This video is a highly efficient distillation of the S1 syllabus that perfectly balances core statistical concepts with practical exam-room tactics. It serves as an essential tactical roadmap for students who need to convert complex theory into high-scoring performance under time pressure.
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A LEVEL MATH - S1 - FULL SYLLABUS RUN THROUGH (NOTES IN DESCRIPTION)Added:
It's time to come everyone.
If you could bring back one old Formula 1 track or rule, what would it be? Uh, no speed limit in the pit line duration of this class. I don't know.
uh a message about why there is so inconsistency. We we will there is no inconsistency.
You just need to look at you just need to look at um the kind of data there is and then decide tea or coffee obviously tea.
I'm an old school person so I prefer tea.
Will you do concepts only? What do you mean? We're just going to do a syllabus run through. That's it.
Yeah, we look at the nature of data whether it's continuous or discrete.
And sometimes even if after finding the upper class boundaries the data is still u the data is still discrete the values are still discrete we take those values.
I'm getting an SSR don't worry.
I've shared my notes in the description.
There is a link in description where you can find my notes.
I thought in the me When did I start watching F1? In grade 8.
We have done the crocodile question already. I did it in the last stream. I did it in the stream last year. So, I've already done the crocodile question. I'm not going to do that again. Okay.
No, I'm not going to solve any questions. I'm not going to solve any questions. I'm not going to explain a topic from scratch. This is a syllabus run through.
Guys, I've already solved the crocodile question. Why do you want me to solve it again? Uh, watch yesterday's stream.
The crocodile question was from May June 2022, paper 5 variant 2.
Okay, let's begin now. The very first chapter measures of central tendency. Okay.
Now the concepts that we're going to cover in measures of central tendency is types of data. That's just a revision.
Mean median mode and interquartile range of ungroup data. Mean median mode and interquartile range of group data.
Variance and standard deviation.
Combined mean variance and standard deviation. Coded mean variance and standard deviation. Okay. Now all right now so let's begin with types of data very similar to O levels.
So so we have two types of data. One is numerical or quantitative. The other is qualitative. Numerical or quantitative for example height, age, mass, time and qualitative could be like blood group, gender, grades, color. Now numerical data is basically the kind of data that we focus on which can be divided into two categories. One is ungrouped or discrete.
So here we have this is ungrouped and or is scattered be also known as discrete and the other is group data. Now group data can further be divided into three categories. Is is volume really a problem? Because I don't think so. I've done everything I can.
>> Yeah. I don't think one year but I think the volume problem I don't see anyone else making that complaint. So I'm afraid the issue is at your end. I've done everything I can at my end. Okay.
Now group data can be divided into three categories.
One is where you have a group with discrete values. Okay. Then you have class intervals which are continuous like for example where one ends the other starts. And then you have class interval which are not continuous. Okay.
which are later turned into continuous by finding upper class boundaries and lower class boundaries. So that is something that we're going to have a look at in so when finding mean of variance do we have to show each value in a working and do the rest in the calculator? Yes, it's you have to show all the working and then you can you can actually work it out in your calculator but you have to show all the values. Okay, a paper attempting technique in I'll explain that towards the end. Okay, now what is mean? Mean is represented by Xbar and it is also known as arithmetic mean. How do you find out the mean of ungrouped data? It's summation X upon N.
Summation X means sum of all the values.
N means divided by the number of terms.
If you have discrete values, you multiply the frequency with the value of X and then you divide by the total frequency. If you have group data in for in terms of class intervals then what you do is you take the midpoint of every class interval. Okay. Okay. Now note mean can be misleading if there are extremely large or extremely small values. For example salaries or score of an exam. Now you know in some questions it asks you to comment which is an appropriate measure and which one isn't.
So you got to watch out for that. Take if you have extremely large or extremely small values then mean can be misleading. Why? Because these values or extremely large values and extremely small values disrupt the mean. If there's an extremely large value in the data set, it will shoot up the mean. If there's an extremely small value in the data set, it will drop the mean. Okay.
So, it is useful when the total value is also of interest. For example, total income or total production. Preferred when values are close to one another or when there is no outlier. Okay.
Now for example a class full of dumb kids and then there is one really smart kid. So what will that smart kid do?
That smart kid will just shoot up the average of the class. Similarly if it's the other way around then the average of the entire class will just drop. Okay.
So that is that is when the average is not a good representation of data when the values are not close to one another.
So if the values are close to one another the average is a good representation. Okay, median. What is median? Median is the middle term when data is arranged in either ascending or descending order. Okay, now if you have ungrouped data, how do you calculate the median? You do n + one upon the 2 term where n is the number of terms. If you have group data with discrete values, you add one and then divide it by two.
Okay, same as ungrouped data. But if you have class intervals, so in class intervals, this is how you calculate the if you want the median interval. What you do is you take the total frequency divide by two and see in which interval is that frequency falling. Or if the question says estimate the median that usually happens when you have a cumulative frequency curve. Okay. So if you've drawn a cumulative frequency curve, the question can ask you to estimate. But if you have let's suppose a histogram or you have data in class intervals and the question is not asking you to draw a cumulative frequency curve. It will ask you to calculate the interval in which the median lies. Okay.
Now it is useful when you have extremely large or extremely small values. Median is resistant to outliers for example salaries. So again a question might ask why is median a good measure? So remember it is useful when you have extremely large or extremely small values. Okay. Now let's talk about quartiles.
So in statistics cortiles are the set of values which has three points dividing the data into four equal parts. Okay. So there is there is Q1, there is Q2, there is Q3. Okay. So cile the data is divided into three parts. All right. Now how do you calculate the median? How do you calculate the lower quartile of ungrouped data? You calculate the median of the first half of the data. So I'm not in favor of using the formula n + 1 upon 4 or n + 1 into 3 upon 4. I'm not in favor of using those formulas. Okay?
I'm talking about ungrouped data. If you have ungrouped data, what you do is you calculate the median of the first half example so that you guys can understand how it works.
And then median we already know. If you have to calculate the upper quartile of ungroup data, then what you do is you calculate the median of the second half of the data because it's very confusing.
Sometimes it's 0.25, 25 sometimes it's 75 so it's very difficult to figure out which value to keep which value to ignore okay so that's why I have another method which is much more simpler now lower quartile what is lower quartile lower quartile is 25% of the total frequency and then we look at the corresponding value and remember lower the actual values of lower cortile median and upper cortile can be calculated through a cumulative frequency curve and even then what we're doing is we're doing an we're basically calculating an estimate estimate of the mean is is basically calculated when you're taking midpoints when you're multiplying frequencies by midpoint that is when you calculate the estimate of the mean. Okay. Now then there is intercortile range which is basically the upper quartile minus the lower cortile. Now here's an example of calculating the median and the lower quartile and the upper quartile of ungrouped data. So notice the first thing that we've done is we've calculated the median. What is the median? The median is 17. Now what the median does is it divides the data into two halves. So this is the first half.
This is the second half. Okay. If let's suppose there are two values combined with the help of which you're calculating median then you will ignore those those two values and you will look at the uh number of values on the left of those two and the number of values on the right of those two. They those will be your two halves. But right now 17 is your median. So that means all the values on the left hand side of 17 are fall into the first half of the data.
All the values on the right hand side of 17 fall into the second half. So if you want to calculate the if you want to calculate the interquartile range you will calculate the median of the first half. You can see that this is the first half. We've calculated the median of the first half that is our interquartile.
Okay. So this is what this is basically that is our lower quartile. Sorry. So this is Q1. And then if you want to calculate the upper quartile you will look at the second half of the data and calculate the median of that and that is basically your upper quartile. Okay. Now it's examples. So we'll do that also.
Don't worry. Now what is mode? Mode is the most frequently occurring term ungrouped if it's ungrouped or it is the value with the highest frequency if it is grouped and discrete or it is the value with the highest frequency when it is it is basically the sorry value.
It is the interval with the highest frequency when the data is grouped with class intervals. All right. Note there can be more than one mode meaning the data can be biodal. Now mode is useful when occurring when the most occurring value is of interest. For example, if if there is a new if if let's say you're in the t-shirt business and you want to start a new design, you will start by stocking the size that sells the most.
Not useful when the data set is small.
Okay. Now, I might as well disable it. Okay, I will enable it after every after the completion of every chapter and then I'll take your questions. Okay.
Okay. Now let's have a look at some examples. So here are some very basic examples of the mean, median and mode. Okay. And these are these are not positive questions. So don't worry about it. These are just some questions from the book. Okay.
Again, some questions from the book just to refresh the basics. Okay. So notice that here when we are calculating the mean, what we're doing is we're multiplying the frequency by X. Okay?
And we have made a separate column for FX. That's what I would recommend. Also here when we're calculating the mean, we're just manually adding up all these values. Okay? And then here when we're calculating the median, what do we do?
The median will be 100 + 1 that's 101 divide by 2 that's the 50.5th term that means we will take the average of the 50th and the 51st term. So here you can see this is this is the column where I've started writing down the cumulative frequencies. So the frequency of this is 15 then you do 15 + 25 that's equal to 40. Then you do 40 + 16 that's equal to 56. That means from the 41st to the 56th term from the 41st to the 56th term they're all equal to what?
they're all equal to three. So if your median is the 50.5th term, which means the average of the 50th and the 51st term, that will also be equal to three.
