 Hello, welcome to this lecture on bio-mathematics. Currently, in this section we are discussing statistics relevant to biology and we discussed about averages, standard deviation and some kind of distributions. So, we said the data, the simplest thing that we can learn from data is average and then standard deviation and in fact, to present the data instead of presenting a set of numbers, we can present it as a distribution which will have more meaning, which will convey much more than a set of numbers would convey. So, now, we discussed various distributions, we discussed distribution and said that typically many things in nature fall into something called normal distribution in quotes. So, now, today we will understand what normal distribution is. So, we just said in the last lecture that normal distribution has the form p of x is equal to a e power minus x minus x average whole square with some constant b. So, this is this particular form, normal distribution as this particular form. We will try and understand more about this in this lecture. We will pretty much understand more, learn more thing about this. So, the title of the lecture is, so this is in statistics itself and the specific title of today's lecture is understanding normal distribution. In this lecture, we will try and understand pretty much everything about normal distribution. Like whatever you need to know, know about normal distribution, we will try and understand this lecture. At the end of this lecture, I hope you will, you will know pretty much, you will be very confident to deal with normal distribution and pretty much learn everything about normal distribution that one should know. So, now, what do we, what is the simplest thing that we should learn about normal distribution? So, let us say you are given a normal distribution, which is like a normal distribution, which has this bell like shape. What do, what can you know from the, from a first look? What do you know from a first look? What all things you can learn from this? You know that the probability is high around 150. So, we had yesterday discussed, let us say this is above the height of a, this is the probability distribution of the height of students in a particular class. So, if x is 150, let us say this is 150 centimeters. So, this then, if you know that the probability has a peak, the maximum number of students will have around, heights around 150. And as we go to either side, the probability of finding students with lesser than 150 and more than 150 kind of decreases. So, this is something, which we can know from the first look as we discussed last time. But what more can we know from just looking at it? Can you tell, we look at here, can you tell what is the average? By looking at the curve, can you tell what is the average or what is the standard deviation? Well, you should be able to do this or you should be able to do pretty. So, you can at least roughly say by just, just looking without doing any measurements. Just by looking at the axis of the graph and then the range of the graph, you can tell and just looking at the graph itself, you can tell roughly what is the standard deviation and pretty much accurately what is the average. So, that is the question here. Can you tell what is the average and what is the standard deviation? At the end of this lecture, you should be able to do this. Now, we know that this normal distribution is a Gaussian function. It is a mathematical function is called a Gaussian. It has this particular form A e power minus B x square. So, this is this distribution, this if you plot this A e power minus B x square for some a given value of A and given value of B, you will get this curve. So, there are just two parameters that describe this normal distribution. One is, so we will discuss about that, but just by plotting this function, you will get this Gaussian, this particular bell shape. By plotting this using a, either you can take numbers, take a value for A and B and for different values of x, you can plot it and you will get this. But in many cases, you must have seen that normal distribution is presented in this particular way. Have a look at here. It might be, you might have seen that it is presented in a way that p of x is 1 by 2, 1 by 2 pi sigma square exponential x minus x 0 whole square by 2 sigma square. You must have seen this. So, what is this relation between this and this? It is clear that A is 1 by 2 pi sigma square and B is 1 over 2 sigma square and x 0 is 0. So, that is what the relation between, by comparing this and this, we can know that 1 by 2 pi sigma square is written as A, 1 over 2 sigma square is written as B and x 0 0. So, why this particular form of A and B? Why is A is 1 by 2? A can be anything, but why is it related to the sigma? Here the sigma is standard deviation. So, for normal distribution, sigma is standard deviation. So, why is A and B related to standard deviation? So, that is, why this particular form? Why is A is 1 by 2 pi sigma square and B is 1 over 2 sigma square? So, you can know this just by comparing. So, let us, let me write it here. So, what one function here is A e power minus B x square and it is also 1 by, so this is the Gaussian function. When normal distribution is written, it is often written