I was waiting for the thing to start and welcome to your final exam preparation session. And today will be the last, last, last, last one because after today, there won't be any other exam preparation sessions that I will be hosting. But hopefully if you participate and take part, we might decide that we will have another one during the week before you go out the exam. But let's see how today progresses and then we will start. I will discuss at the end what's next. So we're going to look at the mock exam, which is one of the past exam paper that I created just to give the students that are in my E-Twitter group some exposure or practice into a timed environment when they do their exam so that you get a feel of getting a totally new question that you've never been exposed to, but that is related to your module and how are you going to be dealing with that? So we're going to look at that just now. So I created this mock, which is one of the past exam paper. So let's see, we're going to do it together. Some of the questions and see how we answer them. So I expect you to participate. Let's see. The first question, the following information is collected from an application form or a call to a certain bank, which variables below are quantitative? So A is total of monthly expenditure in rent. E, street address of an applicant. C, gender, D, mental status. E, types of job. Quantitative variables are those variables that you either can count or measure. Which one? A. Do we all agree? Let's see if it's A. And A is correct. Sorry to be a pain. I think someone's not muted again. Can you just make sure that you're muted, please? It maples with the sound. OK, I see there is a hint and I don't see anyone who's unmuted. Koji, I don't know if I'm pronouncing your name correctly. I think you are the person, your mic is unmuted and there is always sound coming up in your background. Can you make sure that you are muted? It says muted, but. Hi, ma'am. It's Matthew here. I couldn't be able to to see the questions that we have shared. Are you not able to see the question? But I'm sharing my entire screen. And now. Lockout and log in back. Is there anyone who also is experiencing the same challenge? Nobody? No, we can see your screen. You can see the screen. OK, so please log out and log back in. Pickle. OK, so let's move on to the next question. Which of the following statement is not green about the mean? A, it utilized all the values in its calculation. B, it is not affected by extreme outliers. C, the value of the mean times the number of observation equals the sum of all observations. G, in a symmetric distribution, the mean, the median, and the mode are all equal. E, it is the best measure of central tendency when the data is not viewed. A, B, C, D, or E. B. It will be B. It is not affected by extreme outliers, are you sure? Which one is not? So yes, definitely. Because we know that the mean, it is affected by extreme outliers. If your data is viewed, your mean will pull towards the bigger data. If it has very big numbers, huge numbers in terms of big numbers, it will be pulled towards that. So B is those who are in my E-Twitter group. You can take, after we've done with this, you can also go back and try this MOOC exam paper as well. Because every time you try, you will get a new question. It's not only these questions that we are looking at. So you can try as many times as possible to get a variation of questions. And when I talk about variation of questions, it's not the same question like in terms of this. It's totally new question that you will get. So please try to use this to practice as many times as you can. Hi, Lizzie. I have a question on that one. So I think it's possibly my understanding of the word observation, but could C be right too? No. Remember C, because I cannot go back now. What C says is I cannot write there. What C says, it says, if you take the sum of all the observation, I can't even now remember, but let's go back to the mean. So this will be your mean and your observation. It says, if you did it say divide or multiply, I think it says, if you multiply the mean with the sum of the observations, you will get the number, the observation. What it says is if this is the sum of all observation divided by the x bar will give you the mean. That's what C was saying to you. It says the number of observation will be given by taking the observation, dividing them by the mean. Sorry, because how I read it is that it was saying that if you take the mean itself, then you multiply it by the number of the observations, then you will get the sum of all observations. Yes, that's the same thing. If you take the mean, so if I take the mean and multiply, cross multiplication, mean multiply by the number of observations, I will get the sum of all observations. That's what it says. Yes, we get the means by adding everything together and dividing by how many people. So it's basically a reverse. It's the reverse because if I need to make the sum of observation, the subject of the formula, I'll multiply with the answer of the mean. The mean times n will give me the sum of all observations. The sum of all observations divided by n will give me the mean. Okay. Sorry, the question was not correct, right? It was to say the choose one that's not correct. Not the correct answer. We already moved on that one, so I can't go back because I made it linear so that you have the same feeling I can't go back. There is no go back. It's only next page. Okay, so let's look at this one. So if you have a question before I press the next button, please ask, don't wait until I press already on the next button because then we lose that question. A graphical technique applicable to qualitative data is A, is it a scatter plot? B, is it an ogive? So C, is it a histogram? D, is it a stem and leaf display? E, is it the pie chart? Remember, this is categorical data, which is qualitative data. It's to be. It will be E because A scatter plot is when you have two numerical values and ogive is when you have a cumulative frequencies and you put them on a graph. And histogram, it is a pie chart for numerical data. It's a stem and leaf plot is when you have numerical data and it can be a 10th stem and leaf or hundreds stem and leaf or a decimals stem and leaf plot. And E is a pie chart where you break your categorical data into different categories. And the answer here is E, check. Are there any questions before I move? If not, then we move. The following data represent the number of children in the sample of 11 families from a certain community. And there they give us the data. Which one of the following statement is incorrect? A, it says, the median and the second quartile are not equal. B, only the median and the mean are equal. C, only the mode is equal to the median. D, the median is five. E, this distribution is symmetrical. Oh, it's not symmetrical. When it's not symmetrical, it means the mean and the median are not equal. So what do you need to do on this question to answer this? Let's come back to our, let's wanna make it smaller. We go to our data. You need to sort your data. One, sort your data in ascending order. Number two, calculate the median by using the median position first. N plus one divided by two. And find the answer to your median. Three, calculate the mean, which is the sum of all observation divided by N. So let's do that together. Let's sort the data. You have three zeros, one, two, three. Did we have the mode somewhere? Let's see. Oh, yes, we did have the mode as well. So we'll need to also, number four, we'll need to identify which value is the range. So how many zeros we have? Make this bigger so that everybody can see. Three zeros and how many? Three one, one, two, three. And two, two, and two, fours, and five. Okay. So they should be 11, one, two, three, four, five, six, seven, eight, nine, 10, 11. So let's find. Let's find the median position. So they are 11 plus one divided by two, which means it's 12 divided by two. And it's on sixth position. So if it's on sixth position, we count one, two, three, four, five, six. That is your mean. Calculate, sorry, that is the median. Calculate the mean and all of them divide by how many they are. Okay, the mean is 1.8, 1.82. One. 0.8182. Okay, we can just keep it to two decimal. And the mode, I don't even have to worry about that. The mode. It's by model. It is a by model one, but the mode is zero and? One. And one. So now let's answer the question to choose which one is incorrect. Let's do my answer because when I click some way it, okay. So we have all the information we require. So let's answer the mean, the median and the second quartile are not equal. So what is the second quartile? So let's go to number five. We can say quartile two, we find the position by using n plus one divided by two, right? So the median and the second quartile are not equal. So are they? And also remember we're looking for the correct answer, right? So number one is, is it correct or incorrect? In. It will be incorrect. Only the mean and the median are equal. So median and the mean one and one point. So it's also incorrect. Only the mode is equal to the median. That's, do they, are they incorrect? It's incorrect. The median is five. Is it correct? It's incorrect. It's incorrect because the median is five, is one, not five. And the distribution is symmetrical. So we can check. It's not symmetrical. So is the mean equal to the median? The median, no, no. If they are, it will be the median. If they are equal, right? So are they equal? No, they are not. So our correct answer on this one is, E. It's E. Now we need to calculate the standard deviation. So to calculate the standard deviation, the formula to calculate standard deviation, I can put it right here, because this is the sample. So the standard deviation is the square root of the sum of your X observation minus the mean, squared, divide by N minus one. So we already calculated what the mean is and we found that the mean is 1.81. So you can use that information to calculate here. So you can say the square root of two minus, because it's the same data set, two minus 1.81 squared, plus zero minus 1.81 squared, plus until you get to the end, I'm gonna get there, two minus 1.81 squared and divide that by 11 minus one. So you can do that. Or alternatively, you can put your calculator to state mode. So let's do that with a case. You are not muted and it creates an egg. So let's put our calculator to state mode. I hope you're all using the case your calculator. If not, if you are using a sharp calculator, I will also give you the steps to the sharp calculator just now. But let's start with the case show. So we first put our calculator to mode state two and one, because we're only working with one variable, one minus one, it's for one variable. And we have the table. We capture the information by putting the data and pressing the equal sign. Zero, zero is also a data point, right? Don't forget about that one. Zero equal, five equal, one equal, one equal, four equal, zero equal and two equal. They should be 11 dataset captured. Once you have captured all your information, you can use your arrow to check if your data is captured correctly. And once you confirm, you just press the AC button. And you press the shift and you press one first step because we're reaching for that orange button. And you get this because we're interested in the standard deviation. If you're interested in the summation, I just want to demonstrate to those who didn't know the summation are all those sum X. If one time they ask you to calculate the summations, there is the sum X squared and the summation of X. So shift always, step and we interested in four. So you press four and we will understand the deviation. Otherwise we can check the mean, which is number two to see if we did calculate it correctly. So the mean equal 1.8, 1.81, we did calculate it correctly. So let's go to the one that we want. So it's the bar, which is four. And we're looking for the standard deviation. This is the population standard deviation. This is the sample size. So we're interested in the sample standard deviation. You need to be very careful. Sample standard deviation, population standard deviation can also look almost similar. It's sigma equals to the