Okay? Now then we have some examples where we have to calculate the mean, median and mode of group data with continuous class intervals. You can see that since we have to calculate the mean, median, mode of um group data with continuous class intervals, we have to first calculate the midpoint. Okay. So we first calculate the midpoint so that we know exactly what to multiply the frequency with because and in questions like these it will always say calculate an estimate of the mean. Okay. In this question it will not say calculate the estimate of the mean because the mean that you're calculating is the exact value. It will here however it will say calculate an estimate of the mean because you're not multiplying the frequency with the exact value of x. You're multiplying it by the midpoint. Okay.
Okay. Then notice the question says calculate the median interval, calculate the model class. Now median interval means that you will look at the total frequency divided by 2 that's 160. Again you will start calculating the cumulative frequency and see in which interval does the 160th frequency lies.
So 30 + 80 what's that equal to? That's equal to 110. Then 110 + 90 that's equal to 200. So that means from the 111th all the way to the 200th term they're all in this interval. So that means the median interval must also be this interval.
Then the model class model class is basically the class interval which has the highest frequency. So you first look at the highest frequency that is 90.
What class does it belong to? It belongs to this class. So that means this is the model class right here.
Okay. Now notice that you have I know you guys might have some questions about the greatest and the least value of intercortile range. Now I've taken out some cases which we will discuss inshallah I think is chapter I don't know the how the units are in the textbook so I'm not too sure about that but Yeah.
Oops.
Okay. So, where were we?
Yeah.
Now note in order to calculate mean of non-ontinuous data it is important to convert it from it is important to convert it from non-ontinuous into continuous data to you non-ontinuous from non continuous to continuous data There is a disclaimer over here.
Sometimes even if you don't convert it into continuous data and you calculate the midpoint, it remains the same. Okay?
But the problem is when the data is starting from zero. If the data is starting from zero and then if you do not convert it into continuous, you will get your answer wrong. Okay. So first of all realize that this data is not continuous. Why? Because when the first class interval ends the other class interval does not start with that. Okay.
So notice that there is a gap. Okay. So this ends at 119. This starts at 120.
This ends at 129. Starts at 130. This ends at 139. The other starts at 140. So there is a gap. Okay.
So what you do is you basically convert it into continuous to calculate the midpoint. Now in this case in majority of the cases even if you don't you get away with it but the only problem is when the data is starting from zero. Okay. So is standard rule always convert it into continuous before you calculate the midpoint. Now the question is how do you convert it into continuous. So here's how you convert as far as the intercortile and that is concerned that we'll talk about shortly. Okay. First let's understand how to convert it into continuous.
So what you look at is the difference between where one class ends and the other starts. Okay.
So what is the difference between 119 and 120? The difference is one. You divide that by two. To find the upper class boundary you add it. You add 0.5.
And to find the lower class boundary of the next interval you subtract 0.5.
Okay? So I repeat the difference between 190 and 120 is 1. So 120 minus 119 upon 2 that is 1. 1id by 2 0.5. Now what are we going to do with the 0.5?
We're going to add for the upper class boundary.
Okay. and we're going to subtract for the lower class boundary. Okay? Now, this is what you do for every interval.
But here, you don't have to do that every single time because the difference is the same every single time. Okay? So, the difference between 119 and 120 is one. The difference between 129 and 130 is one. The difference between 139 and 140 is one. The difference between 149 and 150. So, it's one throughout. So that is why what you're going to do is you're going to increase this by 0.5 decrease this by 0.5 increase by 0.5 decrease by 0.5 increase decrease increase decrease increase decrease increase decrease increase increase decrease now the question is what to do with the starting value and what to do with the ending value now since throughout we have done five so that's exactly what you're going to do to the starting value the starting value you will subtract by.5 so it becomes 109.5 the ending value you will increase by.5 so that it becomes 189.5 aa Sometimes there is a problem. Sometimes the problem is that the data is starting from zero. Like for example, if it's time, if it's height, if it's length and the values are starting from zero. Now, if the values are starting from zero and there has been a question, let me see if I can find that question where the data is not continuous and it's starting from zero. Yeah. So here for example, here you can see the data is not continuous. There is a gap and it's starting from zero. So here you will add 30.5 to the ending value to find in the in the last interval to find upper class boundary but you will not subtract 0.5 from zero. Okay, you will not subtract 0.5 from zero because length can't be negative. Okay, length can't be negative. Okay, also here do I convert the data into continuous before finding midpoint or do I just find out the midpoint as is? Question for you guys. Yeah. In order to find out the midpoint to calculate the mean, do I convert the data to continuous before calculating the midpoint or do I do it as it is? Question for you guys, we first convert it into continuous.
Okay? Why? Because it's starting from zero. Okay? Because it's starting from zero, we have to convert the data into continuous and then we find out the midpoint. Okay? So here's the thing.
Here's the thing. You know, if you don't convert it into continuous, sometimes you get away with it, sometimes you don't. If you do convert it into continuous, you always get away with it.
So, where is it that you're safe? You're always safe when you convert it into continuous. Okay? So, that is why make a standard rule and always convert it into continuous. If you do not convert it, no, no, both don't work. Okay? Both don't work. If you calculate the midpoint of 0 and 9, what do you get?
You get 4.5. But if you calculate the midpoint of 0 and 9.5 we get 4.2 4.25 4.75 4.75 0 + 9.5 / 2 4.75 okay so you have to convert it into continuous and then calculate the midpoint and there is a question later on where you have to calculate the variance. So you will use the data set where you have converted into continuous and then calculate the midpoint. So long story short, this is long story short.
No, beta are not always 0.5. I've explained that you look at where one class ends and the other starts. Okay, you look at that difference and you divide that difference by two and then you add that to find out the upper class boundary. Subtract it to find out the lower class boundary. Okay? Now to make a histogram or a cumulative frequency curve, we must convert the data into continuous form. Remember that. Okay. To make a histogram or a cumulative frequency curve, we must convert the data into continuous form. Okay, now in the end we'll talk about that. Okay, we'll talk about that when how to find out the greatest and the least.
helps finding out how consistent the values are by telling us how close the values are to the mean. And what is standard deviation? Standard deviation is basically square root variance and it tells us how close the values are to one another. Okay. Then I've done some examples. Now these formulas are given in the formula sheet but still it's important and in the formula sheet I think only one formula is given not all the formulas are given. So beta yoga that you memorize the formula okay or at least memorize the one that is not given. So I've never really referred to the formula sheet.
So now you given it h okay formulas are given so you don't have to memorize but still I hope you guys have solved enough questions that you have all the formulas memorized.
Okay now this is something that we just discussed. I did a whole stream on this.
Now if you have ungrouped if let's say you have ungrouped data and you want to find out the standard deviation or the variance what you do is you square all the terms. So this is the data set you want to find out the standard deviation or the variance you square all the terms you sum up all the square of the terms you divide them by the total number of terms and from this you subtract the square of the mean. Okay that's one way to calculate standard deviation. If you want the standard deviation you take the square root. If you don't want the standard deviation, let it be as it is.
Okay. Then there are some more examples.
Some more examples.
Okay. Now, if you want to calculate the variance of group data with continuous class intervals, again remember what you have to do is you have to take out the midpoint. Okay. So, what is what is it that you're going to square? You're basically going to square the midpoint.
So you can see it's going to be summation f x² upon f minus the mean square. The question is what is x square? x square is basically the square of the midpoint. Okay. Now then some more examples. You can have a look at these examples. This is a question that I solved in the last stream just today only. So you can have a look at that. This is basically combined mean, variance and standard deviation. Important but not tested. I mean it's uh difficult but it's not tested so frequently. But still have a look at it. You never know when it might show up.
Then we have then again this is something that we just did today only. So I'm not going to go into detail of coded mean and variance extreme cover. We'll just go over the important key points. If something is added to all values of X, it does not impact the variance. If something is deducted from all values of X, it does not impact the variance. That means the variance remains the same. Okay. the variance after you add something from in to all values of x given that it's the same value and you've added it to all values of x it will remain the same or even if you subtract something from all values of x it will remain the same all right but as far as the mean is concerned let's have a look at that also if something is added to all values of x the same is added to the mean if something is deducted from all values of x the same is deducted to the mean deducted from the mean so this is an important formula to memorize if you've added a to all values of x then the new mean will decrease by a. If you deduct a from all values of x, then the new mean will also decrease by a. Okay. Now, is there a reason why? Yeah, of course. So, I've done you can have a look at these notes. You will understand what is the reason behind this. Okay. But detail then these examples I solved in today's class only. What I want to discuss is this. Okay. So, let's discuss this. Now this is a question where it says the time taken by 200 players to solve a computer puzzle are summarized in the following table.
Yeah, coded mean assume mean they're the same.
Find the greatest possible value of the interquartile range of these times. Now notice what has the question done? The question has done 60 minus 10. A here there's nothing to worry about. In order to find out the greatest possible value of the intercortile range, we look at the interval in which the upper cile lies. So what is the interval in which the upper cile lies? The upper cile will be 75% of 200. What is that equal to?
That's equal to 150 I believe. Okay, let's calculate the cumulative frequency also. So this is 16.
16 + 54 that's equal to 70. 70 + 78 that's equal to 148 + 32 that's equal to 180 and then 200. So that means the upper quartile lies in which interval?
This is the interval in which the upper cile lies. Okay. And the value is t is greater than or equal to 40 lesser than or equal to 60. Okay. So this is the interval in which the upper cortile lies. Now as far as the lower quartile is concerned for the lower quartile you will do 25% of the total frequency. So 25% of 25% of 200 that's equal to 50.
But how else will you calculate in which interval the upper quartile lies and which interval the lower quartile lies?