that if you look at any book, normal distribution is written such a way that x minus sometime it is even written x average whole square by 2 sigma square. So, if you compare this and this, it is clear that this is A. So, A is 1 over, sorry this is a sigma square here, 1 over 2 pi sigma square and B this is 1 by 2 sigma square and x average in this case is 0. So, this is the, if you compare this and this, just you get this particular thing. Now, x average can be anything, but A is always 1 by 2 sigma, 1 over 2, B is 1 over 2 sigma square. So, why this particular, how is sigma is coming here? Why is it suddenly sigma coming here? Why this particular form for A and B? We will understand this in this lecture. So, let us start with the simplest case. So, the simplest question is, so if you have this Gaussian function, so if you have a Gaussian function P of x is A e power minus B x square, how do we find A? So, this is the function of normal distribution. If you plot this function, you will get the normal distribution of bell shape curve. Now, from this, how do we find A? So, that is the question. So, let us think a bit. So, this is probability distribution P of x. What does this mean? This means that, the probability of finding, having a value of x is this. So, probability is this, that what is the total probability? Total probability is always 1. Why? Because, like if you have, you know that in the case of head and tail, half, half is the probability. So, half, half plus half, probability of finding head is half, probability of finding tail is half, half plus half is 1. This is something that you know. Total probability has to be 1. In the case of marks that we discussed in the last lecture, if you have a probability of finding 1 mark, probability of finding 2 marks, probability of finding 3 marks, probability of finding 4 marks, up to probability of finding 100 marks, the total probability has to be 1. Because, you will surely find a student with probability, you will find all, even if all students got 100 marks, probability of finding 100 is, the sum will be 1. Let us say, all students get 53 marks, all students get 60 marks, still the sum, total probability of finding all the marks is 1. So, the total probability has to be always 1. That is the definition of probability. So, when we defined probability the other day, yesterday, in the previous lecture, we said that P of x, when we wrote, we had written that, in the case of H we had, like in the case of P of H i, we had written that, this is, can be written as the, the, the function. So, you had, you had n of H i into sum over i, sum over i n of H i. So, sum over i, if you divide this. So, now, if you do, now sum over of P of H i, this has to be 1. So, total probability always has to be 1. So, knowing this, knowing the total probability has always to be 1, we have this function P of x is a e power minus b x square integral minus infinity to infinity P of x d x has to be 1, minus infinity to infinity P of x d x has to be 1. What does this mean? This means that, minus infinity to infinity P of x is this. So, a e power minus b x square d x has to be 1, a e power minus. So, this is, if this is the case, a is, a can take a outside. If a is outside, this would imply that, a is equal to 1 over minus infinity to infinity e power minus b x square d x. So, just by knowing the integral of this function, we get a. a is 1 over e power minus infinity to infinity e power minus b x square d x. This is a. So, we get a. Now, how do we, how do we calculate this integral? What is the value of this integral? So, let us, quickly find out the value of this integral. So, at this, to begin with, I will just tell you the answer that, minus infinity to infinity e power minus b x square d x is root of pi by b. How do we get it? Later, we will discuss, how do we get it? How do we do this integral? But, for the moment, just understand that, the answer of this integral is root of pi by b. If you have a function e power minus b x square and if you integrate minus infinity to infinity, the answer is e power minus b x square is root of pi by b. How, we will show this is the answer indeed. But, for the moment, just take my word. This is the answer. If this is the answer, a is 1 over root of pi by b. It is root of b by pi. This is a. So, our probability distribution, if this is a, the probability distribution p of x, basically is, can be, can be written as p of x is equal to root of b by pi into e power minus b x square. So, this is our p x and surely we can phi c that minus infinity to infinity p of x d x is 1. So, this is the Gaussian function and a, we found already a. So, this is the full form of the Gaussian. Probability distribution, normal distribution has this particular form. Well, so now, this b by root pi is written such a way that minus, the total integral is 1. Well, now, what do we want to find out? So, we want to, we have, we had this first person. What is standard deviation? So, for the moment, we will consider, to begin with, we will consider this particular form for the probability distribution p of x e power minus b x, b by root pi e power minus b x square. Now, then we will find out for this particular case, what is the standard average and standard deviation? So, we know the