square root of the variant, which is your sum populations, your X minus your population mean divided by capital letter N, which is the population. You can see that the only difference is the minus one. So you need to pay close attention. Don't choose the incorrect one. So let's choose our standard deviation, which is four and equal. And the answer is option number C. 1.7787. Okay, I've done that for you. Those who are using Casio or Financial Calculator, just give me a second. That's the problem. I needed to open all the calculators that I have. Like I said, I cannot share my entire screen as I mitigate. Just give me a second. You can use the Shop Calculator. So with Shop, you also do the same whether it's Financial Calculator. So those who are using Financial Calculator, you will use the Enter button. And those who are using the Shop Scientific Calculator, you will use the M plus. So we first put our calculator to stick one mode and we press zero for SD. And it will look like this. Your calculator will be on state zero. And to capture the data, you're going to use the value and Enter if you are using a Financial Calculator. If you are using a Scientific Calculator, you're going to use M plus. Scientific, you will use M plus button, which is on the same position as those who are using Shop Financial Calculator. Calculator, you will use the Enter button. They are on the same location, right? They are above the close bracket. So to capture your data, you say to minimize you first, because I can't see my data, you will press two and you will press M plus. And you continue zero M plus. So you will say to enter, zero and four M plus. One M plus, zero M plus, five M plus, one M plus, one M plus, four M plus, zero M plus and two M plus. And it will say data set 11. If you make a mistake, not a problem, you just press second function mode, which will clear your data and you can start capturing again. On your case, if you make a mistake, you just go mode, shift, and you follow the same steps and capture the data again. Okay, so now you are ready.  To calculate the mean, you will press button number four to calculate the sample. Standard deviation, you will press button number five to calculate the population. Standard deviation, you will press number six. If for some reason they ask you to calculate the variance, right, which is something that we didn't touch with the previous one. So the standard deviation, or your S on your calculator, you will use sigma X or SX. If you want to calculate the variance, this is the standard deviation. To calculate the variance of the population or the sample, you will just press the X squared button. That will give you the variance of either one of them, right? So it's easy. You will press this X squared button. On the case, you will press the X squared button. It will give you, for example, if I press the X squared, it will give me the value underneath the square root, which is the variance. Okay, so now let's calculate the variance, which is five. So because it's written in green, you will press alpha and button number five and press equal and that is one comma 7787. If I want the variance, I just press the X squared equal and it will give me the variance. If I want to go back to the standard deviation, you just take the square root of the answer and that will give you back. So it's just moving around and knowing which one is which. Okay, so that is correct. Are there any questions, any comments, any query, anything you wanna ask before I move, before I press the next button. Are we good? Are we happy? Let's see, are there any? No, nothing on the chat. Okay, so let's move to the next question. So the next one, I'm gonna give it to you to answer without me giving you the answers. The following data represent the salaries of a sample of dating employees in a firm and they are in decimals. As you can see, calculate the coefficient of variation. So the coefficient of variation, I can just give you the formula, coefficient of variation. Let's put it here. CV is your standard deviation divided by the mean multiply by 100. So all you can do is put your calculator to state mode and then calculate your standard deviation and the mean. You can do that together and put the data bigger so that everybody can see them clearly. Shift and mode, I need to go back. Mode, stat, it's two and one. And I can keep capturing the data, 26.5, 23.5, 29.7, 24.8, 21.1, 24.3, 20.4, 22.7, 27.2, 23.7, 24.1, 24.8, and 28.2, there should be 30 of them. And I can just double check. That I've captured all the values correctly. 2.7, 20.4, 24.3, 21.1, 24.8, 29.7, and 23.5, and 26.1. Okay, I'm gonna give you a chance while I also capture using the other calculator. Second function C8 should clear out all the numbers, 26.5, and plus 23.5, and plus 29.7, 24.8, and plus 21.1, 21, 21, and plus 24.3, and plus 12345620, 24, and plus. The challenge with the sharp calculator is you don't know where you are. If you make a mistake, you start from scratch. 22.7, and plus 27.2, and plus, so you will need to be very careful with how you capture your data with the sharp calculator. I hope I did capture them correctly. 24.1, and plus 24.1, and plus 28.2, and plus, because I cannot see the data or go back to the data. Whereas with the Casio, if I'm looking at the Casio, if I double check my data and I see that number one, I captured it incorrectly, it's still fine, I can replace the value by just pressing what it should be, and press equal, it will replace the same value there, and so on. With the sharp calculator, you cannot. So I hope you all have your information in, so let's calculate the coefficient of variation. I will start with the Casio calculator, so I'm gonna clear. On the Casio, we will shift, start, and remember, let's go back to the formula. Remember, our formula is standard deviation divided by the mean multiplied by 100. So we'll do the same. So four, we're going to get the standard deviation, which is button number four, divide by shift, that, four, and button number two, and I can say equal, multiply the answer by 100%, and that gives us 10.8, 10.7891, so which, nine, two, if we leave it to four decimal, into three decimal, it will be 10.789, 10. So the only one is D. So let's look at the Casio, should give us the same answer. So with the Casio, we also go on and off, and we go alpha, five, divide by alpha, four, equals, and multiplied by 100, equals, and that gives us 10.78919, which is nine, two, which is D. You need to practice, in order for you to know how to use your calculator properly, you will need to practice. If you do have a template, Excel template that summarizes or calculates this, you can use that, but you cannot wake up on Wednesday and hope that on Thursday, when practice on Wednesday and hope that on Thursday, when you write your exam, you will know what is happening. You want, it needs you to start practicing now, find more questions, especially in study, you need three to calculate using your calculator, especially for the mean and the standard deviation. You will have to do that. We go to next. Are there any questions? Hi, Lili, just a quick question. Regarding what your statement was earlier regarding Excel spreadsheets and calculating the formulas, is there a lot, I'm just assuming, because I've done a lot of Excel formulas for everything from the critical values to Poseidon and everything else in Excel. So if I can use that, that'll help me a lot in trying to calculate those formulas. Yeah, if you do have a template that has all that calculating everything else, you can use that. Fantastic, thank you very much. All right. And if you would like to also share that with the rest of the group, it will also help because what we're trying to do is to share as much information with everyone as possible so that everyone is equipped because we don't want to leave anyone behind. All right? As much as I support everyone, I also want everyone to support everyone, to support each other and not rely only on me. Okay, so. So it was a one more question. Yes. In the exams, I haven't tried this on the mock exam, but I have obviously multiple screens where I do my work. You know, one screen is for answering questions on multiple choice, and otherwise we have my Excel speech open. I've used IRIS before and IRIS obviously requests only one screen share, but the mock exams I see, it doesn't do anything with screen recordings, only your camera that it does everything with. All right, so what I will suggest you do is check with the mock exam that the lecturer has given you. Yeah. Fly with that and see if there are some discrepancies or anything that you pick up when you are doing, when you move across the screen as well, right? If not, then why not? So you can always even minimize and create a split screen like this, if I can put it this way. You can do it that way, where you have, the challenge is only this piece of other thing that is popping up, but you can do it this way because your camera is only to always stay on the side, right? So your camera will be somewhere on the navigation side or at the bottom, if you minimize your screen like this, it will show right here at the bottom, which as long as your face is not blocked, I don't think it should be any problem. But yeah. But you must try with the mock exam and see, because I think the more you try and see different ways that you can use your platform, the better to get prepared for your exam so that you don't get surprises during the exam. If you get a notification to say your screen needs to be maximized all the time, then you just make sure that you don't repeat that. Your screen needs to be maximized. Yeah. So let's move on to the next question, which is probabilities. Also remember that your questions in the exam will follow the structure of your study unit. So the first few questions will be from study unit one, two, three, and then you move to study unit four, study unit five, study unit six. There won't be mix and match where you've answered first the study unit 10 and then study unit, it will follow the structure. So if you know your structure of your module, you will know when you add different questions. So let's look at the basic probabilities. Suppose that A and B are mutually exclusive, such that the probability of A is 0.3 and the probability of B is 0.2. Which one of the following statement is incorrect? So you need to validate each statement. You cannot get away from that. You will have to test and validate each and every statement to see if it's correct. So let's do that. We'll start with A and move to other statements. Let's start with A. A says A and B are independent. So the first thing that we need to also take into consideration is what is given in the statement. So I'm gonna write there what is given in the statement. So the first step, we are told that the probability, oh, sorry, even A and B are mutually exclusive. So in your mind, somewhere you should know that if it's, A and B are mutually exclusive, therefore the probability of both of them being together, the joint probability of A and B will be 0. What else have they given you? They told you that the probability of A is 0.3. They also gave you that the probability of B is 0.2. So those are the information provided to you. So now let's use that information to select which one of this is incorrect in the exam. As soon as you get your answer, you move on. So we're looking for the incorrect so that I always remember what we're looking for. So A, how do we test that A and B are independent? So we need to check that. Let's use another. So we need to check that the probability of A given B is the same as the probability of A. If we can find that they are the same, then A and B are independent. So let's check that. Let's start with the probability of A given B, which is the probability of A and B divided by the probability of B. So in order for us to test that, we know that the formula, the probability of A given B will be given by the probability of A and B divided by the probability of B. So what is the probability of A and B is 0 divided by the probability of B, which is 0.2, which is equals to zero because any value divided, zero divided by any value will just be zero. So they are not independent. So number one is incorrect. Probability