How else will you calculate that? So this is equal to 50. Now what is the interval of the 50th frequency? The interval in which the 50th frequency lies is this one. T is greater than or equal to 10 less than 20. uh this is the interval in which the lower cortile lies. Now since we have to calculate the greatest possible difference that means we will pick the greatest value of the upper quartile and the least value of the lower cile. So we will do 60 - 10 that's equal to 50. Okay, pretty straightforward. Right now, this is where it gets a little complicated where so that was one example. Here is another example. Example be closer.
So this is a question where what is it that we're given? We are given let me make some room for some working.
Yeah.
So what is it that we're given? We are given the population of 150 villages.
Notice that the data is not continuous.
Okay. Notice that the data is not continuous. But of course upper cortile is 75% lower quartile is 75%. Lower cortile is 25%.
Understood the formula.
Okay. Now so the question here says find the greatest possible value of the intercortile range for the population of the 150 villages. Okay. Now yeah now notice that here we're talking about population that means people and the data is not continuous. Okay the data is not continuous.
So what do we do? We first have to convert the data into continuous form.
Okay. So 50 and this will be equal to 850.
This will be 850. This will be 1250.
Okay. And this will be 1250. And this will be 2050.
This will be 2050.
And this will be 3150.
This will be No, wait a minute. Aub difference. 2050. This will be 3250.
Sorry. 3250. And this will also be 3250.
And then this will be 4850.
Okay. Now continuous and not continuous basically means that if there is a gap in between it's not continuous. Okay.
And this nature of data what is the nature of data? The nature of data is discrete. Okay. We're talking about population that means the nature of data is discrete. Okay. That means you can only take whole numbers. You can't take you can't take 0.5 or 0.25. Okay. So to find out the greatest possible value of the intercortile range once again you will start by listing the cumulative frequency. So 8 8 + 12 20 then 70. So 70 + 48 that's equal to 118 118 + 32 that's equal to 150. So 75% 75% of 150 what is that equal to? 75 of 150 that's equal to 112.5.
So 112.5 means that the interval in which the upper cortile lies is 2050 2050 to 3250. This is the interval in which the upper cile lies. What about the lower quartile? So lower quartile we will do 25% of the total frequency. That means 25% of 150. So 0.25 into 150 what's that equal to? That's equal to 37.5.
37.5 means this interval which is from 1 1250 to 2050. Okay.
Now please note that what we have done is we have basically we have basically converted the data to find out the upper class boundary and the lower class boundary. Okay.
So in the previous part, why 60 - 10 shouldn't be 59 - 10 because the interval is not equal to 60. Okay, good point. But they what is it that we're dealing with?
We're dealing with time. You're you're right that the data is not equal to 60.
So it has to be less than 60. But the problem is what is the value that you're going to take that's less than 60? Are you going to take 59.5? Are you going to take 59.99?
Are you going to take 59.95? Are you going to take 59.99? Which value are you going to take? Because what is it that we're dealing with? We're dealing with time. We know time is a continuous variable. Time is a continuous variable which basically means that it can take any values between two integer values.
Correct? No. See where is 59.5 coming from? 59.5 does not make sense. Why not 59.99?
Why not 59.99?
So here we can't really pick a value that's less than 60. So that's why we go with 60. Okay. Why? Because these are continuous values. Here the nature of data is discrete.
Correct? The nature of data is discrete.
Now since the nature of data is discrete that means we will take since since we have to find out the intercortile range which basically means the upper quartile minus the lower quartile.
So what value will we take for the upper quartile? Will we take 3250? No, we will not take 3250. Why? Because this is the lower class boundary and this is the upper class boundary. Upper class boundary is what it must be less than.
So that means we will not take 3250. We will actually take 32 49. Okay. So we will take 3249.
So if you take 3250, the question has accepted it. Okay. If you do take 3250, the question has accepted it. And then from this you will subtract the lowest value of the lower cile. And what's that? That is 1250.
Okay. So 3249 -250 that's equal to 1,999.
Okay. But if the question if let's say you take 3250 - 150 the question has accepted it. Okay. Why did we not take 1 to 51? Please understand. Okay. Whenever you work out the lower class boundary and the upper class boundary, it's like limits of accuracy. Take a limits of accuracy of upper bound lower bound. So the value is always greater than or equal to the lower class boundary and the value is always lesser than the upper class boundary. Okay. And to answer your counter question that in that case why did we not take 59.5 over here? Because this variable is such that it's continuous. So there is no fixed value that we can take that's less than 60. So that's why we go with 60. This variable since we're talking about population, it only takes discrete values. So we have to take a value which is less than 3250. And what's the greatest value less than 3250 that we can take is 3 to 49. So that's why we take 3 to 49. Okay. Why? Because the data is always when you convert it into continuous, it's always less than the upper class boundary. And as far as the lower class boundary is concerned, we can take 1 to 5 0 because it can actually be equal to 1 to 5 0. Okay.
This is the rule. This is the rule.
Okay.
All right. Now then there is another example. Let's have a look at that also.
Let's have a look at another example.
Now exactly sir math prep with Ahmed Bal recently video in which he has summarized everything about when to convert the data into continuous and when to uh and how to calculate the greatest intercortile range. So, make sure to check that out. I uploaded it on my story also.
So, you guys should check it out. The name of his channel is Math Math with Ahmed Bal. Sorry, not math. Yeah, Math Prep with Ahmed Bal.
Yeah. Okay. So, make sure to check it out. Math prep with Ahmed Bilal. Okay.
So he made a nice summary of this and he also made a chart where he showed when with check marks when should you convert when should you not. Okay. So make sure to check it out. All right. Now let's have a look at this question. What is the greatest possible value of the intercortile range of data? What are we dealing with? We're dealing with number of chocolate bars. Now quick question.
Chocolate bars are discrete or continuous. We're talking about chocolate bars. Are chocolate bars discrete or continuous?
frequency curve frequency say no you if you're taking median from frequency you don't don't do plus one how to calculate percentile percentile levels you treat it as percentage so if you want to calculate 70th percentile 70 upon the total 70 upon 100 into total frequency then you look at the corresponding value okay it's discrete okay chocolate bars are discreet you can't cut it into half I mean you can okay But as it is they're not in half as it is they're whole numbers. Okay. So that means the nature of data is discrete.
Okay. Now check number two. Check number two is is the data continuous or is it not? No it's not continuous. So we will convert it into continuous. Will we convert it into continuous? Yes or no?
No we will not convert it into continuous. Okay. Why? Because when we do convert it into continuous problem.
If we do convert this data into continuous and we take those values, we run into a problem because after we convert this data into continuous, the problem that we run into is that the values have 0.5 in them. And that's a problem. You can't have 0.5 of a chocolate, can you?
So when you convert the data set into continuous which you might have to if you if you're making a histogram or a cumulative frequency but the problem that we run into now in order to calculate let's suppose now we calculate the upper quartile and the lower quartile. So let's calculate the cumitive frequency we have 18 + 24 that's 42 + 30 that's 72 + 20 that's 92 + 8 that's 100. Okay. So let's calculate the range. Let's calculate the interval in which the upper cile lies. Upper cile will be 25% of 100 which is equal to 25. That means it lies in this interval 11 to 15. No wait this is the lower quartile not the upper quartile. This is the lower quartile.
The upper cile will be 75% of the total frequency which is equal to what? Which is equal to 75. Okay. and that lies in the interval from 31 to 50.
Okay. Now, if you're thinking that in order to calculate the intercortile range, we're going to do the greatest possible value of the upper quartile range and we take that as 50.5 and the lowest possible value of the lower the range in which the lower cortile lies and take that as 10.5. That is wrong.
Why? Because the nature of data is discrete. So, you can't use 0.5. Okay?
So instead the correct way of calculating the interquartile range is to do 50 minus 11 and what's that equal to? That's equal to 39.
of the question. The context of the question is are the values discrete or are the values continuous? If the values are continuous, you take them with the decimal. If the values are discrete and if after converting them into continuous they have 0.5 in them, you don't take 0.5. If after converting them into continuous they don't have 0.5 in them like for example here even though the values were discrete the nature of the variable was discrete but and we took we took what we got after converting them into continuous why because even after converting this data into continuous we got whole numbers that is why not 49 because yeah actual values these are class intervals these are not class boundaries these are class intervals Okay, these are class intervals. These are not class boundaries that we take 49 they go. We took 3249 because these were class boundaries.
These were class boundaries. Okay. And these were class intervals. So that's why I took 3249. Here I am taking the original data set. That's why I'm taking the values as it is.
Okay.
So please give examples of continuous that are continuous like I understand how chocolates are discrete. So here is here I give an example of continuous.
Here is an example of continuous example convert continuous second example can't seem to find it but I will strongly recommend that you guys check out the video that sir bilal posted it's pretty Okay. No.
So that's it as far as hang on. This is a question. Uh summer camp.
Okay. So you guys will find the answer to your question here.
Okay. Now here to give the interval in which the median lies will we convert it into continuous? Yes or no? Question for you guys. Here to give the interval in which the median lies will we convert it into continuous? Yes or no?
Question for you guys.
Yes, we will. We will. Why? What is it that we're measuring over here? We will children.
Okay, you guys are not reading the question. We are measuring the time.
We're not talking about children here.
Okay, we're talking about time. We're not measuring children here. We're measuring the time that the number of children take. Okay. So, this is the variable and the nature of the variable is time which we know is continuous.
Okay. So we're not measuring children.