definition of average and standard deviation. What is the definition of average and standard deviation? So, the average, as always. So, in the, in the previous lectures, we had defined average, x average is defined as minus infinity to infinity x p of x d x. So, just remember, this is always the definition of x average. If you are given a p of x, the probability distribution, the way to find x average is this. You multiply with x and integrate. So, just remember, this is the definition of x average. If p of x, p of x has to be, the integral of p of x is 1. So, the process, we just did a few minutes ago to calculate integral of a, this find out a is called normalization. So, this process, we had here, finding of a, this process is called normalization. So, this is the p of x is a normalized function. If p of x is normalized function, this is the definition of x average. So, we have x average and definition of x square average is minus infinity to infinity x square p of x d x. So, this is the definition of x, x square average. And we know the standard deviation sigma is nothing but, x square average minus x average square, square root. So, this is the standard deviation. So, we have to find out x average. We have to find out x square average and then, we can find out standard deviation. So, we have a p of x here. And our p of x is root of b by pi e power minus b x square. Now, we, you know what is x average, x e power minus b x square and root of v by d x, root of v by pi is a constant. So, I can write it outside. Now, it turns out that, what is the answer of this? Well, without doing this real calculation, just by looking the function plotted, you might be able to know the answer. So, let us just see how the function is. So, this is 0, this is plus x and this is minus x. So, this is minus 1, minus 2, this is 1, this is 2. And this is some function which is symmetric on both sides. It is this way is the same, this way is the same. So, we have a function which is exactly symmetric on both sides, exactly same symmetric on both sides. That means, whatever the value of plus 1 is the same value at minus 1. So, the e power minus b x square, if you find out at plus 1 is same as the value at minus 1. So, the only thing, but differs is x. So, when you do this integral, for every you multiply with 1, first let us say x will have a some value positive value 0.1 and integral is nothing but a sum. So, you multiply with 0.1 and multiply with minus 0.1 and sum of them. So, same value you multiply with 0.1 and minus 0.1 and you sum, you sum, you are bound to get 0. So, since it is symmetric and since it is from minus infinity to plus infinity, the answer of this is 0, because for every positive value there will be a negative value which is cancelling. So, if you take one term in this, if you expand this, if you integrate this integral, if you write it as a sum. So, let us write this quickly. So, integral minus infinity to infinity x e power minus b x square can be written as sum over i x i. So, let us call i is equal to minus infinity to infinity x i e power minus b x i square delta x. So, this is d x, if you wish. So, this can be written. So, now, let us have a look at this function when x i is. So, there are many values. So, let us call this minus 1, 1, minus 2, 2. So, if x i is. So, let us look at this value 2 and minus 2. So, at this particular 2, what do you get? 2 into e power minus b x square x 2 square which is 4 plus minus 2 e power minus b 4. So, this is 0. So, just by summing both sides, you are multiplying with plus 2, you are multiplying minus 2. So, for every plus value here, there is a minus value multiplying on the other side. So, the total of this integral is 0. So, this is one way of saying it. So, it turns out that the x average is 0. So, x average which is integral x root of b by pi e power minus b x square is 0. Now, there is a rigorous way of showing this which we will… This is actually not very difficult at all. So, let us… If you want, let us do that the rigorous way. So, what we want? We want root of b by pi e power minus b x square x dx. I can write this as root of b by pi del by del x of… So, let us look at this. So, what is del by del x of e power minus b x square? This is e power minus b x square into derivative of minus b x square. So, this is e power minus b x square minus 2 b x. So, just by knowing this, just by knowing that that e power minus b x square into minus 2 b x, this is into times… This is the derivative. We immediately see, we can immediately see that just from this, I can say that del by… Just by this, you can see that del by del x of e power minus b x square divided by 1 over 2 b. So, I take 1 over 2 b the other side is equal to x e power minus b x square. So, x e power minus b x square can be written as minus 1 over 2 b del by del x of e power minus b x square. So, you substitute this there and do the integral yourself. You will find the answer that it is indeed 0. So, I will not do this clearly. You can do this yourself just by substituting the fact that x e power minus b x square is del by del x of e power minus b x square minus 1 over 2 b. Because, this you know, if you do this, you will get this. So, let us do that. So, let us substitute this here. So, what we want? Del by del x 1 over 