of A given B is not the same as the probability of A because the probability of A is 0.3 and the probability of A given B is 0. So event A and B are dependent. So we're looking for the incorrect answer. So therefore number A is the incorrect answer in terms of this. Number B, it says you need to find the probability of A given B is the same as the probability of B given A. We know that the probability of A given B will be equals to zero. Similar, the probability of B given A will be the joint probability divided by B. And we know that the joint probability is zero. So this one will be correct, right? Because the probability of B given A is given by the probability of a joint probability of A and B divided by the probability of A. The probability of A given A and B is zero divided by the probability of A of 0.3. Therefore, it will be equals to zero. So both of them will be equal. The probability of A given B, it's the same as the probability of B given A based on the information given. So that is correct. The next one, four. It's correct also. Yes, because the probability of A or B is given by the probability of A plus the probability of B minus the probability of A and B. And we know that this is zero and this will be zero comma three plus zero comma two, which is equals to zero comma five based on the information given, right? So number C is correct. Number G, we answered it because based on the information given, right? We did do that. The probability of A and B is zero. Okay, number E, what is the complement? It's correct also. One minus. The complement will be one minus the probability of A, which is one minus zero point three, which is equals to zero point seven. So the only answer that is incorrect is A. In the exam, if you find that number A is your answer that you are looking for, you just choose it and move on to the next question, right? Are there any questions? Before I move to next, you see no hands, no question in the chat so we can move on. Don't be shy to ask. Then we move on to the next question. Question eight, it's also probabilities. Also remember, you have 11 study units so you're writing out of 25 question, I think. If there are 25 questions, therefore it means every study unit might have at least two questions per study unit with the exception of few study units that might have three questions. So expect at least a minimum of two question pairs, study unit or every study unit. So there should be study units that you know better than the others as well. Okay, so the following table describes the breakdown of roller skaters or roller skate sale by age and sex. So age at the top and sex on gender. So they have given us. So what do you need to take into consideration? What are these venues that they have given you? Probabilities. There are probabilities. When you are working with probabilities and they give you decimals, you must know that those are probabilities. If there were whole numbers, you will know that those are your events. Yes, your events. Now on this table, what is missing? Is your totals, right? Yes, is your totals because on your totals, that's the way you calculate the probability of a simple event. So if our question there, let's go there. Just wanna move this to the side so I can have space. So if our question is like this and they tell you that you need to find the probability of F or O, you just need to rewrite this in the formula. The probability of A or B is equals to the probability of A plus the probability of B minus the probability of A, not O, N, the probability of a joint. NB, so we can rewrite this in this format. So let's rewrite it in this format. So the probability of F or B, F or O. So it means we can say if we take ASF, so it will be the probability of F plus the probability and we replace B with O minus the joint probability of F and O, see how easy it is. So because we don't have the total here, you will have to add the totals. So what is the probability of F? So you go to not F complement but F, female. You will take 0.109 and 0.112. What is that? Do the calculation, the total? 0.221. 0.221 and the probability of O. So you go to O, not the O complement, the O. It's 0.4130. Plus 0.413, okay? Minus the probability of a joint of O and F, which is 0.109. 109, so calculate that and see which one is correct. E. And the answer would be 0.525, which is E. That's how you will find the answers. Is there one more? We have a question. I've already pressed next, yes? No, it's fine, you don't have to go back. I just wanted to find out the probability of O. Why did you choose the one on the first? Because on the table they gave it to you like that, right?  So you just look at the letters that they've given you, the way included in there. All right, thank you, I include the O complement. Yeah, now we are in discrete probabilities. Which one of the following is not an assumption of a binomial? So you need to know also theory on some of this. So binomial, we're looking for the one that is not. So A, the number of successes in the trial are counted. You must remember that binomial is one of the discrete processes. So it comes from a discrete process. So you need to know what a discrete process looks like or a discrete variable looks like. All trials are independent. Think about a coin. If you toss a coin, when a coin falls on a head, does it have an influence on when it falls on a tail? Think about it in that way. Each trial must be classified as a success or failure. I don't even have to explain that. The name says it all. It's a buy, means two outcomes. All trials must be identical. And E, the probability of a success, is equals to 0.5 in all trials. Think about a dice. Think about if I have a survey that looks at how many number of people own a car in a household. Or how many cars every household owns. Will the probability of success in all the different scenarios I gave you be 0.5? Think about it. OK, so which one? Oh, the following are not properties of a binomial distribution. It's E. E equals E, because A, the trials comes from a discrete probability of process, which means discrete means counted processor. The trials are independent, because one outcome does not affect what the other, or does not have any influence on the other. And there are always two outcomes, a success and a failure. All trials are identically able. And E is the only option that does not fit the binomial distribution property. The number of cars owned by each of the families in the city is shown in the following table. So we are given a table with the number, which will be our cars, or will be our x value. The number of families, our frequencies, what they didn't give you is the probabilities. You will need to calculate the probability of each and every one of them. So we'll suggest what you do is your x will be given by 0, 1, 2, 3, 4. It will be our table, and you will just need to calculate the probabilities. In the exam, you don't have the time to calculate the probabilities. Go and look at what the question is asking, and only calculate those events that they are asking you to calculate. Calculate the probability that any family selected at random owns a minimum of one car. What is minimum in a mathematical symbol? It's less than. It's greater than or equal to. Less than or equal to. Minimum is less than or equal to. So if they say a family owns a minimum of one car, therefore it means we are only interested in those ones in this category. So we need to add all of them. So what you will do is to find the probability of x less than or equals to 1, you will just add, but you will need to add all of this value so that you can calculate your total. You need to have the total. Sorry, that is the other thing that I missed. So you will need this value here, which is the total, and you will need those two values and add them together. Then you will need to calculate the probability of x is equals to 0, plus the probability that x is equals to 1. So go ahead and add all of the values and tell me how many they are. 487. Wait, wait, wait. 6487. That is the total. So you can say 27. Divide by 6487, plus 1422. Divide by 6487. Or you could just say 27, plus 1422. Divide by 6487. And I hope everyone has calculated and you found that that is the total. The answer is D. Are we happy with B? Moving on to the next question. My system is very slow. The law enforcement agency claims that the number of times that a patrol car passes through a particular neighborhood follows a poison process with the mean of three times. Nightly shift, let x denote the number of times that a patrol car passes through the neighborhood during night shift. Which one of the following statement is incorrect? So it means we need to validate or evaluate every statement. So we need to do that. There are a couple of things that we will need to have handy.  We need to go to Poison table. You will need to go and use your Poison. And those are the things that you will require to answer the question. So let's start. The first two questions that we need to evaluate, we're looking for the incorrect answer. The first two, we need the Poison table and the last one. So let's use the Poison table to answer one A, B, and E. So our lambda from the data, it says it's three times. It's given to you the mean, the lambda history, which is the same as your variance and so on. I'm going to put it that way. So let's go find the probability that x is less than 2. So that is the probability of x less than 2 will be the probability that x is equals to 0 plus the probability that x is equals to 1 plus the probability that x is equals to 2. So we'll have to add three things. So let's go to the table. We're going to look for lambda equals to 3. And there is our lambda equals to 3. We're looking for 0, 1, and 2. So we're looking for the first three. Add them together. Question 0.4232. Is it 4, 3, 2, or 4, 1, 3, 2? It's 4, 2, 3, 2. 4, 2, 3, 2. 0.4232. So let's see if that is our answer, which is correct. Now let's go to x is equals to 5. x is equals to 5. It's 0 comma 1, 0, 0, 8, which is correct. Let's go to e. x is equals to 3. x is equals to 3. It's 0 comma 2240. 2240. So a, b, c, a, b, and e are correct. Now let's see. I'm going to pull up my, the variance is equals to 3. Yes. And the average is equals to 3. Then there is no incorrect answer here. No, there is. Sorry, number 2 is incorrect. Sorry, I read it wrong. Number 2 is it's 0 comma 1, 0, 0, 8. I already draw. Sorry, my bad. This is the incorrect, b is incorrect. b is the incorrect value. And that will happen in the exam. So you need to pay attention to all the numbers, because they might be a little bit tricky. That's why when you write your exam, your stats exam, especially because it's in the morning, you need to have a well-groomed interest so that you don't make mistakes, silly mistakes. It's a game of concentration. Okay, we only have one hour. We are left with one, two, three, four, five, six, seven, eight, nine, 10, 14 questions that we still need to go through. Let's see how well we can do. I should have stopped the session at 4.30 so that it doesn't become long, but it's fine. I will split it before I upload it. Question 12. We are now in normal probabilities. Now, when you are in normal probability and when you move into sampling distribution probabilities, you will need to understand the key weights and key things that are given in the question as well, because the two look almost exactly the same. So a max on the chemistry test follows a normal distribution with the mean of 65 and the standard deviation of nine. Approximately what percentage of students have a score below 50? What is below 50? Less or equal to 50? Below? Less than 50. It's less than. So they're asking you to find the probability that X is less than 50. They have given you the standard deviation. They have given you the mean. The mean is 65. The standard deviation is 12. And you just need to find the probability that Z, we standardize that. Z is less than X minus the mean divided by the standard deviation. That's what you need to do. That is less than X is 50 minus 65 divided by the standard deviation of 12. Do the calculation. It's negative 1.25. Now, the other thing you need to remember always, like I always say, we will go to the table, right? The cumulative standardized normal distribution table. The probability of Z less than a value, which is a value in this case minus 1.25 will be the value we find on the table. The probability that Z is greater than a value, we will say 1 minus the value we find on the table, right? Will be 1 minus the table value. If it's between, if Z lies between two values, A and B, we say the table value for B minus the table value for A. Those are the things that you always need to remember, right? When we work with probabilities. So let's go to the table, cumulative standardized normal distribution table. The one with negative and positive values. So we're looking for minus 1.25. So we go there. We look for minus 1.2 and the five at the top, where they both meet, that's where we are. The answer is 0.1056, let's see. The answer is 0.1056, but it says what percentage? So you will need to multiply that by a hundred. So when you multiply that by a hundred, you get 10.56, right? The answer will be 10.56%. And if we round it up to one decimal, it will be 11%. 