We're basically measuring the time that they take to complete an arithmetic test. All right. So now in order to state in which class interval the median lies, we will first convert this to continuous. So 0.5 30.5 30.5 45.5 45.5 to 65.5 then 65.5 to 75.5 then 75.5 to 100.5 okay now state which class interval contains the median. So median basically means 50% of the total frequency. 50% of 250 what's that equal to? That's equal to 125. Now let's make a list of cumulative frequencies. So 21 21 + 30 that's 51. 51 + 68 that's 119. 119 + 86 that's 205.
That's good enough. We'll get the job done. So the interval in which the median lies is 66 to 75. So will we give our answer as 66 to 75 or will we give our answer as 66.5 to 75.5? Question for you guys. Will we give our answer as 66 to 75 or as 66.5 to 75.5?
Question for you guys.
Are you guys sure? 66 to 675 wrong.
Second say 66 to $75.
Just to give you guys a disclaimer, even if you give the class interval with 65.5 and 75.5 acceptable here acceptable.
This is from 2022. This is before 2022.
2024 onwards. You will notice that they follow. Okay. Now, condone means tolerate. Okay. Condone means tolerate.
2024 onwards you will notice that they have they are strictly following a single rule. Okay. And that rule is I'll explain that rule is if the class intervals are not continuous if the class intervals are not continuous 50. The question said median that's why we calculated 50%. Okay. So if the class intervals are not continuous then you convert them into continuous and if after converting them into continuous the values that you get are acceptable as in the variable is such that it can take those values like for example values chocolate bars can't people humans can't be halfed but chocolate bars can't be halfed but time can be halfed so values time case may we can take the continuous values and you can see that here also they have accepted this one as well okay so I'll I'll not put a cross here I will put a check mark over here okay I know it is confusing okay this is the only part of graphical representation of data that is confusing. Okay.
So yeah, for example, for example, the median, for example, if the question said, for example, if the question said, find out the greatest possible intercortile range, what are you going to do then?
If, let's suppose the question said, find out the greatest possible intercortile range. It's part and let's say the question said that you have to now find out the greatest possible interquartile range.
Let's suppose if the question said find out the greatest possible intercortile range. So where does the 75% of data lies? 75% of 250 that's equal to 187.5 that means this interval.
Okay. So if the question said find out the greatest possible interquartile range in that case we will do 75.5 minus 25% of 250 that's equal to 62.5 that means in this interval so - 45.5 if let's suppose the question said that okay so 75.5 - 45.5 = to 30 why are we taking the values in decimals why are we taking the values and decimals here. And why did we not take the values and decimals over here? Why? Look at the look at what it is that we're measuring. We're measuring chocolate bars here. Here we are measuring time. Time we know can take decimal values. Chocolate bars since they're discrete by nature, they cannot take decimal values. That is why we are going to take 0.5. Okay? Why not 75.4?
No. See why not 75.4? I know why you're saying 75.4 4 because it must be less than 75.5. No to why settle at 75.4 why not 74.999 the Q is we can't really figure out the greatest possible value to take we take the actual value now there are no more questions that follow this rule. So we have done all of them and there.
Let's take a break. Let's take an Amas break. What do you guys think? Let's take an Amaz55.
Yeah, let's take an Amas break.
Z table.
Again, if you're here with the intention of learning that from scratch, we're not going to do that. Okay? We're not going to learn it from scratch. We're just going to go have because I've already taught normal distribution in detail.
I've taught discrete random variable in detail. I've taught binomial, geometric, everything in detail. So I'm not going to do it from scratch. And I've taught how to use the calculator as well. Okay?
So watch the normal distribution stream in which I've explained how to use your calculator. Watch the binomial stream in which I've explained how to use a calculator. Okay? So let's take an amas break. Let's take a 5m minute break. Five actually 8:55.
So let's take a 15-minut break. Take 15 to 10 minute break and then we will continue.
Okay.
All right guys, welcome back.
So let's continue.
So we were looking at graphical representation of data speed so that we can finish before 10 or 10:30 max.
So the concepts that we have to cover over here are box and whisker, stem and leaf, histogram, cumulative frequency curve. Okay. So first let's have a look at box and whisker. Now what are all the values that you need to make a box and whisker? You need the least value, you need the lower cortile, the median, the upper cortile and the greatest value.
Okay. So this is how you make a box and whisker. In the marking scheme you will notice that in the end they have drawn a line. Now not that the line is part of box and whisker. It's just that it makes it easy to read where the lowest value is and where the greatest value is.
Okay. Now what is the benefit of using a box and whisker? Let's have a look at that. The benefit of using a box and whisker is so that it's easy it's basically easy to look at a box and whisker and get an overview of the data. Okay. It's it's useful if you want to have a look at the summary of the data. Okay.
And if you want to have a look at the all the cortiles, the lower quartile, the upper quartile and the median. Okay.
Then what is the drawback? The drawback of using a box and whisker is that you really can't see all the values because box and whisker only contains some specific values. So the biggest drawback of using box and whisker is so that you cannot see individual values in a box and whisker. Okay. Now box and whisker makes it easy to see the spread of the data. The lesser the interquartile range, the more reliable and consistent the values are. The results are close to one another. So please understand that the interquartile range, the lesser it is, the better it is. Why the lesser the better? Because the intercortile range is tells you the difference between the upper quartile of the data and the lower quartile. Okay? So come in quartile range the the more consistent the values are. Okay? Now skewess of the data. So our data is symmetrical when the mean is equals to the median which is equal to the mode. And in that case the difference between the upper the Q3 minus Q2 is equals to Q2 minus Q1. Okay.
This is basically what happens when data is normally distributed. Why is it called box and whisker? Because it's literally a box and whisker. So this is a box. Okay. And this is the whisker.
Okay. That's why it's called box and whisker.
So lesser the median. No I meant lesser the intercortile range. The lesser the intercortile range, the better it is because intercortile range tells you consistency of the values. What if no unit when labeling the axis example for population? Uh, yes, I'll do P3 stream.
Sara, what do you mean by that? What if no unit when labeling the axis? I'm not sure if I understand that. Okay. Then you have positively skewed. So positively skewed is when basically this is what it looks like. So the difference between the upper cortile and the median is greater than the difference between the median and the lower cortile. Okay.
And it's the examples are a difficult test wealth distribution number of pets people own. So you will find you will find the frequency is higher on the lower side when so you will basically find that when the values on x are low there is a high frequency and the frequency with greater x values are less. So for example, how many people do you expect to own number of pets? Okay.
So you will expect a lot of people to own few pets and very few people to own a lot of pets. Okay. So here on the x-axis you have the number of pets.
Okay. And on the y- axis you have the frequency. So how many people do you expect to own a lot of pets? You would expect very few people to own a lot of pets but you'd expect quite a few people to own few pets.
Okay. So this is an example test. This is just something that I've written to understand the purpose of skewess. But this is not something that's tested.
Okay. And then you have negatively skewed. Okay. This is not something that's really tested. So we can skip that for now. Okay. Then you have stem and leaf. So stem leaf is something that's that was part of O levels in IGCSC also. So basically what you do is you look at the data and stem and leaf you have two types of stem and leaf.
Back to back backto-back stem and leave diagram is useful when you have to compare two sets of data like for example if you want to compare uh marks of two clauses if for example you want to compare the performance in or if for example you want to compare the masses of two data sets so that is when it's useful to draw a backto-back stem and leaf diagram for example there was a question about comparing batteries of two different mobile phone companies okay so because we're comparing the batteries of two different mobile phone companies. It's a good idea to make a backto-back stem and leaf diagram so that we have we have all the values in front of us. Okay. So what is the usefulness of stem and leaf diagram? It is useful to compare we're talking about backtoback it's useful to compare two sets of data. Con since we list out all the data they are unsuitable for large data sets. So a stem and leaf diagram doesn't matter whether it's backtoback or a single stem and leaf diagram. It's not practical if you have lot lots and lots of values. Okay.
Now, here are some questions. So, in this question, the question said, find the median and intercortile range of the pipes produced by company A. So, how did we find out the median? Median was basically, so altogether you have 19 terms. Okay, we're talking about company A. So 19 + 1 is 20. 20 upon 2 is the 10th term. So, what's the 10th term? The 10th term is make sure that you have a look at the key so that you understand how to read the stem and leaf diagram.
So in between you basically have the first two decimal places after zero and then on the leaves you have the final decimal place. Okay. So 334 basically means 0.334 for company A and 331 basically means 0.331. Okay. So always look at the key. The key will help you unlock this stem and leaf diagram. Now notice how did we calculate the lower cile? So five divided the data into two halves. Okay. So we have 1 2 3 4 5 6 7 8 9. We have nine values before five and we have nine values after five. So in order to calculate the lower quartile we took the median of the first half of the data. So in the first half we have nine values. So 9 + 1 upon 2 that's the fifth term. What's the fifth term? 1 2 3 4 5.
So 0.348. You can see that's the median.
Then for the upper quartile we did the exact same thing. The fifth term on of the second half of the data. So 1 2 3 4 5. So that's 0.366.
Okay. Now then you have a question where you're given the data set of company B as in you're given the lower cile, the median, the upper cortile and you then have to draw box and whisker for both a then after that you have to make one comparison between the diameters of the pipes produced by companies A and B. So what comparison can be made? There are plenty of comparisons that can be made.
So you can see that company B has a wider spread, okay? Has a greater in intercortile range, which means company B is more inconsistent than company A or company A is more consistent than company B. Here you go. Or you can make a comment on skewess that company B is more positively skewed compared to company A. Okay? Or another com another comparison that can be made is by looking at the extreme values. Company B has a smaller least value compared to company A and a greater greatest value compared to company A which means that company B has more extreme values. Okay.
