution for our sample mean okay like I said it's three marks these calculations here were very easy but figuring out the possible values for x bar was the hard part and the question you first look at it and you're like what I have not done this before so now you have done it if this pops up again in the future exam you'll be ready for it okay whereas part two of the question is determined the approximate distribution of x bar for n equals to 50 including all relevant parameters stating any assumptions you make well what we can say is because n is equal to 50 we can now use the central limit theorem to approximate the distribution of x bar okay so must write that out you must identify for that for n equals 50 that that is large enough for us to use the central limit theorem to approximate the distribution of x bar okay you get a mark for that I mean you could even say we're gonna you know normally distribute it you could say that you could say that but central limit theorem kind of implies implies that okay so the first thing what we need to do is find our expected value for x bar okay and we know that the expected value of just say one of them is equal to 0.6 times 1 plus 0.3 times 2 plus 0.1 times 3 and this is going to give us 1.5 there I mean what I'm basically just doing is taking each of these values multiplying it by the probability and adding them all up together okay then what we have is the variance of x bar this one is a little bit more complicated what we're gonna be having it's you know that whole situation where it's the x i's minus you know the whole x bars squared kind of kind of vibe going on so what we're gonna be having you know divided by then all of our ends so we're gonna be having a situation where we have 1 minus 1.5 squared times 0.6 plus 2 minus 1.5 squared times 0.3 plus 3 minus 1.5 squared times 0.1 defining that whole thing over our 50 and we see that we're gonna get 0.45 divided by 50 which is equal to 0.009 okay and now what we can do is we can use this and this as our two parameters so we can say the approximate distribution of x bar is normally distributed with 1.5 as the mean and 0.009 as our variance and tada there you are question done and you can see the second part of it was a lot easier than the first part but still overall I would say it is a fairly simple question like I said the tricky part was just understanding what to do or how to handle it but if you've got any questions please feel free to ask me in the comment section below thanks so much for watching guys cheers