11.56. So we're looking for minus 1.25. It will be 11%. 11.56. 11% is E. We can forever to respond. Let's go to 12. It's still connected. Let me swap my connection, right? Which one of the following statement is incorrect? Is the session still recording? Yes. Which one of the following statement is incorrect with regards to standard normal distribution? A, the standard normal distribution has the mean of zero and the standard deviation of one. A normal distributed random variable X can be standardized according to the Z score. C, the standard normal distribution is symmetrical or is symmetric around the mean. D, the normal distribution is a discrete. E, the area underneath the curve is always equals to zero. One, which one is incorrect? Think about normal distribution where we talk about, you can see that we are adding numbers together, right? So from normal distribution distributed with the mean of zero and the standard deviation of one. That is correct, correct? And this is symmetrical. That should be correct. And the area underneath the curve, we always say this side is 0.5 and this side is 0.5 to the left and the area to the right is 0.05. Therefore, it means the sum of all probabilities will be equals to one. So A, B, C and E are correct. So therefore it means normal distribution, it is a continuous distribution. Agree? So if that is the case, then the only incorrect statement here would be option number would be D. So you always need to remember that normal distribution is A, continuous variable or a continuous distribution. Come, come, come, come. I don't know why it's so slow. You might be experiencing load shedding maybe around your area maybe or next to you maybe. I was load shedding at that point. Now load, it came back, it just came back not so long ago. I shouldn't be load shedding. I'm back from load shedding. And I don't know why, maybe because I'm connected via my phone and it's holding with only two bus and my Wi-Fi doesn't wanna connect. It's its own type. 50% of all college students attend within 50 mile of their home. In a sample of 500 college students, the probability that the sample proportion will be between 0.5 and 0.55 E. So we need to find the probability of the sample proportion being between 0.5 and that. So our population, based on this information, what is our population proportion? It's 0.5. It's 0.5 because it's given in the statement right there. So we need to find the probability that the sample proportion lies between two values, 0.45 and 0.55. So you can calculate them separately. So we can start with 0. We can start with the second part, which is z of, let's calculate it together anyway. So then we do all of them together. I'm just gonna make this bigger because I need to use the probability of z lies between your sample proportion minus your population proportion divided by their standard error, which is the population proportion one minus the population proportion divided by N. And we do the same on the other side. Let's substitute the values. Sample, 0.45 minus our population proportion, 0.5 divided by the square root of 0.5 times one minus 0.5 divided by, and our sample size was 500 divided by 500. So you do the calculations. 0.55 minus 0.5 divided by the square root of 0.5 times one minus 0.5. 0.45. Sorry, I'm not gonna stop at 0.55. What is 0.55? 0.55. 0.55 minus 0.5, then the second one is gonna be 0.45. Remember, I'm just converting the standardized. So I'm still on the standardized. I'm standardizing this side, which is this side. Standardize that side. I haven't gotten to the minus the other. So once you give me your values, so I'm still having them as in this way. I'm still having them in the format, the probability that Z lies between A and B. I'm still having them in that format. But this is 0.45, right? And this is 0.55, right? I've substituted them correctly. 0.5 is my population proportion. 0.5 is your population proportion. So give me the answer to the 0.45. It's negative 2.236. Only two decimals. Oh, negative 2.24. And on the 55. 2.24. It's a positive 2.24. Now I have A and B. I can do the probability of Z less than 2.24 minus the probability of Z less than minus 2.24. So let's go to the table and find on the positive side, 2.24. The positive side, we're looking for 2.24, not 2.24, 2.24. And sure this, so let's see, I think it's supposed to be 2.23 because the next number is two, not bigger than five. Yeah, let's go there. Let me double check, since I get two different answers. So let's use me as the break, the person who breaks the tie, which is 0.5 times one minus 0.5 divided by 500 equals, it's 2.4, the answer. Can you see? Negative 2.24. Yes, yes, the first one. Yes, and then let's do the second one. The positive. Let's do this, second one is minus 4.24. Must check how you calculated the second one. Oh, okay. Okay, so let's go to the table, 2.24 is 0.9875. I keep on forgetting, I'm using the white box, 0.9875, I hope, 9875. And then we go to the negative side, negative 2.24, negative 2.022, negative 2.2 and four at the top where they meet, which is 0.0125 minus 0.001.25. 0.0125, okay? And that is, do the calculations. 0.975. 0.297, that is the answer, so let's see, which one? 0.975, which is option C, happiness. Are we good? Especially, are we good? Especially with my method that I'm using, you could have, immediately at this point, split it into two from there, and did the probability of 0.55 minus 0.5 divided by the standard error, minus the probability of, minus the probability of 0.45 minus 0.55, it will still work it out the same way, because the answer will be like this, because this will fit into that. So it's the same process that I followed. They didn't break any rules. Okay, so let's move on to the next question. Today we didn't even take a break. Oh, gosh. Now it's even worse. Taking forever. And I just double check something quickly. It's gonna disconnect. You see, are you back? Am I back? Are you able to see my screen? Yes. Yes. I have two network connectivity, and it still refuses to connect with my Wi-Fi. I don't know why. Lizzie? Yes. Would it be possible, the file that you shared earlier, would it be possible to also add it to the eTutor site? Yes, I will add it to the eTutor site once the session is done. Did I create a folder called exam? I think so. Yes. I'll edit another exam preparation folder. Yes. I will add that, and I will also add the other document that another lady shared on the WhatsApp group as well. Okay, perfect. So moving on to the next one, after some couple of minutes of waiting, the proportion of eligible voters in the next election who will vote for the ANC is assumed to be 0.55. That is our population proportion. What is the probability that a random sample of 500 voters less than 0.49 will say they will vote for the ANC? So what they are asking you is to find out of the way and out what they are asking us. So we know what the population proportion is, is 0.55, and our sample size is 500 voters, and our P, they want us to find the probability that a random sample, not a sample, not X, but the proportion, which is P, is less than, less than is just a less than 0.49. We need to calculate that. So we normalize the probability of Z, less than P minus the proportion divided by the standard error, which is the population proportion, one minus population proportion. And this question is unfair because then I've got two population proportion questions. But when you do the trial on the actual paper, you might get a different question as well altogether. So in sampling distribution, we also have the standard deviations, right? And the means is not only the proportions. So it's just that on this, on our options that we got now, it only gives us the proportion P of Z, less than our P is always in the question is 0.49 minus our population proportion of 0.55 divided by the square root of 0.55 times 1 minus 0.55 divided everything by 500. Let's calculate and do the calculations. What is the answer? Minus 2.697. Yeah, round it off to 2 decimals. Minus 2.710. We don't have negative 2.70. So we need to go to the negative side of the table to go find our answer. So let's do that. So the negative side, we're looking for 2.70 and the answer is 0.035, right? 0.035. Is that what we will get? Because it's less than. Always remember that, right? For the less than, the value you find on the table will be that probability that you are looking for. I'm going to ask that we stop now and take some few minutes so that then I can sort out my network and also stop the recording and then we come back and we do the rest of the questions. As long as how long it will take us, are you guys good with that proposal? We take only a five-minute break. So let's do that. Before I move to the next question, I'm just going to leave it at this point. Let's stop the recording. Because I might switch off as well. I'm just going to stop this.


So with error control, we need to retransmit. So we need a mechanism for detecting when to retransmit. And with stop and wait ARQ, it's very similar to stop and wait flow control. Send data, receive the act. Send data, receive the act. Of course, we care about what happens when things go wrong. Say when we lose data, yesterday we got to the case of, if I bring it up, it was yesterday. Yesterday we covered this case of A transmits data. The first one successful, everything's OK. And the second one, we lose data. The second data frame transmitted didn't get to B. The way that A knows it doesn't get there is it doesn't get an act within time. So it's waiting for an act, it's waiting, waiting. No act comes back after some predefined time. So we must know what this interval should be. Our timer expires and that triggers a retransmission. The same data frame, data frame two is retransmitted. And in this case, it gets delivered successfully. So this was an example of lost data. Then we move to another case when there's a lost act. Same initial case, data, the first piece of data is delivered. B gets the first piece of data. The second one is transmitted, gets delivered. B gets the second piece. The act comes back, but something goes wrong. Any questions on stop and wait ARQ? The act comes back, but doesn't get to A. Something goes wrong with our link. There's an error. So that again, from A's perspective, we're waiting for the act. Waiting for the act. Act doesn't come within our time out period. So that triggers a retransmission of data two. And the problem that we get is that when that data two arrives at B, the question we raised yesterday is what do we do with it? What should B do with that data? And the right response is we should ignore it. Because if B accepts this data, it gets the data, the message in there, and it uses it in some way, then that would be wrong because it's the same data. It's just a retransmission of the previous one. So we need some way for B to realise, ah, this piece of data I just received is in fact a retransmission of the previous one, and therefore I should ignore the second case. If we didn't ignore it, then, depending upon what the application was that was transferring the data, but things will go wrong. The example I used yesterday, if the data was saying, transfer 10,000 baht to Steve's account, the bank receives