So that was an example of a stem and leaf with box and viscera. Then let's talk about histogram. Again histogram is something that's part of O levels in IGCSC. So there's not much to discuss except that we take class intervals on the horizontal axis and we take frequency density on the vertical axis and frequency density is basically calculated using this formula which is frequency upon class width. So here's the frequency 320. What's the class width? The class width is basically the difference between the two intervals.
Very easy, nothing too complicated.
Okay. Now if you want to calculate the mean remember to calculate the mean we always take the midpoint before we multiply it by the frequency. Okay. So this is the same example that we have done regarding calculating the mean.
Then let's talk about the cumulative frequency. A cumulative frequency graph is used with data that has been organized into a grouped frequency table. The cumulative frequency graph can be used to find estimates of percentiles and cortiles. As the data is grouped, it is not possible to find the actual values of these statistics, but you can find an estimate with the help of a cumitive frequency graph. An important thing to remember is that if the question says less than a value on the horizontal axis, you look at the corresponding value on the vertical axis. If the question says greater than a value on the horizontal axis, then you do total minus the corresponding value on the vertical axis. Like for example, if the question says find the number of students who scored more than 50 marks, so you look at the corresponding value of 50. But since the question says more than 50 marks, you'll subtract it from the total. If the question says find out the number of students who scored less than 50 marks, you will look at the you look at the corresponding value of 50 and whatever that value is, that's going to be your answer. Okay. Okay. Then the heights in cm of 160 sunflower plants were measured. The results are summarized on the cumulative frequency curve. Okay, the question says use the graph to estimate the number of plants with height less than 100. So you look at the corresponding value of 100.
Whatever is the corresponding value that's your answer. But if the question said more than you would have subtracted the corresponding value from the total frequency in which case it would be 100 minus 60. Now then the question says use the graph to estimate the 65th percentile of the distribution. So you will look at 65% of the total frequency.
What is that? That's 104. You will look at the corresponding value of 104. And whatever is the corresponding value, that's your answer. Remember before you solve questions like these, make sure you get familiar with this scale. Okay.
Use your graph to estimate the interquartile range of the heights of these plants. So intercortile range is calculated by first calculating the upper cortile. Since we're using a cumulative frequency curve, we don't have to worry about continuous or anything of that sort. Whatever is the corresponding value of the upper quartile of 75% of the total frequency that's your upper cortile then 25% of the total frequency that's your lower cortile subtract them and that's basically your intercortile range okay then some more examples a backto-back stem and leaf diagram once state an advantage of using a stem and leaf diagram compared to a box and whisker to illustrate this information so we can see all the values on a stem and leaf diagram. That is an advantage.
Whereas a box and whisker plot only takes a few specific values. There you go.
Again, intercortile range. Okay. This is something that we just discussed. Okay.
So, will line touch? No, the line is not supposed to touch the origin, but the line is supposed to touch the least value. Like for example, frequency example.
Let me see I can if I can find it in a past paper question.
Hang on.
terror. Let me come across. Let me look at an example and tell you.
Just give me a minute.
In order to calculate the mean, you need to find out the midpoint after converting the data into continuous if it's not and then you're going to calculate the mean. Okay, that's the rule. If you want to calculate the mean and the data is in class interval form and if it's not continuous before you calculate the midpoint you have to make sure that the data is continuous and then you can go ahead and calculate the Then we'll move on.
I know. I know. Screen is gone. I know.
I know. I'm aware of that. Don't worry.
Just give me a minute.
Okay.
Speech. Speech.
here here.
So basically this is a question where you have to draw a cumulative frequency curve.
So notice that you will first find out the cumulative frequency.
Okay. So the cumulative frequency will be 4 12 + 14. What's that equal to? This is from February March 2023 paper 5 variant. Obviously in February March there's only one variant. So 26 then 51 then 51 + 7 that's 58 58 + 2 that's equal to 60. Okay. Of note that the values that we mark when we are making a cumulative frequency curve is always the upper class limit the upper class boundary. Okay. So these are the points that we will mark when making a cumulative frequency curve. Okay. And then once we have marked these values, okay, once we have marked these values, then what will we join the cumulative frequency curve with?
Okay to go cumive frequency curve end. So we will join it with another point and what will that point be? That point will be x is 30 and y is zero.
Okay.
Just give me a minute. Let me just solve this question and show you guys how it works.
Okay. Now, so let's pick a suitable scale. The scale that we will pick is notice that X is starting from X is starting from 30. So we don't necessarily have to start from zero. We the values that we need to start marking actually start from 60. So that means we can start let's pick a nice suitable scale. Okay. For the horizontal axis and for the vertical axis as well. So what's a nice scale that we can pick? Let's put some thought into it. A nice scale that can be chosen is 2030.
Greatest value 240. Okay. Now, so maybe if I do 40, 40, 80, 120, 160, 200, 240, 280. Yeah, that will work. So 40 80 120 160 20 200 240 280.
Okay. On the vertical axis let's take 10 20 30 40 50 60. Okay. Okay. So what's the corresponding value of 60? The corresponding value of 60 is 4. So where will we find 60? in between 40 and 60.
So that the corresponding value is four.
Okay. Then what's the corresponding value of 90? That is 12. But I see roughly values but that's not the point. Point is close 110 26. So 110 26. So I'm assuming somewhere over here.
26. No.
Okay. Then we have 140 and 51. So 140 and 51 will be somewhere over here. Then we have 180 and 58. So 180 and 58 means somewhere over here I believe. Yeah. And then we finally have 240 and 60. So 240 and 60. That's right here. Okay. Notice that the points that we have marked are these. Now, this is a very very rough drawing. Okay? But we're not going to leave it here. What is the value from where the data is starting? The data is starting from 30. So, that means we will join it to 30, 0. So, we don't necessarily have to join it to the origin, but we have to join it to the starting point of the horizontal axis.
Okay? Or whatever it is, whatever the variable is. So we have to join it with the starting value of the variable. So what's the starting value of the variable? That is 30. Okay. Now there's no need to join the there's no need to make a jagged line because I've started from zero. So this is 0 40 80. So there's no need to make a jagged line or anything. Okay. Now so can we show upper cile, lower cile and median from the diagram? If it's zero then we join to the origin. Yes. From the diagram itself or not allowed?
Yeah, you will. If you've made the diagram then the question will ask you to do your calculations from the diagram.
Okay. Okay guys. So that is it as far as graphical representation of data is concerned. Now let's continue to chapter number three which is probability. Okay.
Now probability is another area where the examiner is basically one of two areas where the examiner can get really creative. So I believe the difficult questions that we can come across are questions from topics such as probability and permutation and combination. So there are three concepts. There's normal probability which involves sample spacing tree diagram. There is conditional probability and then there is independent and mutually exclusive.
Okay. So what is probability? You guys don't need an introduction. I think we all know the basics. Okay. So if let's suppose if let's suppose the question involves two dice or if let's suppose the question involves tossing a coin and then rolling a dice. So it's always a good idea to first do sample spacing so that you have all the possible outcomes in front of you. Okay. Now if it's without replacement that means when you take something out you put you don't put it back. So what happens is the color that you take out or whatever it is that you take out decreases and the total also decreases. Okay.
And remember if the question if the order is not specified then we consider all possibilities like for example if the question says one is red the other is green then this can be fulfilled two ways red green or green red. So whenever there are multiple ways of achieving the same outcome what do we do we add the probabilities like for example here whenever the question says and/ both/all we multiply the probabilities sum of probability of total probability is always equal to one probability of winning is equals to 1 minus the probability of losing probability of same is equals to 1 minus the probability of different probability of pass equals to 1 minus the probability of fail so oppos if prob the sum of the probabilities of opposite events is always equal to one okay So probability of an event not occurring is basically equal to 1 minus the probability of that event occurring. So whenever the question says and or both or all we multiply the probabilities whenever the question says or we add the probabilities and whenever there are multiple ways of obtaining a certain outcome usually happens when order/antity is not specified then we find probabilities of all possible orders and then add them up. Like for example, if the question says one is red, the other is blue, that means blue, red, red, blue. If the question says that at least one is blue, that means one blue, two blue, three blue depending on whatever the total is.
Jump conditional probability.
So conditional probability is it means to find the number of favorable outcomes under a specific condition. So number of favorable outcomes with the given condition divided by the probability of the given condition. So probability of A given B basically means the probability of A intersection B which means the probability of both the events occurring together divided by the probability of B and whatever is written after the word given that is what comes in the denominator. Find the probability that the game is delayed given that it rains.
So the probability of rain comes in the denominator and in the numerator it is basically delayed and it rains. Find the probability that the student is a male given that the student is left-handed.
In the denominator, we put the probability of the student being left-handed. In the numerator, we put the probability of a student being a male and a left-handed. Okay? Then there are some examples that you guys can go through.
So then there is independent.
Yeah, independent events. So two events are set to be independent when the outcome of the first does not change the outcome of the second event. So remember the test for independence. If the probability of A into the probability of B is equal to the probability of A intersection B that means they are independent. If they're not that means they're not independent. Okay. And we've done uh if you watch the probability stream you will see that I've done multiple examples of this. When do we have to use factorial and probability?
That is when you have multiple ways of achieving the same outcome. Like for example, if you're tossing four coins and you want to find out the probability of getting one head. So one head can be head, tail, tail.