2 b with a minus sign integral e power minus b x square d x. So, this integral has to be written outside. So, this is 0. You can convince yourself that minus infinity to infinity, this is 0. Do it yourself and convince yourself that this is indeed 0. .. The next question is, how do we calculate x square average? So, we have p of x is root of b by pi e power minus b x square. So, for most of the time, I will write, sometime I might write root of b by pi as a for convenience. So, what we want to find out is that x square average, which is nothing but x square e power minus b x square minus infinity to infinity d x. So, we want to find out x square average now. So, we know, p of x is root of b by pi e power minus b x square and x square average is a into a is root of b by pi, which is a. I write it as a sometime minus infinity to infinity x square e power minus b x square d x. So, this is the definition of x square average. Now, we want to find out, what is x square? So, first let us do this particular integral. We know that a is root of b by pi. So, let us calculate, what is integral x square e power minus b x square d x. So, let us, let us do this calculation. So, what do you want to calculate? We want to calculate integral minus infinity to infinity x square e power minus b x square d x. To do this, we do a standard trick, which is, we can write this particular function x square e power minus b x square as minus del by del b of e power minus b x square. So, if you, if you find the derivative with respect to b and multiply with a minus sign, you indeed get this. If you can, if you know, del by del b of e power minus b x square is e power minus b x square times the derivative of this, which is x square and with a minus sign. So, this, just by using the fact that x square e power minus b x square is minus del by del b, derivative of this function e power minus b x square. We can, instead of x square e power minus b x square, we can substitute this back and write it as minus infinity to infinity with a minus sign here and del by del b of e power minus b x square d x. So, del by del b is independent of x, because b and x are completely different variables. So, I can take this outside. So, this is minus del by del b of minus infinity to infinity e power minus b x square d x. So, what did we do? We wrote integral x square e power minus b x square d x as minus del b of del by del b of e power minus b x square d x minus infinity to plus infinity. We know the answer to this. We just told some time ago that answer to this particular integral minus infinity to infinity e power minus b x square d x is root of pi by b. So, this is the root of pi by b. So, the answer, the full answer, that is, if that is root of pi by b integral minus infinity to infinity x square e power minus b x square d x, there is nothing but minus del by del b of root of pi by b. What is this? There is a root pi there. Then, del by del b of root 1 over root b. 1 over root b can be written as b power minus half. So, this is nothing but root pi into derivative of b power minus half is minus half is a minus sign. So, this is a minus half. So, this is a half and minus minus cancels. So, plus half b power minus 3 by 2. So, this is the answer of this. So, this can be, if you wish, can be written as root pi by 2 into b power 3 by 2. b power minus 3 by 2 can be written as 1 over b power 3 by 2 and there is a half and there is a pi. So, integral x square e power minus b x square d x, there is nothing but root of pi by 2 b power 3 by 2. Now, what do we want? We want x square average. x square average is defined as, as we said, you have to multiply this with root of b by pi, which is our a integral minus infinity to infinity e power minus b x square d x, x square d x. So, this part we just found. This as, so this is equal to root of b by pi, root of b by pi into, this just we found. This as root pi by 2 b power 3 by 2 by 2 b power 3 by 2. So, b power 3 by 2 is what? So, b into b power 3 by 2. So, let us write b power 3 by 2 here. b power 3 by 2 can be written as b into root b, if you wish. So, this can be written as b into root b. So, root pi root pi cancels, b root b and b power 3 by 2, they together produce. So, if you wish root pi and root pi cancels, this can be written as root b by 2 b power 3 by 2. Root b is b power half. So, if you write root b as b power half, this is 1 over 2 b. So, x square average is 1 over 2 b. So, what did we find? We found two things. We found that, e x average of a Gaussian function of this particular probable normal distribution, if you look at here. If you look at the slide, you will see that x square. So, what we just found that x square average is 1 over 2 b and we had found that x average is 0. So, we found two things. We had found x average 0, x square average is 1 over 2 b. And what do we have? We have p of x is root of pi by b e power minus b x square. So, if you, this is what written the slide roughly here, that x average is x p power p x d x, which is 0 and x square average is x square p of x d x is 1 over 2 b. So, and what is the standard deviation, if you know this two? If you know this two, the standard deviation is x square average minus x average squared. So, the standard deviation, here x square average is, this x average