it here, transfers 10,000 baht into my account, but then receives a copy of that same message again. And the bank doesn't know, is this a second amount of 10,000 baht that should be transferred or is this just a copy of the first one? So it gets confused. So we must have some way to detect that this data is a retransmission of this one. Then we can ignore it. We still send the act. So B receives the data, realises, ah, this is a retransmission. Maybe A didn't receive the act, so they retransmitted. So let's send them the act again and let's hope it gets through. So we still send the act. How does B know to ignore this one? How does B know to ignore this data message? What is B gonna do to realise that this is the retransmission of the previous one? We can't just check the data. The data is a sequence of bits. It could be that the sequence of bits we receive is identical to the previous one. That could have been intentional. We don't know. So we need some other way for B to know that this is a retransmission. What can we do? How can B know that this data message is the same as the previous one? Sequence numbers. Give each data message a sequence number. So a retransmission will have the same sequence number as before. And in that way, B will know, ah, this is a retransmission. So in fact, if you recall back to stop and wait flow control, there were no sequence numbers. We only introduced sequence numbers with sliding window. But in fact, with stop and wait ARQ, we actually need sequence numbers. And I'll start to draw them. And with stop and wait, we only need a one bit sequence number. Zero, one, zero, one, zero, one. So the first one, let's say this is sequence number zero. Even though it was the first piece of data, the sequence number we gave it was zero. That was included in the frame. When B sends the act back, what acknowledgement number does it include? This is a sequence number. The sequence number is what we call the number we include with the data. And when we send an act back, we include an acknowledgement number, the act number, putting out a space. So each frame now includes a number inside the header. Remember, the act says, what's the next number I expect? B received data with sequence number zero, therefore sends an act saying, I now expect data with sequence number one. A receives that act, sends the next piece of data, which of course has sequence number one. B receives data with sequence number one, sends an act with act number, what's the act number here? Zero, the number after one is zero when we use just one bit. With one bit, it goes zero, one, zero, one, so on. So the next one, expect it is zero. It doesn't get to be though. Oh, sorry, it doesn't get to A. So after a timeout, A retransmits the previous one. So it's the same sequence number. It's still sequence number one. Now when B receives this data frame with sequence number one, it realizes this is not in order. B received data with sequence number zero, the first one, and then received sequence number one. B is expecting sequence number zero. That's what it said in the act. I expect to get zero. If I receive one, then I know that that's not correct. And that's how B knows to ignore this data. B is expecting sequence number zero, but it gets one, so it ignores that, but still sends back an act and what act number? What is B expecting? What did B just receive? Look at the sequence numbers, the purple ones. B received sequence number one, it's expecting zero. It receives sequence number one again, it ignores that, realizing this is out of sequence. It should be zero, one, zero, one. Here it's one, one, ignore this one, and still tell A, I'm still waiting for zero. I'm still waiting for data with sequence number zero. So the next piece of data, when A receives that, A realizes, okay, my data with sequence number one was successfully delivered, I can move on. Zero means that one was delivered. The previous number to zero is one. So now the next data we send, the next one we send, the third frame, the third piece of data has sequence number, what's sequence number? 50% chance of being correct. Zero, when we send back an act saying, I expect to receive zero, then of course, if everything's okay, the next one we'll send is zero. Assuming it's okay, it gets delivered, data gets delivered, eventually act comes back. Act number, data zero received, say thank you, I now expect data with sequence number one. So the act number is one in that case. And upon reception of this data frame, the third data frame, B receives and processes that data. So we haven't shown the processing time, but data three has been successfully delivered here. So we do have sequence numbers in stop and wait ARQ, and you'll go back and you'll fix the example from yesterday and add sequence numbers. So I didn't draw them yesterday, but the reason we need them is because of this case, the lost act, just see what happened. Sequence number, actually we'll go back to the original case first, then we'll go through that one. Here in this diagram, I included the sequence numbers. In the normal case, you'll see data sequence numbers, zero, one, zero, one, zero, one, just keep going like that if everything's okay. And the act, one, zero, one, zero, and so on. So that's the normal case. In the case of a lost act, the same, zero, data one, but when we retransmit, of course, we don't change the sequence number. Original data, retransmission of that previous one, and we know that by the same sequence number. And that's how B knows to ignore that data frame. Any questions on stop and wait ARQ? And the example, look at the example you drew yesterday. Does it always begin with zero? The sequence number, no, not necessarily. It must start somewhere a sequence number. It makes sense to start at zero, okay? But in theory, it could start at one. But it would be agreed upon by the actual devices which are using this protocol. So similar with sliding window in the previous topic, remember we had sequence numbers say zero to three or zero to seven. It's natural to think that they'll start at zero. But in practice, they don't have to. We can start anywhere in that range. And in fact, in practice, in some protocols, they don't start at zero. But in the examples that I use, I usually start at zero because when we start counting, we start there when we use binary at least, okay? When the ACQ is, the sequence numbers, of course, we increment for each new data frame, okay? We start, first data frame is zero. So just think of the sequence number we're incrementing. First data frame is zero. Next one is one. The next one is, well, we don't have two with just one bit. So we just got a one bit sequence number. So the next one, instead of two, we go back to zero. We wrap around to zero. And then we go back to one, increment to one, increment, we don't have two, so back to zero. So it's really just alternating. And the ACQ is just saying, okay, be received data with sequence number zero, send back an ACQ saying, I expect, in the next data delivery, sequence number one. So the ACQ number is what's expected. Does that address your question? So this is like sequence numbers in sliding window, but just one bit. Sometimes this is called an alternating bit protocol, that is the data frames that just alternate, zero, one, zero, one. Maybe just go back to the one from yesterday, where I didn't draw sequence numbers, but we can add them just for completeness. Same as before, sequence number, let's, sorry, sequence number zero, ACQ number one. Next data, sequence number one. Data was lost. Next data, what's the sequence number here? One, because it's a retransmission of the previous one. When B receives data one, it sends back ACQ number zero. Next one I expect is zero and so on. But we need sequence numbers to keep track, and especially if there's a lost ACQ for B to know that there is a retransmission. Any questions on stop and wait? That's the basic stop and wait, ARQ. ARQ means automatic repeat request. It's the general name for these mechanisms for retransmissions. We automatically repeat the data transmissions when we detect a loss. In this case, we detect a loss from a timeout. A few issues which is related to this and the next protocol and in fact in general to these types of data link protocols. One is the timer. So the way it works is after A sends the data, it starts its timer and once it reaches some value, we say the timer expires or a timeout occurs and that triggers a retransmission. What value should this be? How long should we wait for the ACQ to come back? So that's a parameter that we need to have. What is this period here? What do you think it should be? How long should A wait for the acknowledgement to come back? Any ideas? Compare with the previous round. You mean what happened in the previous successful transfer? Okay, and how would you compare compared to what value? Propagation for the data to get to B plus the ACQ to be transmitted and get back, okay? So that's, we should wait long enough to give B a chance to get the ACQ back to us. So we send data. In the normal case, we know there's going to be some propagation delay, some processing possibly at B, some ACQ transmission delay, plus some propagation back.  So I need to wait long enough to give a chance for that ACQ to get back. So the timeout interval must be large enough such that we can get an ACQ back in the normal case if there was nothing lost. How long should that be? How would you calculate that? If you're computer A, how do you calculate that time? Round what? But what do you need to know to calculate that? Let's say it's your phone. Your phone is communicating to the base station using some mobile technology. There's a link. You need to, when you transmit data, you need to know how long it takes for the ACQ to get back. What does it depend upon? Well, propagation delay depends upon the distance. Does your phone know how far away you are from the base station? Not very accurately. It may be able to estimate, but generally not. Sometimes we don't know what the propagation delay will be. Now it might be one microsecond, tomorrow maybe 10 microseconds. So in different situations, the propagation delay may not be known in advance. That's a bit of a problem. In this example, I didn't draw the processing delay, but there is a processing delay from when it's received until when the ACQ comes back. How long does it take a computer to process a frame? Estimates, anyone? How long does it take a computer to process, say, a thousand-byte data frame? Is it the same for every computer in the world? No. The processing delay if my laptop is computer B, it may be very short. If your phone is computer B, it may be very long. And A doesn't know that. So we cannot easily predict what the processing delay will be. So in fact, it's quite hard, in some cases, for A to know how long it takes to get the ACQ back. So usually we need to estimate what the maximum time would be. So under the worst conditions and set the timeout interval to be larger or slightly larger than that, we need to give some time for B to process and send the ACQ back. So let's say the propagation delay was 100 milliseconds. The transmission of the ACQ was 10 and the propagation back was 100. What should the timeout be? 100 there, 10 to transmit, 100 back. How long should I wait? At least two propagations plus the ACQ. So at least 210 in that example. Larger because sometimes that maybe there's a processing delay which we cannot predict. So usually we make it slightly larger. Why not? Okay, instead of 210, why not make it 1,000 or 1 million milliseconds? I expect the propagation to be 100. Maybe the ACQ transmission changes. Why not make the timeout interval very, very long? What's the problem? Let's say it takes about 210 milliseconds, normal case to get the ACQ back. But I wait for say 1,000 milliseconds before I retransmit. What's the problem with that? Why not have a large timeout interval? Look at our efficiency when there is a loss. In this case, send data. All this time when I'm waiting for my timer to expire, I'm being inefficient, I'm not doing anything. I'm not transferring data. The longer I wait before I resend, the lower the efficiency. So if there's a loss, I really want the timeout to be as short as possible. So that, okay, as a last timer expires, let's resend again and hopefully it gets there the second time. So we want the timeout interval to be short in the case there's a loss, so we are efficient. But we want it to be long enough such that in the case that there's no loss, we get a chance for the ACQ to come back.  