Let's say you're tossing three coins and you want to find out the probability of getting one head. So it could be head, tail, tail. It could be tail, head, tail. It could be tail, tail, head. So what instead of working this out, you find out the number of ways in which these three letters can be arranged. You find out the probability of one and you multiply it by the number of arrangements. Okay. Now, mutually exclusive. If two events cannot occur together, they're considered mutually exclusive. Okay. Like for example, getting a head and tails at the same time. They're mutually exclusive. Okay.
So, if the probability of a intersection b is equals to zero, that means they cannot occur together. They are mutually exclusive. If it's not, that means they're not mutually exclusive. Okay.
Then there are I've done some examples about mutually exclusive and independent actually which I've solved in the probability stream. So make sure to check it out. Okay. This was actually the most difficult independent question there has been so far. So you'll find it in the probability stream. Make sure to check it out. Okay. Now let's talk about discrete random variable. So again discrete random variable is something that I've streamed recently. Let's go over it.
It's one that can only take a number of countable distinct values such as 012.
For example, the number of heads or tails when flipping a coin is a discrete random variable. So in discrete random variable, the number of trials are fixed. For example, a coin is tossed three times. A dice is rolled four times. Three balls are drawn at random.
Probability distribution table is drawn to find probabilities. A coin is tossed three times. Draw a probability distribution table for the number of times head is obtained. So if you're tossing a coin three times, you could get zero heads. You could get one head.
You could get two heads or all three heads and then you find out their probabilities. That's basically how you make a probability distribution table.
Okay. Now, sum of all the probabilities is equal to one. If you want to find out the mean or the expected value of X, you basically take X multiply it by its respective probability and you add the values. If you want to find out the variance, you take the square of each value, multiply by the respective probabilities and subtract the mean. So, subtract the square of the mean. Okay.
Then some more examples. Again, something that I've streamed recently.
So if you want to learn this topic, you can do that. Although I really hope that's not the situation. So one thing to keep in mind when solving coin related questions in discrete random variable is this. Notice the question says Jacob has four coins. One of the coins is biased. So such that the probability of obtaining a head is seven upon 10. The other three coins are fair.
Jacob throws all four coins once. So whenever you have a coin related question what you do is you fix the order in which the coins are thrown and then you look at the different outcomes that you can get. All four coins are thrown once. Therefore different orders are not to be considered. However, different outcomes will still be considered. So we will fix the order in which the coins are tossed and then in those coins we will consider all the different possible outcomes. Okay.
Okay.
What time is the stream till? Probably another hour, I guess.
Okay. So, some examples that I've solved.
Yeah. All right. Now, then we have binomial distribution.
Okay. So, what is binomial distribution?
In binomial by means two. Okay. In binomial we only are interested in two outcomes favorable and unfavorable.
Whatever outcome is considered favorable the other automatically becomes unfavorable. Okay? Even if there are multiple. So in binomial distribution remember the number of trials are fixed.
How do we identify that a question is of binomial distribution? By looking at the number of trials. The question will always say 10 days are selected at random or the question will say 12 people are selected at random. The question will say okay uh seven weeks are selected at random.
Okay. So always look for a fixed number of trials. A coin is tossed 10 times.
Five objects are drawn at random. A dice is rolled five times. Okay? And there must only be two outcomes. If there are more than two outcomes, we group them into two. So for example, we're interested in winning. So whatever is not winning, we consider that as unfavorable. We're interested in red. So whatever is not red, we classify that as not red. So for example, we're interested in a train being late. So we'll just look at the probability of late that is favorable and everything else becomes unfavorable. Okay. Now each outcome must be independent of the other. That means head and tail probability of one event should not affect the probability of other events.
So that means in those questions in without replacement questions you cannot you cannot use binomial. Okay.
Probability of favorable outcome plus unfavorable outcome should be fixed. P + Q equals to 1. Trials must not be more than 30. Okay, so here's a question which says a manufacturer makes two sizes of elastic bands, 40% of the bands produced are large bands, 60% are small bands assuming that each pack of these elastic band contained a random selection. Calculate the probability that in a pack containing 20 bands. So you can see that the number of trials are fixed which means this is binomial distribution. You look at the number of trials which means the number you look at the trials are fixed which means this is binomial. Okay.
And then it says more than 17 bands.
This means 18 19 20. Okay. Sir histogram will not connect the bars if range or is not connected. What to do then? You convert it into continuous. Okay. Okay, if the range is not if the class intervals are not continuous, you convert them into continuous by finding upper class boundaries and lower class boundaries. Okay, now there are some questions which are here also see the question says 14 cars are chosen randomly pointing towards that this is binomial.
Okay, so no please don't watch this for admins.
This is totally irrelevant for admins. I said then there was a question. Yeah, there are these questions where you have to find out the value of n. So to find out the value of n, we use we can use logs if you're a P3 student or instead what you can do is you can do trial and error. Again, I've solved questions like these in the binomial stream. If this is something that you're seeing for the first time, make sure to check them out. Okay.
Okay. Now let's do geometric distribution.
Okay. So, geometric distribution, how is it different from binomial distribution?
In geometric distribution, the number of trials are not fixed and they continue until a desired outcome or success is achieved. If you want to use logs, go ahead, use logs. Examples, the number of times you have taken an exam until you pass or the number of job applications before first acceptance. The number of times calling before someone answers.
So, always look for until so and so is obtained. Okay. Or until a head is obtained until until until until a tail is obtained until a six is obtained. So look for look for until so and so is obtained that is pointing towards that this is geometric distribution. Now how do you find out mean in geometric distribution? You do one minus the one upon the probability of favorable outcome. That's how you find out the mean. Now for the first success to occur on earth trial R minus one trials must fail. That means if let's suppose for you to clear the test for the first time it takes you six attempts that means the first five attempts must have resulted in failure.
Or if let's say it takes you it takes you three attempts to get the first head that means the first two attempts did not get ahead. Okay we basically resulted in failure.
Then there are some rules to follow which you can either memorize or you can use your common sense. Okay. So these are some standard rules. This is plan B.
Otherwise you can use common sense. Like for example the question says greater than or equal to three. Greater than. Let's suppose the question says that it takes at least three attempts to clear the test. Now taking at least three attempts to clear the test is the same as not clearing the test in the first two. Okay. So again, you can memorize the standard rules or you can use your common sense. Okay, I always prefer that you use your common sense.
So if let's say the question says that it takes less than eight throws to obtain a six. So less than eight could be what? Could be 7 6 5 4 3 2 1. Now that will be lengthy because you'll have to work out 7 6 5 4 3 2 1. You'll have to work out seven different probabilities. So instead what you can do is you can find out the opposite and subtract it from one. What would be the opposite of less than 8? The opposite of less than 8 would be greater than or equal to 8. That's why we're subtracting from one. So the probability of taking eight attempts to achieve your first success means not achieving success in the first seven. So instead of finding out greater than or equal to 8, you find out the probability of failing seven times. You'll get the same answer. Okay?
So then I've done a worst case scenario which is basically plan Z. In plan Z what you can do is you can use the sum of GP. Okay, which I do not recommend. That's why I've written it in red or plan Z.
Okay, I do not recommend it. Okay, so I always recommend that you either use your common sense or you memorize the standard rules. If you can't do any of that, then you use the sum of GP. Why?
Because all these terms will form a geometric progression and you can use the sum of n terms formula. But that's that's plan Z. Plan Z. Okay.
Okay. Now some past paper questions that you guys can have a look at later. There are some questions like for example over here find the probability that Rajes is successful for the second time on his 10th attempt. Okay.
So this means that the second success should be on his 10th attempt. As far as the first success is concerned that can happen any time in his first nine attempts. Okay. So in the first nine attempts we just have to make sure that we have one success and that can happen at any attempt it has but it has to be in the first nine. So that is why we're using binomial over here. So this is for the first nine attempts. In the first nine attempts we just want one success and the remaining eight will be failures. And then when it comes to the 10th attempt we multiply it by 0.3 because the 10th attempt must be a success and it has to be a second success. Okay. Now let's now have a look at normal distribution.
Normal distribution may normal distribution is also another very straightforward type of a question formula.
Okay. Now let's suppose the question says the distribution of the heights of some plants is normal and has a mean of 40 and standard deviation of two. Now remember there are two important parameters for normal distribution that is the mean and the variance. If the question says under 42 then what you need to do is you need to first convert these values to second example. Sorry wait wait wait.
Yeah, let's do this. Yeah, so this is an example question which says the distribution of masses of some baby parrots is normal and has a mean of 60 g and a standard deviation of 5 g. Find the probability that a randomly selected bird is under 63. So this means that you have to find out the probability of X being less than 63. So the first thing you will do is you'll find out the zcore. How do you find out the zcore? By using the formula x - mu upon sigma. Now whenever it says less than you always take five of the ZV valueue. So if it's less than 0.6 you'll take five of 0.6.
That's your answer. If the question says more than what you do is once again you find out the zcore but because the question says more than you always do 1 minus five of that value. Okay. So whenever it's greater than you always subtract it from one. And whenever it's minus like for example if you want to calculate Z of less than minus A which means you will basically calculate 1 - 5 of A because you cannot calculate PH of a negative value. So in order to calculate PH of a negative value what you do is you instead do 1 minus PH of positive A. Okay. So some rules that you guys can go over. If let's say you want to calculate the probability of Z being greater than A less than B you will do PH of B minus PH of A. Okay.
Then then there are questions where you do normal approximation.
Yeah.
I've taught how to use your calculator to convert to basically solve normal distribution related questions and I've done a stream of normal distribution recently only. That's why I'm not going to go in a lot of detail. Okay.
But let's talk about let's talk about normal approximation. If I inverse is less than five. Okay, if you want to read the table backwards and f inverse is less than five less than 0.5.