is 0. So, standard deviation is x square average root itself. So, the standard deviation is… So, the standard deviation sigma is x square average minus x average square root. So, this is 1 over 2 b minus 0. So, it is like 1 over 2 b. What does this imply? This implies that b is equal to 2 sigma square. Sorry, b is equal to 1 over 2 sigma square. b is equal to 2 sigma square inverse. So, that is I sigma square is equal to 1 over 2 b, b is equal to 1 over 2 sigma square. So, we get a relation between b and sigma. This gives you a relation between b and sigma. So, b is 1 over 2 sigma square. This is b. b is 1 over 2 sigma square. So, once you know the relation between b and sigma, we can look back our function. Our function was, what is the p of x? p of x was root of pi by b e power minus b x square and we just found that b is equal to 1 over 2 sigma square. So, I substitute this back here. Sorry, this is wrongly written. So, this is the distribution function p of x is root of b by pi e power minus p x square and we found that b is equal to 1 over 2 sigma square. So, if I substitute this for here, what do you get? You get p of x is equal to root of b is 1 over 2 sigma square. So, you get 1 over 2 pi sigma square e power minus x square by 2 sigma square. This is 2. x square by 2 sigma square. This is 2. This is 2. So, p of x is equal to 1 over root of 1 over 1 over 2 pi sigma square e power minus x square by 2 sigma square. So, this is what precisely written here. So, we had asked this question. Why this particular a and b? We found that a is. So, there is a square root here. Sorry, there is a square root here, which is forgotten here. So, the correct form as it should be, it is p of x is root of 1 by 2 pi sigma square into e power minus x square by 2 sigma square. So, this should be the correct form for. So, this is the precisely the reason why sigma square is put here, because if you just call it a e power minus b x square. And we found that the, if you calculate from the, you take this function and if you take this function and calculate the sigma, that is x square average minus x average square, what we get is 1 over 2 b. So, that means root of 1 over 2 b. So, what means, what is that? Sorry, sigma is root of x square average minus x average square and if you found this, we found that it is root of 1 over 2 b. So, that gives a relation between b and sigma. So, that gives p of x this particular form. So, this is the reason why p of x is this particular form, where sigma is coming here and here. So, we found x average and x square average. Now, this is the function we took was, if the function we took was a e power minus b x square, if you had taken the function p of x is equal to a e power minus b x minus x 0 whole square. Same answer you would have got except x average, you would have got x 0 and x square average minus x average square, you would have got 1 over 2 b. So, this you can see yourself by just finding that x average is x 0 in this case and the standard deviation is the variance x square average minus x average square is a variance and variance is 1 over 2 b. So, the full function, the full function the probability distribution function p of x can be written as 1 by root of 2 pi sigma square e power minus x minus x 0 by 2 sigma square whole square. Now, given thus this particular form just by looking at this particular function just by a graph of this function, what all can you know just by looking at the graph of this function, what all can you know. So, let me take some particular value sigma is equal to 10. So, sigma square is 100 and let me take x 0 equal to 150, just some two values and just plotting this particular function by taking these two values you get this, what we had plotted earlier we get this. So, what does this tell us, so have a look at the slide here, what does this tell us. So, if you look at the slide here, this tells us that, so this is exactly the what is plotted here. So, what do you know from this loop by just looking at this, the peak is at 150. So, the peak is the point where is average value. So, x 0 is 150, so x average is 150. So, x average is 150. So, what does this mean, x 0, so peak is at 150. So, just by looking at the peak we can know what is the average value. So, average value is easy to find just look at the peak, the position of the peak is 150 that will be the average of this distribution. Now, how do we find the standard deviation, how do we find the sigma. So, just by look at this slide here, so you look at this paper here. So, we know that p of x is 1 by p of x is 1 by 2 pi sigma square e power minus x minus x 0 whole square by 2 sigma square and x average is 150 which is x 0. So, just note couple of things, if you know x 0 and if you know the sigma you know everything about p of x. So, you have to know only two things, you have to know x 0 and you have to know sigma then you can know you know everything about p of x. So, I told you if I find x 0 by looking at the peak we can find x 0 and how do we find sigma, we want to find the sigma. So, it turns out that sigma is related to the width of the distribution. So, if you look at here, if you look at the slide the width this distance. So, this is this are related. So, how do how is this related? You can measure