We don't wait for too long. So generally we try and, if we can estimate how long it takes to get an ACQ back, we set it slightly larger than that. There's no easy way to calculate in some cases. Maybe we can try and draw those cases, roughly draw what happens. Let's say we have a case, we'll consider two cases. One, when we send the data, we set the timeout to be too small. Normally it takes some time to propagate, the ACQ comes back and it's going to propagate back. The ACQ will arrive here, but if I start my timer at this time and if I set it too small and it expires here, what happens? I'll resend. When I didn't really have to, I resend, but if I waited a little bit longer, I would have got the ACQ back and I wouldn't have needed to resend. So this is the problem of, if you set the timeout interval to be too small, you'll retransmit when you don't need to. That's a waste. And the other case, if we set it to be too large, we send our data, in the case that that data is lost, we'd normally expect it to come back. Let's say we set the timeout to be very long, then it times out and we retransmit. In this case, with a very long timeout, we spend a lot of time waiting before we resend and that's very inefficient. We want to spend as much time as possible transmitting data. In this case, a long timeout can lead to lower efficiency. So ideally, we'd like the timeout to be a little bit larger than or at least the time it takes to get the ACQ back. In practice, we usually give it a little bit of freedom and make it a bit larger so that if there's some variation, we'll take that into account. But it should be at least the time for the data to propagate, the ACQ to be transmitted and to propagate back. Not too small, not too large. Any questions about the timeout interval? In some systems and in some protocols, it's defined in advance. That is, the value is given. In others, the software try and estimates the best value over time. So it measures how long it takes the previous one and then adapts the timeout interval to take that into account. But we will not go into those protocols. That's on one of the slides here, I think. How long should the timeout interval be? What's another thing that we skipped over? How related to both sliding window, stop and wait flow control and the ARQ protocols, how big should our data frame be? Another design issue. In these examples, we didn't say, but in previous examples, we gave an example, we gave a value of, I think we calculated 8,000... 1,000 bytes was the data frame plus 20 byte of header. How big should our data frame be? In general. If we go back...  I don't have the picture, but when we calculated even for stop and wait, we said we had 1,000 byte messages. A data frame contained 1,000 bytes plus 20 bytes of header. And we calculated the efficiency. And we said, what about the data frame? Should we make it smaller or larger for higher efficiency? With the data frame, regard to efficiency, should we make the data frame or the amount of data larger or smaller? If you go back to your notes, you should see that we said, larger data frame leads to higher efficiency. It doesn't matter if you're on this slide, but around here, we did some examples. And one of the cases was, okay, when it's 1,000 bytes, we got an efficiency of about 96%. Then we reduced the data frame from 1,000 bytes down to 100 bytes. We made it smaller and our efficiency went down. And the trend we said was higher efficiency is achieved when we have a larger data frame. So, and that applies with the others as well. Larger data frame can lead to higher efficiency. So, why not make the data frame a million bytes? We said, a larger data frame gives us a larger efficiency. Why not make it a million or 10 million or a gigabyte? Higher waiting time, who? The efficiency will be higher in the normal case. That is, if we have a larger frame, we'll get a higher efficiency with stop and wait. What are some problems with a large data frame? If you think of stop and wait ARQ, the problem with a large data frame is that we need to rescind a lot if there's an error. If we have, if we lose a frame, if we transmit a frame and it's lost, then we lose a lot all at once. And we need to retransmit a lot. But if we have a smaller frame and we lose just one frame, we only lose a small amount of data and only need to retransmit a small amount of data. I'll show you a slide that mentions that, but try and draw that. Let's say we have a data frame. We've got two options, one large data frame or maybe several smaller data frames. So we consider two cases. A case one where we have one big data frame and case two where we have four small data frames. Same amount of data in total. We're still sending the same number of bits, but we have different size frames containing those bits. Let's say that when we transmit that, that there's a small error, a single bit error in the transmission. And that occurs randomly. So at some point in time, maybe this is a bit different from our other diagrams, but think of time going in this way. That is, I start transmitting the frame. So I've drawn it on a different axis here. I start transmitting the frame and I finish transmitting here. And if I use four frames, I start transmitting here and finish transmitting here. I'm not gonna show it going from A to B. Just look at the frames. And let's say that there's one bit error randomly in that time. And I'll choose a random time point. Let's say randomly we choose a time and it's here. What I'm trying to show is that we transmit all our bits and at some point in time, there's an error on the link and just one of those bits sent is in error. And it's the bit at this point in time, this red line. When B receives, look at the top case. When B receives this data frame, what does it do? It ignores it or discards it because this data frame, the top one has an error in it. And we said if there's an error, there's an error detected, we discard that frame and eventually A will retransmit. So in the top case, we essentially lose the entire frame due to one bit. And what we retransmit, eventually sometime later, what will be retransmitted, well, that entire frame. Sometime later, A will retransmit that frame because it had an error in it. Now in the second case, A transmitted four frames. There's one bit error. What does B do when it receives these four frames? And to make it a bit easier, let's put numbers in them. B receives these four frames, frame number two has an error in it. So we'll discard frame number two and eventually that will need to be retransmitted. It depends upon the scheme, we'll see some other schemes later, but if we don't receive this piece of data, it will need to be retransmitted at some time later. So of those four frames, one of them, the one that contains the error, will need to be reset. What I'm trying to show is that with smaller frames, in the case of errors, smaller frames lead to less retransmissions, which is better. The second case is better because we only have to resend a little amount of data. In the first case, we need to resend everything. So in the case of errors, smaller frames are better. In the case of no errors, we've said larger frames are better because they can give us higher efficiency. Also, larger frames are better because the header is a less proportion of the total frame, a large amount of data. Another factor regarding frame size, let's say our receiver buffer, the amount of memory at the receiver is 4,500 bytes. That's the size of the buffer at the receiver. Let's see, given different frame sizes, how much of that buffer space we can utilize. What if we transmit frames which are always 4,000 bytes in length? How many frames can we fit in our buffer at the receiver? One, okay? We can't split them into smaller chunks. So if we transmit a 4,000 byte frame, our buffer contains 4,500 bytes. So we can fit at the receiver one frame and we effectively waste 500 bytes because we use up 4,000 of that buffer and the other 500 bytes, we've got nothing in there. What if we used a different size frame? What if we used a 3,000 byte frame? How many can we fit in the buffer? Only one at a time. And we still waste, in this case, even more. That is, we've got 4,500 bytes of memory but we only use 3,000 bytes at a time. It's no use having that extra 1,500 bytes. Try a smaller size frame. Transmit frames of 2,000 bytes each. How many can we fit in the receiver buffer? We can fit two at a time, wasting just 500 bytes. So that's better than the first and second case. We don't waste as much as the second case, the same as the first, but we can fit two frames. Same amount of data though, 2,000, 4,000 bytes in total. Try a smaller frame. Four frames we can fit in but we still waste some space in the buffer. So there's no use having 4,500 bytes if we never use that 500 bytes there. Try 500 bytes. How many can we fit in the buffer? Nine frames. How much do we waste? That's the best case. That is, if we have smaller frames for a given size buffer in general, we'll be able to utilize that buffer in a better manner. We'll waste less space. It's the idea of you've got a box and the smaller the objects you put in there, the more efficiently you can utilize that storage space, that box. So smaller frames are better with respect to using that buffer space that we have available. And the previous case, smaller frames are better in the case of errors, less to retransmit, but in the case of no errors, larger frames are better because we have higher efficiency, less overhead. More time transmitting. Somewhere there's a slide that says that. Here, what size frames do you use? In practice, the technologies usually impose a limit on the frame size. If you're using Wi-Fi, wireless LAN, there's a limit of the frame size, I think 1,500 bytes, or a LAN 1,500 bytes plus some header. Why do they have limits? What are the good values of the frame size? Well, larger frames, less overheads due to header. The more data in there, the smaller the header is compared to the total size. Smaller frames can utilize the buffer space better like we just saw. If we have to retransmit, smaller frames are better, less to retransmit. There are a few other reasons we will not cover this one, there's the third one yet, efficient sharing. So there's a trade-off. There's no optimal or best frame size. You need to consider these different factors. Large frame size and small frames, we can usually compare under different conditions to see which one is best for a particular scenario, but there's no one best value. So that two issues that are relevant for all of these protocols, flow control and error control. What's the frame size? And what's the timeout interval? What's a good timeout interval? Any questions on those two issues, general design issues? So some of those reasons are why when you say download a large file across the internet, that the file is not sent all at once, it's sent in chunks, in frames, or in general packets, send part at a time. We've got two more to go through, but they're quite similar to what we've seen already. Go back to stop and wait. So with error control, there are three general approaches. Stop and wait ARQ we've gone through. Then the next two, go back in and selective reject. They are similar to sliding window. Stop and wait, send one frame, get an act back.  