Let's have a look at that.
Yeah, when reading the table backwards if the value so for example here when we take when we take five across the equals to sign it becomes five inverse of 0.2.
So if we have to take five inverse of a value which is less than 0.5 we simply subtract it from one and multiply the other side of the equals to sign by minus one. Now it does not matter whether you multiply the other side of the equals to sign by minus one or the same side. Take care both of them work.
You just have to make sure that you multiply one side of the equals to sign by minus. Okay. So f inverse of 0.2 can't we can't figure it out. So instead we take f inverse of 0.8 and make sure that you multiply either the other side of the equals to sign by minus or the same side doesn't matter. multiply the other side or the same side of the equals to sign by minus doesn't matter.
Okay.
Okay. Now normal approximation. How do you convert from binomial to normal? You do that whenever the number of trials is really large. Like for example, if it's greater than 30. So you will find this either most likely you will find this in a binomial question and then all of a sudden it says a random sample of more than 30 values. Okay? could be 125, could be 200 or whatever. So if the question says a random sample of 125 vehicles, notice that previously it was talking about 11 vehicles. We could use binomial. But now all of a sudden it's talking about 125, which is a hint that we have to convert from binomial to normal. Now what are the conditions? The number of trials must be greater than 30. So can you take questions where they say the value has to be I've solved those questions in the normal distribution stream. Okay. If n is equals to 29, do I still use binomial?
Well, theoretically, yes, you still use binomial, but you will not come across questions where n is 29. Okay? I've done questions where it says within so and so in the normal distribution stream.
Please do check it out.
So, let me see if I can find one. Yeah, see it says that is within 0.5 cm of the mean. That means more than 0.5 of the mean and less than 0.5 of the mean. So you add 0.5 and you subtract 0.5. Now let's suppose if the question says within so and so standard deviations of the mean. Okay. If let's suppose the question says within so and so standard deviations of the mean then whatever standard deviation it is talking about you will take that value only.
Like for example if the question says within two standard deviation of the mean. So that means your zcore will be 2 and minus 2.
Why? Because when you find out the zcore, what are you going to do? You're going to take that value, subtract the mean from it, divide it by the standard deviation.
Let me see if I can find a question where it says okay. So here for example if the question says find the probability that these three adults have spk h what is sp okay bio students cheese.
Three adults.
Adults key.
Just a minute.
Okay, three adults are chosen at random from this population. That's okay. Since the question is talking about adults, that means we're looking at the adults crata.
Okay.
Find the probability that each of these three adults has S bp within 1.5 standard deviations of the mean. That means it will be mean + 1.5 sigma or it will be mean min - 1.5 sigma.
Okay. Notice that when you convert them to zcores notice what happens. Okay.
When you calculate their zcores notice what happens. So how will I calculate the zcore of this value? I will do mu + 1.5 minus mu sorry mu + 1.5 sigma minus mu upon sigma. So what happens? Mu and mu cancel. Sigma and sigma cancel and here I will do mu - 1.5 sigma - mu upon sigma. So what happens now? Mu and mu will cancel.
Okay? And sigma and sigma will cancel.
So you don't really have to work out the values because the question says within 1.5 standard deviations of the mean and the zcore basically tells you how many standard deviations away from the mean the value is. So simply take 1.5 positive and 1.5 negative. That's it.
Okay. So yeah.
Okay. Now then whenever you convert from binomial to normal what you do is you apply continuity correction. Okay.
And how do you convert from binomial to normal? Basically the number of trials times the probability of favorable outcome becomes the mean and the number of trials times the favorable outcome times the unfavorable outcome becomes the variance. Okay.
So I repeat the number of trials times the favorable outcome becomes the mean and the number of trials times the fable outcome times the unfavorable outcome becomes the variance. Okay. And since we are using standard deviation in normal distribution, so don't forget to always take the square root of this value when using to calculate the zcore. Okay. Now, calculator tricks, I've already made a video on it. Okay. I've already made a video on it. Repeat the 1.5 thing. They go, if the question says, calculate the probability that it is within 1.5 standard deviations of the mean, that means it could be more than 1.5, it could be less than 1.5.
So whatever standard deviations it says 1.5 standard deviations of the mean. So that means the zcore will be 1.5 only. Why? Because the zcore tells you how how far away it is from the mean or how many standard deviations away it is from the mean. That's what the zcore tells you. So you don't really have to find out what the mean is and then add 1.5 times the standard deviations.
Notice that when we convert these x values to z values, what do you end up with? you end up with 1.5 and minus 1.5.
Why? Because what is Z? Z tells you how far away is the value, how many standard deviations away is the value from the mean. So if it says 1.5 standard deviations away from the mean, that means it could be positive 1.5 standard deviations or it could be negative 1.5.
So you take Z as positive 1.5 and negative 1.5. Okay? If it says within the mean then you take mean plus that value and mean minus that value. Okay.
All right. Now don't forget to apply continuity correction when converting from binomial to normal. Again something that we have gone over in the last stream. You can do that also or you can you can either memorize the rules or you can use your common sense to figure out how to convert it. Okay. Okay. Now let's have a look at the final final chapter which is permutations and combinations. Okay.
Permutation combination detailed stream key.
So we'll just go over it briefly.
Okay. So is there any case where it's greater than or equal to for a prox? No, no, there is no case when there is greater than or equal to. Uh, no actually there is a case when there is greater than or equal to. Greater than or equal to means at least in which case you subtract 0.5.
Okay.
Now with permutations and combinations, let's just go over the important bit.
Okay. The important bit is this bit.
Yeah.
So if the question says if there are two objects and the question says they're not together, what can we do the bit? This is this is what we do in a syllabus run through. We don't solve anything. We just do a syllabus run through. I think today is your first syllabus run through. I said that I'm not going to solve anything. I'm only going to go over the concepts. Okay? So not if you want to if you want to see how it's done then you can watch the stream okay you can watch the stream in which I've covered it in detail. So not together and there are two objects we can use the fence post method or you can do total minus together doesn't matter.
If the question says no two are together and there are more than two objects then you can only separate them using the fence post method. Like for example there are three fence and all of them must be separated in a in an arrangement then you can only do fence post method.
Okay you cannot do total minus together.
But if let's say there are more than two people or more than two objects like for example three friends and the question says not all are together then you cannot use fence post. You can only use total minus together. And those of you are asking what fence post is please watch the stream that I did on permutations and combinations. I did that uh yesterday only. Okay. So the question says no two together that means completely separated. Or if the question says not all together that means at least one is separated. Okay. And here you can see what it means. So here you can see all are together. Here you can see no two are together. And these are cases that not all are together. Okay.
So not all are together. For this you do total minus together.
And for no two are together the only way you can do this if there are more than three pe more than two people is fence post. Okay.
Now then there are some examples which I've solved combinations which import Like for example, probability with combination. That's an important part which I want you guys to have a look at.
Wait, before that have a look at this also. Nine people are to be divided into a group of four, a group of three, and a group of two. If you're dividing into groups and the groups are not identical, then you do not divide by factorial.
Okay? But if you're dividing into groups and the groups are identical then what you do is you divide by factorial. Like for example 12 people are placed at random in three groups of four people each. Find the number of ways this can be done. So because every group is a group of four. So and there are three groups. So you divide it by three factorial. Okay. I did a 4hour stream yesterday covering permutations and combinations. Okay. So don't be lazy.
All you have to do is do some clicks and you can watch that stream where I've covered literally every concept there is. Okay? So I'm not going to solve a permutation combination question because there's no point. I'll just be wasting everybody's time who and majority of the people have seen that stream. So I'll just be wasting their time. Okay?
Which questions probability which requires selection? Like yeah for example here when using combinations with probability we have to consider repetition as well. Basically we have to consider repetition as well means we have to treat repeating letters as unique. Okay. So this was a question that we solved in yesterday's stream as well or maybe a question similar to this. I don't know. The question said four letters are selected at random from the nine letters in the word andromeda.
So total number of A's you can select is 9 C4 which is equal to 126. Okay.
Find the probability that this selection contains at least one D and exactly one A. So at least one D means that the number of Ds has to be greater than or equal to one and exactly one A means that A has to be one only. So what is the probability of selecting one D and 1 A. So notice that since there are two D's to choose from, we have done 2 C1 for the D and we have 2 C1 for the A and we have done 5 C2 for the rest. Why?
Because once we cross out the A's and D's, we're down to 1 2 3 4 5 letters and we have to choose two. So that's 5C2. Uh 90,720, you're right.
So what if you have to what if we select two Ds and one A? So we'll do 2 C2 for the D's, 2C1 for the A and then 5 C1 for the remaining letter and we get 10.
Okay. So this is basically an important thing that you must consider and that is that whenever we're using combinations with probability, we have to consider repetition as well. That basically means we have to treat treat repeating letters as unique. Okay.
Okay. Uh the question that was most difficult, it was a February March question.
It was a February March 2023 paper 5 variant 2. Okay.
By at least I meant if the question is such that you have to convert it from binomial to normal and the question says okay find out the probability that it's find out the probability that the length is at least five. Okay, five means greater than or equal to five which means we will not take five. will actually take 4.5. Okay, that's a rule of continuity correction. Rule of continuity correction says that if it says at least which means greater than or equal to in that case what we do is we subtract 0.5 rules.
So at least means greater than or equal to in that case we subtract. Okay, if it says greater than in that case we add.
If it says lesser than in that case we subtract. But if it says less than or equal to then we add. Okay, crocodile question I solved yesterday.