the width here, the width here, the width here, you can measure the width of the distribution at various places. Which of this width will give you sigma, will this distance give you sigma or will this distance give you sigma or will this distance give you sigma. So, this is the question which of this distance will give you sigma. So, let us look at this again the Gaussian function. So, Gaussian function is a e power minus b x square, the Gaussian function is a e power minus b x square or p of x if you wish. Now, this we know that this is nothing but 1 by 2 pi sigma square square root into e power minus x square by 2 sigma square. So, for when will it let us say. So, when will it sigma will be equal to. So, what we want? We want just by looking at the width by measuring this distance. We want to get the sigma, but where will you measure the distance that is our question. We measure it here or will you measure it here. So, what we want? We want this distance. So, this is if this is x, this distance is 2 x. So, we want 2 x to be sigma. That is what we want for a particular value of y. So, if we if you look at this what we want? We want for a particular value of y for a for a given we want this distance that is this is x and this is also x. So, the 2 x we want a sigma for a value of y. So, given a y there is a place of there is a particular value of y for which 2 x is sigma. What is the value of y for which 2 x is sigma? So, let us first put a sigma is equal to 2 x and then calculate. So, what we have? We have p of x is 1 over 2 pi sigma square square root e power minus x square by 2 sigma square and substitute sigma is equal to 2 x. So, what do we get? 1 over root of 2 pi. So, this what we want is just. So, this particular part let us look at this particular part. If you look at this what you get e power minus x square by 4 2 into 2 into 2 x square. So, 2 x square is 4 x square and again 2 into 2 into 2. So, what we get? This particular part. So, we get a let me call this a itself this part a e power minus x square by 8 x square. So, we will get p of x is a e power minus a x square by 8 x square x square and x square goes. So, what you get? You get a e power minus 1 by 8. So, what is a? When x is equal to 0, if you put x is equal to 0, p of x is a. So, if you e power minus 1 by a if you multiply a with e power minus 1 by 8 at that particular value a is this as a constant is a maximum value which we know a is the peak. So, a we know just by looking at it because if you have a function this values a this is the peak when x is equal to 0 what is the value of y that is a. So, if you multiply a with e power minus 1 by 8 that will be the y where sigma is 2 x if the width is equal to sigma. So, if you look at the distribution if you look at this particular distribution and multiply this peak with e power minus 1 by 8. So, what is e power minus 1 by 8? So, it turns out that e power minus 1 by 8 is approximately 88 dot some percentage. So, it is approximately 88 percentage. So, if you multiply p when p of x is a e power minus 1 by 8 if what did we find we found that if p of x is a e power minus 1 by 8 sigma is equal to 2 x or in other words we found that if sigma is equal to 2 x p of x is a e power minus 1 by 8. So, if you look at that 8 if I look at 88 percent from this peak. So, you multiply e e power the peak by this particular number which turns out to be 88 percent because this is e 1 by 8 is around 0.125. So, 0. e power minus a e power minus 0.125. So, this is coming 0.88. So, this is something like 0.88 of a approximately. So, if you look at the 88 percent of this peak somewhere here if you measure the width at the 88 percent at this particular point you will get the standard deviation. So, the width so the peak at the peak the x value of the peak is average and the 2 x the width the width at e power minus 1 by 8 of the peak is the standard deviation. So, just by looking at it we can know standard deviation as well as the average. So, to summarize what did we learn. So, we learned that we learned that x average if we is. So, we what we found out we learn how to calculate x average we found out how to calculate x square average we found out how to calculate sigma. So, if you have a particular function of the if a Gaussian function we found that sigma is related to if you have a function of the form a e power minus b x square sigma is 1 over 2 b square root. And we found that the peak is x average and the width at e power minus 1 by 8 times of the peak maximum value is 2 x which is at the 2 x the width at this particular point will give you the standard deviation. So, we learned you many things about normal distribution. So, use this information use this ideas next time when you see a Gaussian distribution just look at where is the peak and the width at 88 percent of the peak. And also do yourself what is the width the full width at half of the maximum we found that the width at 88 percent of the maximum is sigma what is the width at half of the maximum you find out yourself. So, keeping keeping this things that we will learn in mind that integral of e power minus b x square and average and x square average this will be very useful for various things in the future. So, discussing this we will stop today's lecture. Bye.