These two are based on sliding window and just small differences between them. Go back in, let's use the example. It comes up a bit small there. I have it bigger somewhere. This is go back in and the slide or the picture in the slide is slightly different from what we've been drawing it. We try to simplify instead of drawing the rectangles for the frames, it's just showing the timing of the exchange of those frames. Remember with sliding window, we can send multiple frames before we have to wait for an act. It's not data, act, data, act. Here we have multiple data frames. So frame where it says frame, it's a data frame. Zero, one, two, three, four and it sends multiple. And then acts in this case, they're called receive ready. B is saying I'm ready to receive more at different times. So this is the normal case, sending data frames, getting acts back. We wanna consider what happens when something goes wrong. Maybe we can write on there what's been received by B. B has received frame zero, zero and one. At this time it's received zero, one and two. Then it receives frame three, so zero, one, two, three. Then from B's perspective, a little bit later, it receives frame five, that's a problem. B has received zero, one, two and three. We expect to receive frames in order. We get zero, one, two and three and then we're waiting, we receive frame five, frame with sequence number five. That indicates something's gone wrong at B because if A is sending them in order, I should have received four next. But I get five. So that tells B there's been an error. And in this go back end protocol, what it does, it sends back a special act message saying, there's been an error, I'm waiting for frame number four. It's called a reject message. The name's not so important, but think of it as an act saying, I'm still expecting frame number four. I've got zero, one, two and three. I must receive them in order. Don't send them to me out of order. If there are errors, then you must retransmit. So, and what go back end does, is when B receives frame five, because it's out of order, it discards it, throws it away. And similar, a little bit later, B receives frame six. Also out of order, discards that frame. Why did it receive them? Because with the window in this example, A was allowed to send frame zero through to six. The window was seven, so it was allowed to send them, but one of them was lost. Frame four we see was lost. A didn't know that, but B detected something went wrong because it received five before it got four. So it sends back an act saying, I'm waiting for frame four, and it discards any subsequent frames, five and six in this case. So, when this reject message comes back to A, it realizes, ah, something's gone wrong. I have sent frames zero through to six, but B is expecting frame four. Therefore, I must go back and retransmit frame four, and five and six. I go back and retransmit N frames, where N is the number of frames since, and including the one that was rejected. So we see what happens. B is expecting four. That triggers A to retransmit frame four, five and six. Four, because that's what B is looking for. That's what it's expecting. Five and six, because A knows, since sending frame four, I've also sent five and six. Therefore, we'll retransmit those three frames. Let's go back N. When we have a sliding window, we keep track of the frames, and if we, in this case, B detects a loss, because it receives out of order, the source, A, will need to go back and retransmit multiple frames. In this case, five and six, as well as the lost one. Why? Well, because it keeps it simple for the receiver. Even though it did receive five and six, okay, it just ignores them. I must get them in order, and it doesn't need to buffer these two. So it only stores the ones that it's received in order. It discards five and six. The problem being is that A will need to retransmit five and six. So it makes it easy for the receiver, but it's a little bit of a waste in that we need to send five and six, even though they were successfully received. A variation of that is selective reject. We'll drop back to go back in in a moment, but let's just see the variation. B receives frame zero, has zero and one. Zero, one and two, then it receives three. Everything's okay. It's waiting, it's waiting. So this is selective reject, the next protocol. Four is not received, it was lost. It receives five. So what B does, again, it detects an error. I last received three, I received five, four's missing. So it'll send back a special act saying, effectively, please retransmit four. It's a reject message or a selective reject. I reject or I expect to receive frame four. But what's different from go back in is that we have zero, one, two, three. We're missing four, but we also have five. We buffer five as well. So we do store frames out of order with selective reject. It's a little bit more complex for the receiver because it must have some buffer space for this frame and keep track of the ones which are missing. It sounds simple, but in some implementations that adds too much complexity. And when it receives six, it's got zero, one, two, three. It's still missing four, five and six. When A receives that selective reject message, it knows, I'll need to retransmit frame four. So I'll retransmit frame four, but not five or six. So this is where it differs from go back in. We only retransmit the one that was rejected. And when frame four's received, then we have all of the frames in order. What is B expecting next? The next number is seven. So it sends back a receive ready or an act saying I now expect seven. So selective reject is more efficient in that we don't retransmit so much, but a little bit more complex at the receiver in that we need to buffer those frames received out of order and keep track of them. So that's the real difference between these two.  I'll go back to go back in and then we'll see a couple of other points. Do we have it? Go back in. At this point, we have just zero, one, two and three. We've discarded frame five. When we receive six, we don't do anything, just discard it. So we just have zero, one, two and three. When A receives the reject message, it realizes it needs to retransmit four as well as subsequent frames it's already sent, five and six. And we'll receive four, send receive ready waiting for five. Then we eventually receive five and then six. And send a receive ready saying I'm waiting for or an act saying I'm waiting for frame number seven. So that's the same point as selective reject. Go back in, more retransmissions but simpler, selective reject only retransmit the lost message but more complex. In this case, when we have the window, the receiver can detect an error. See, by receiving frames out of order, B knows there's been an error and can send this special act back saying really there's been an error, this reject message. On this example, the response messages are called reject messages, receive ready messages. And in selective reject, there's also a selective reject message. But in general, they're all acknowledgements. They're all just acts coming back using different act numbers and having different meanings. This example considers, what if the act is lost as well? Similar to stop and wait, we have a timeout. If we don't receive an act within time, we realize something's gone wrong. So in this case, A has transmitted five, six, seven. After transmitting each frame, it sets a timer. If it doesn't receive the act in time, that'll trigger a timeout and trigger it to, in this case, slightly different, send a special message to B saying, what have you received? A received request. And then it sends back an act. So if we lose an act, instead of retransmitting data, we send a special message to B saying, please send me an act again. That's what the meaning. And it sends the act again. And once we get the act, then we know at what number we're up to. We see that, okay, act with sequence number seven was lost. We timed out and that triggered this special message to be sent. Don't worry too much detail about the P-bit. This is just a message saying, please send me an act. Let's see what, just finish this example. What has been received? Zero up until seven has been received. Zero up until seven. And then the next zero has been received at this point, the data frame. A hasn't received an act for some time, so it times out. It asks B, please send me an act again. Which is this one. Saying what are you expecting? And B is expecting frame with sequence number one. Therefore, the next frame sent is sequence number one. And then we receive the frames received all the way through to two. And it would keep going. The main thing I want you to pick up from go back in is, and also selective reject, is how they detect that they've lost a packet at the receiver and what is retransmitted. So this is the part I want you to understand. The lost act is not so common, but lost data, and the difference between go back in and selective reject is important to know. Lose the data triggers a special act saying, please send again, send four, five and six again. In selective reject, lose the data. In selective reject, we lose the data. Please send, this is B telling A, please send frame four again. And frames five and six are buffered. Once we receive frame four, we've got everything back in order and we can move on to frame seven and so on. That's the main points of these two and the differences. The lost act not so important because it's not so common. Any questions on go back in and selective reject? Just understand that they use a sliding window and they have retransmissions in different manners. Any questions? Where do we get to in our slides? These describe how go back in works. Also about this special act that comes is when we lose an act. Again, just understand the difference between go back in and selective reject. That's the main point there. The details are not so important there. Selective reject minimizes retransmissions compared to go back in, but the destination needs a larger buffer and it's more complex at the destination. So that's bad. Turns out because of that extra complexity, go back in is more widely used in especially simpler devices. One thing that we've missed and it comes up in these slides and we've seen it before, haven't explained it. What is the maximum window size that we have available? There you go, just remember that. In sliding window, including go back in, two to the power of K minus one where K is the number of bits in the sequence number. I think it's on a slide somewhere, but if we have a K-bit sequence number, we're almost finished just to summarize. When we've got a K-bit sequence number, what's the maximum window size that we should use? Remember in sliding window, we used an example of, I think, the window size was seven. We had sequence numbers from zero through to seven. There's a three-bit sequence number. The maximum window size was seven. Well, the general rule is that with sliding window and with go back in, sliding window flow control, go back in error control, the same mechanism, the maximum window size is two to the power of K all minus one. So if K is three, it becomes seven. If K is four, it becomes 15. So that's the maximum window size. You can have smaller, but generally we set it to the maximum. With selective reject, it's slightly different. It's two to the power of, and the exponent is all K minus one. It's slightly different. So if K is three, with selective reject, what's the maximum window size? If K is three, then selective reject, we should have a maximum window size of four. Two to the power of two. If K is four, then the maximum window size is eight for selective reject. It's smaller than the other two cases. It's due to the way that the retransmissions and we keep track of the sequence numbers. If we set it larger than these maximum values, we'll get some other problems with the protocols. It's nice to study them, but we do not have time to explain why it's those cases. Go back in and sliding window, two to the power of K minus one. Selective reject, two to the power of all K minus one. We've done timeouts. Where are they used? You can have a look. Many protocols in many different networks and links make use of these concepts. Wireless LANs, LANs, dedicated links, say from your home ADSL router to your ISP, older links between devices. And in fact, the concepts of these protocols are used in the internet every day. When you download a file, you send an email, you use TCP as a protocol and TCP uses these mechanisms. We'll mention that protocol towards the end of the course, but flow control and error controller widely used in the internet and in communication systems. That ends this topic and next week we'll move on to the next topic of multiplexing.