Crocodile question I solved yesterday only. Okay guys, now last and not the least, let's go over a bit of the paper attempting pattern.
Paper attempting pattern to discuss and then that's it.
Now these are all the topics that we have. Okay, we watch yesterday's stream.
I solved plenty of questions in yesterday's stream of permutations and combinations where we did questions of different groups, same groups. Okay, we did a lot of questions in yesterday's stream. Okay.
So, if question says selection, do we use C only? And when we use if you're talking about selection, then you use combination. If the question says arrangement, then you use permutation.
Okay.
What's that?
Could you show how to find the median of stem and leaf diagram? They go find out the median. That median divides your data into half. Now, if you look at the first half and calculate the median of that, that is the lower cile. If you look at the second half and calculate the median of that, that's your upper quarter. Simple as that. Okay? Now, so these are the kind of questions that you might expect. Okay?
Since I'm discussing the paper attempting pattern, I want you guys to pay attention.
So that's one way to make you guys pay attention.
Now there are some topics of which the question can't really get too difficult.
Okay. So for example measures of central tendency/spread is questions majority of the times are very easy. Same goes for graphical representation of data. Now they may be lengthy. Okay. But the concepts are very straightforward in the sense that the examiner can't really surprise you.
Okay. The areas where I think the examiner can get really creative is probability and permutations and combinations. Okay, these are the areas where I think the examiner can get really creative. The rest of the topics, given that you guys have solved enough past papers, you will notice they're very repetitive. Okay, if you do 10 questions of normal distribution, you will notice that it's become very predictable. If you do 10 questions of binomial, you'll notice that it's become predictable. Sometimes very rarely geometric distribution is also a surprise but it's very rare. Same goes for discrete random variable. Very rarely is a question of discrete random variable also complicated. Majority of the times they're easy. Okay. When when it's scenario based you know when you have like coins being tossed that is when it gets complicated. Other than that it's very simple. Okay. So where I'm when I'm drawing a circle these are the areas where the examiner can really test you. Other than that it's pretty simple. So when you start first make sure that you tackle questions of measures of central tendency/spread then questions related to graphical representation of data then you do normal distribution. Why normal distribution? Because normal distribution honestly I think is the most straightforward chapter there is in S1. Okay normal distribution is rather straightforward.
Even if it's inverse normal, even if you have to read the table backwards, those questions are also very repetitive. Yes, you have to watch out for continuity correction. You have to watch out for for example uh being careful about reading the table. So this is the kind of stuff that you have to watch out for or you have to watch out for like for example if you worked out the probability of one and let's say the question is talking about a sample of 500 then to find out how many of them are of 500 greater than so and so whatever it is that you've worked out you multiply it by 500. So there are things that you can forget to do a normal distribution but conceptually it's very straightforward. Then there is binomial distribution which is usually followed by normal distribution normal approximation question where please do not forget to use continuity correction.
Okay. So remember if you come across a binomial distribution question which then follows normal approximation the continuity correction use the binomial distribution. And if it follows normal approximation, do not forget to use continuity correction.
Okay. All right. And then do geometric distribution.
Okay. Now the paper is how long the paper is? 1 hour and 15 minutes.
Correct? Every paper is 1 hour and 15 minutes. So you don't need to think of it as 1 hour and 15 minutes. You need to think of it as 75 minutes. Okay?
You guys should be done with these discrete random variable.
So discrete once you're done with this round then in the next round attempt discrete random variable and rounds divide to make it make it simpler.
Whoops.
So when you start the paper let's start from the beginning. When you start the paper, I want you to attempt the question of measures of central tendency/ spread, then graphical representation of data, then normal distribution. Okay? Do these three questions and then in the next round do questions of binomial distribution.
Okay? Then after binomial distribution do geometric distribution then Then after geometric distribution do discrete random variable about the final round attempt probability and permutation and combination. Okay. Okay. Now the paper is worth the paper is uh will give you you'll have 75 minutes but for the last two chapters make sure that you give yourself at least 30 minutes.
Okay. You have 30 minutes to solve a probability question and the permutation and combination question. Okay. Now it is possible. Why? Because the topics that we have mentioned they're lengthy.
Yes. Binomial distribution might require a lot of working. Geometrical geometric distribution might require a lot of working. Same goes for normal, but they're not that difficult. So in S1, you literally don't even have a single minute to waste. So just make sure that for 75 minutes, you are 100% locked in.
Doesn't matter even if a dinosaur walks in, it should not take your attention away. Okay? Make sure that you're 100% locked in. Doesn't matter what happens in the examination hall, you should not lose track of your paper. Why? The problem is like for example let's suppose you're working out all the possible scenarios and permutations and combinations and then you know you just get distracted even for let's say 5 seconds you will lose every you will basically lose all your chain of thoughts and you will have to start again. Same goes for probability. Let's say the probability question is a complicated question where you have to think of multiple scenarios. Okay you get distracted for 5 seconds you lose all your chain of thoughts. Okay. So do not get distracted even for a single second. Now there are there are some concepts there are some topics of which you can use your calculator to check your answer. So for example you can use your calculator to check answers of mean variance and standard deviation. Watch videos of it. Even if even if you don't find a video on my channel just write the model number of your calculator and write the concept it is that you trying to find the video of video will do the job. Same goes for binomial distribution. I made in the streams I've taught how to use a calculator for binomial distribution. Same goes for normal distribution either be up and you can actually use it for discrete random variable as well. Okay. So these are four areas where you can use your calculator to check your answers. So make sure that you do that and please don't think of a calculator as a substitute of working. It's your exam.
It's not your calculator's exam. Okay?
So you're only going to use your calculator to check your answer. Okay?
And I strongly advise you to do it especially when it comes to normal distribution and binomial distribution.
In fact, in all these areas, I strongly advise you to use your calculator to check your answer. Emphasis on the word check. Okay, so I'll enable the chat now. This is it. I wish you guys the best of luck for your exam. I will be waiting to hear from you guys how it goes.
Okay, you will you'd always want to do more. You'd always want the answer to this question, that question, this scenario, that scenario. My advice, just pull the plug. Okay, pull the plug and just focus on whatever it is that you've done so far, not what can be done. Because after a certain point, you can only make it worse. Okay? and you've reached that point. After this point, you can only make it worse. You've reached that point. So, just go into spectator mode. Okay? Just pull the plug now and go into spectator mode. Don't solve anything. Don't don't pick a pass paper to solve. Just just go into spectator mode. Okay.
Okay.
Thank you, sir. Will you remember me as the F1? Yeah, I'll probably sir. Please keep P1 streams of I'll keep P1 mocks but not P1 streams. Sir, hope you're well. Could you please confirm if you take values to be discrete when we have to find out probability and permutation questions?
What does that mean? Values to be discreet when we have to find out probability and permutation questions.
I'm not sure I understand. What do you mean by that? Uh, get a good night's sleep. I hear you guys have your exam in now. When's your exam? Is it AM or PM?
Math PM exam P.M. Okay.
So, don't wake up too early that you start feeling sleepy again because you won't have time to take your afternoon nap. So my advice wake up around 10:11.
Okay?
Can we do chronologically? Well, you can do it however way you want, but what's important is that you get your confidence level high. And you will get your confidence level high by attempting the easier questions by securing marks.
See, when do you find yourself in a position to take a risk?
When you know that you can afford to lose. So again, I'm not saying okay, you get those you you put yourself in a position where even if you get those questions wrong, there there's nothing to lose. But if you solve the difficult questions first and god forbid you're unable to do them, then your confidence level will drop. Okay? And if your confidence level drops in a math exam, trust me, it can only go downhill from there. Okay? So don't let that happen.
Start with the easy questions. Get your confidence level high so that God forbid even if you get the difficult question wrong. And let me tell you when it comes to permutation and combination question look at all the difficult questions that we have solved. We've solved the crocodile question. We've solved the alligator question. The happiness question which took away our happiness clearly. And we have solved those dart competition question. There was always one part that was difficult.
Okay. There's always one part that's difficult, not the entire question. The crocodile question had one difficult part. Same goes for alligators. Same goes for uh what was that? The happiness question. So there's one part that's difficult, not the entire question. So don't let that one part dictate your the performance of the entire paper. Just treat it as one part only.
Okay?
So that's it fellas. I will stop here.
I'll wait to hear from you guys tomorrow.
Moa and please guys whatever leak that you come across whether it's genuine or not just report let's have a look at the leak and let's see if if this actually comes in the paper or not and then if it does come then you come back and report no whether you think it's legit or not I can already see that there are very um a lot of people have started scamming already and there are some completely nonsensical questions s that are circulating. Okay? So don't fall for that. Okay? But whatever it is that you do come across, whatever it is that you do come across, report it beforehand. Not after the exam, just report it beforehand. Okay.
So yeah, pull the plug. the Nikico Rosberg method. That's you're right, Acha. Thank you for making these streams lively.
Thank you for always tuning in. I appreciate that. And I noticed that some of you were asking, will you remember me? So, yes, of course, I will remember you inshallah. So, feel free to drop by in if I'm streaming next year and even though if it's if you've moved on and you don't you're not studying math anymore. So, feel free to drop by and say hello.
Uh yes wana I do remember I do try to remember as many students that I can take online guy drawback you can't really put a face next to a name so you have to put in a lot of effort to remember the name so with all the plus points of online this is a major drawback of studying online so let's Stop here guys. Finally. I'll see you guys inshallah in another stream in another P1 stream maybe. So take care and have a good night's sleep. Enjoy the paper tomorrow. Allah hop.
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