The paper I am about to present regards the project material traces of pre-historical human activity in Campinas Forest carried out by State Ecological Museum in Warsaw. The project is developed on the base of cooperation agreement signed on 4 July 2017 between Campinas National Park and State Ecological Museum in Warsaw. Its main objective is the archeological server of Campinas National Park grounds based on possibility of wide query using state of art technologies but most of archeological site verification. Matagals, or at least part of them collected from the sites in Campinas National Forest, are starting the State Ecological Museum will be elaborated as well. In the future there is a plan to elaborate on the object collected in the Museum of Mazawian Ancient metallurgy in Pruszków. Campinas National Park is situated in center of Poland and practically borrows Warsaw acclimation to the north. It is located on central Mazawian lowland between the left bank of Wistula river and Zera river. The park covers over 38,000 hectares. The area of the Campinas National Park includes the Campinas Forest. Campinas National Park is one of the major Polish lowland farm sanctuary. It is estimated that the park is a natural habitat of half mainly of 16,500 native animal species. The biggest among the 50 mammals species is the most which is also a symbol of Campinas National Park. This animal were hunted to extinction in the middle of 19th century to be later reintroduced in the forest in 1951. Today their population totals 300 individuals. The most valuable park areas are under strict protection in the 22 protection zone. The biggest area of Zeraków is over 1,200 hectares. It is located just a few kilometers away from the Warsaw border. Campinas National Park was later founded on 16 January 1959 due to natural values and social importance in the year of 2000. Campinas National Park totals the protection zone was recognized by UNESCO as World Biospherical Reserve called Campinas Forest. The forest forms are clearly distinctive in the natural system. The forest territory compared to neighboring plains areas is characteristic for its big morphological variety. Moving from the north to the south there are current Vistula riverbed and interflute bank area which is marked with red color. Vistula flute plains, northern dune zone, northern pitbox zone, southern dune zone, southern pitbox zone, and ice and marginal valley bank and Wojcko-Pońska plains to the west and Warsaw plains to the east. The highest point is situated near Opalen next to Eastern Park border and it has 106 meters above sea level. The lowest is located in the western part of the northern pitbox zone which is 68 meters above sea level. The differences in relative height between dunes, summit and almost flat pitbox zones amount to 30 meters. Having made this introduction I would like to bring forward to the project guidelines. Its main objective is to carry out archeological server on Campinas National Forest based on possibility white quarry using state of art technologies. First stage of the survey is field verification of archeological sites located in Campinas National Park, grounds and registered in state archeological museum. Objects from these sites were collected during survey done in interwar period or they were finds reported in post-war period. Very often the location of these sites is limited to a very general description which makes it very inaccurate. The aim of our work is to locate sites known from the archives. Further state would be analysis archives collected from research of other centers and museums. Since 1978 a research program called archeological register of Poland systematical field working has been implemented in Poland. The survey of this area was carried out by Stefan Wojda team discovered in mid 80s of 20th century. Discovering a few hundred sites from different periods. Objects found in these sites joined the collection of the museum of Mazovian ancient metallurgy in Kruszkow. At the point only open land habitats filter meadows were explored while forests were ignored. This was due to methodology guidelines of the project. A part of the discovered sites is now covered by forest. An important query supplement is the LiDAR data analysis. We use already made visualization made for two big projects. Country protection against extreme hazards called ESOC and Campinas wetlands. They are planning to recalculate the clouds of points for our project especially areas not included in Campinas wetlands project. With these tools we do not only want to find characteristic feature but also predict places which might have been attractive for settlements of perfectly historical communities. The very important part of project especially for us museum professionals is science popularization. This is full fight during the number of excursion guided by archaeologists and a research worker from the park. The theme of each trip refers to a different historical period or in some cases to historical monuments remaining in a narrative around historical parks and places. Emphasizing and showing the relation between nature and human activities is particularly important. As well as showing changes in Campinas forest natural landscape over the years. Both natural changes and those influenced by human activities. As the next part of my report I shall present field work process in few cases. The field work last for one year and was focused on the unisones and their surroundings mainly in eastern part of Campinas forest. The first group of sites to be precisely located during the surveys known from archives or museum collections. Case one is Łóża d'Una. It is a group of sites located around Łóża d'Una. In state archaeological museum archives in Łomianki folder there is a private letter signed by letter escape probably Stefan Krukowski. Mark with date 13 January 1944 and closing sketch in the moment here is the plan of Łóża near Łomianki site. In state archaeological museum there is no data concerning flint material mainly outers interest. But there is free flint collections registered as Dąbrówka. But the time period between letters and the sites discovery is 19 years. So there is no certainly whatever this that was the same same sites. On the other hand highly improbable that such field work was carried out in 1944. Today Łóża d'Una is completely wooded and ground observation can be done only on small rooted fragments. Despite of few attempts over the time no material was find. And here we have the sketch which I was talking about. Here we have archive maps. Here we have matched the names of these villages which is on the sketch. We have to north is as you see on the other part of the sketch. And here is probably places which we were talking about in this letter. Case two is another dune. It's called Góra Ojca and in state archaeological museum collection there is published. There is set of one hundred and seven flints object named from the name of Laskiewicz. The site register card says Góra Ojca dune 1955 a gift from a doctor from Laskiewicz. However there's some additional information which I found during this study this collection on the labels. And first one say Laskiewicz district Góra Ojca dune behind the institution for visually impaired site two. So I thought that is where is site one in this cases. So I find the second label on the same box of object that changed our view of the analysed set. Laskiewicz district dune near a bus stop institution for visually impressed site one. Okay so I knew there is another site. We have here only one bus stop and this time when this collection come to museum. So we know there is not one site but two sites. So ultimately I try many times go on Góra Ojca dune and another dune near the bus stop. No of flint artefacts was found. And another case is biggest flint collection from Campinas Forest from Grohala. And it consists almost two thousand flint objects. It's collection dating back to final parallelity. Unfortunately based on archive data it is not possible to locate the site precisely. The only thing we are able to do based on labour description is discrepancy relation between particular clusters. In addition there is some value information on the relation between the site and the unspecified dune or wetlands. The survey gives a great news for complex study of highly interesting flint artefacts. The next project objective is verified sites known from the previous archaeological survey located in Campinas National Park. A great number of sites registered in mid-80s century were situated on fields or meadows. As a result of land acquisition policy of the park followed by afforestation of acquired areas. Today large number of these sites are covered by forest of mixed age. I want to present two examples of our field verification. First case is Cherakov. An example might be a case of the Nodific site in Cherakov. Mid-80s was located on a clear foreland of medium height dunes. Neighbouring ascending a path on a fallow land. Today it is covered by a young forest. No objects have been found on this site. Another case is the site of Chisova. It's known from the literature. The verification in mid-80s resulted in collecting more artefacts. Back then part of the site was covered by forest and remaining been a fallow land. Today a young mixed forest grows here including a great number of oaks. During the second visit of sandy clearing we found a few ceramics fragments dating to Bronze Age. Another object from this site. And these discoveries were made by surface. In the forest area studies almost integral tools as the leader analyzes. In our case we have comfort of using two open access visualizations. First is ISOC and another one is visualizations made for Campinas Wetland project. Even a view analysis showed a great number of World War II monuments and post-war military structures. Examples are shelters or trenches. But using a ladder we also have great visualizations show details of already known facilities. Like ammunition depot in Palmyra or Zamčesko's stronghold from 13th century. Also a number of mouths have been distinguished. Some of them probably barrels but without verification it's not possible to determine their functions. It is planned as I said before to recalculate the data obtained using a scanning especially for this project. Popularization as I mentioned previously is crucial to us. It is implemented during Saturday's excursion. Organized for all interest and participation. The idea is to show things that cannot be observed in the first glance as well as combined in nature. Related and historical, archeological problems. So far two events of this kind have been organized. And their claims deliberated chosen. There were two periods chronologically far from each other. First one is emesolithic and another was about interwar. The emesolithic excursion led around Bužova Góra Parabellikdjón. And similar in name people during what emesolithic archeology in the park was discussed. Including Palo.. Environment, fauna, house structures and their traces. As well as object dating to this period. We have also archery and flint mapping presentation. As the second team we have chosen ammunition depot in Palmyra active between 1929 and 1939. It's remaining unknown. But our idea was to show them as a group of dependent facilities to cooperate with us. We invited professor Zbigniew Tucholski from Institute of Science of History from Polish Academy of Science. An expert on railway and contemporary history. And Dr. Koloneł Krzysztof Gajku presented a course of the Tank Battle of Poczeka in 1939. The excursion route led around the railway and bank of ammunition depot against in Palmyra. In further future survey continuation is planned. With time we are planning to verification using method detector especially the Arias. With previously registered notification for example made by foresters. We have also intent to carry out test ranches or grills. Although this might be required consulting with the park staff responsible for environment protection. We should also be willing to cooperate with paleobotanics geologists and geomorphologists. Finally I would like to emphasize this excellent cooperation with Campinas National Park. Both research scientists and field workers who are keenly interested in our project. But also provide us with valuable information. Thank you for your attention.