 to compute that double integral I make use of polar coordinates that is I make the transformation x is equal to r cos theta y is equal to r sin theta then you know the d x d y the element of area in x y plane gets transformed to r d r d theta that is the element of this then your r varies from 0 to infinity because r is a non-negative variable and theta varies from 0 to 2 pi this is your polar coordinate transformation. So, therefore, I square then becomes 1 upon 2 pi 0 to infinity 0 to 2 pi and e raise to minus half r square because this was x square plus y square will become r square cos square theta plus r square sin square theta since cos square theta plus sin square theta is 1. So, it reduces to r square and this is r d r d theta and again this is in a nice form because now you see you have r here and you have e raise to minus half r square. So, again we will make this transformation that r square is equal to t that is what I am doing here and so your 2 d r d r will be d t. So, r d r will be half d t and so that is what I have written here half d t and this will go to e raise to minus half t. So, this is and your right and see the limits are 0 to 2 pi 0 to this thing. Now, here what I have done is already because theta does not appear here anywhere. So, therefore, this is simply 1. So, with respect to theta this just integrates to 2 pi which I have written here and then it is only integration with respect to r and to integrate this I make the transformation r square is equal to t and so this helps me to reduce it further and this is you know minus 2 times e raise to minus half t 0 to infinity. So, this reduces to this which is equal to 1. So, we have verified that the normal p d f that we have defined is valid p d f and then now the second step is to compute the expectation and instead of computing for e x we will compute expectation of x minus mu by sigma that will be easier because we know that the expectation of x is actually mu. So, we will show that this expectation is 0 and therefore, I get the answer immediately. So, here the only in this integral you get this term x minus mu by sigma and so again I make the substitution x minus mu by sigma is equal to y and that reduces the integral to this again it is of the same form. You see that is why the expressions may look cumbersome, but the working is not very difficult and it is just a question of little patience and you start seeing where the calculations are going. So, now you have this. So, again you make the transformation y square equal to t the same steps and you get this and so. So, now what I am saying is that before I make this substitution I will break up this integral to minus infinity. See one way I can immediately from here conclude that this integral is 0 because this is an odd integral. Why is here? This is here the change of sign will not matter, but here the change of sign will matter since this is from minus infinity to infinity this integral will be 0. So, I just thought that I will show you the steps. So, what I am doing is I am breaking it breaking up this integral from minus infinity to 0 plus 0 to infinity. And then when you make the transformation that y square is t then you see this becomes plus infinity because when y is minus infinity square will be plus infinity. So, this is from. So, this will be infinity to 0 and the integral will be e minus half t just as we did here and then this will be 0 to infinity. So, now it is a same integral except that here the limits are upside down. So, when you change the limits it will become minus sign. So, there will be minus sign here and the same integral with a plus sign. So, when you add the result is 0. So, I just showed you the steps in case you were not very sure, but otherwise we could have concluded our computation at this step only and said that this is equal to 0. So, since this is 0 expectation of x minus mu by sigma. Sigma is a constant it goes out. Therefore, this implies that expectation of x is equal to mu. So, I had shown you that the normal pdf is symmetric about mu and this is also the mean and in fact, mu has all the properties. Later on I will show some more properties of mu. Similarly, to compute the variance of x which is actually since mu is the mean it is actually expectation of x minus mu whole square and this I will write in this form. So, now here what you have to do is again first make the transformation that x minus mu by sigma is t. So, d x by sigma can be replaced by d t and the limits remain the same. So, this is a simple integral and this becomes sigma square because you have x minus mu whole square. So, this is sigma square t square this is it. Now, what I do here is the t square I break up into t and t because this integral we have handled already while computing this thing. Therefore, we will do integration by parts. This will be my first function. That means, I take the integral of this multiplied by this and then the derivative of this into the integral of this. This is your formula. So, this integral I have shown you is e raise to minus half t square this t into this thing. I am actually computing this for you. So, I am saying t e raise to minus half t square d t is half when I make the transformation t square is s then 2 d t is d s. So, this reduces to this and therefore, to this. Now, applying integration by parts the integral of this I have already computed for you is this minus e raise to minus 1 by 2 s. So, this would be t into when I transform back again substitute for s from here this is t square. So, then t into e raise to minus half t square minus infinity to infinity you can see that this is 0 because this is t square and so in the denominator t square e raise to infinity is much larger than t in the numerator. So, therefore, this will be 0 you will be left with this and then here again this is what this is your normal pdf where your mu is 0 and sigma is 1. So, the 1 upon root 2 pi is missing. So, therefore, this integral will be equal to root 2 pi because with 1 upon root 2 pi this will become 1. So, therefore, this whole thing is 1. So, this integral is equal to root 2 pi. So, I write root 2 pi which cancels with this and this is sigma square. So, therefore, the variance of random variable. So, therefore, the parameters are now it is very clear what the parameters denote mu is the mean when we say now normal mu sigma square mu is the mean and sigma square is the variance. So, we just saw that normal distribution the parameter mu is the mean and sigma square is the variance. Now, suppose x is n mu sigma square and we consider the random variable y equal to alpha x plus beta where alpha is a positive number and alpha and beta are some real numbers. So, I mean I am continuing with the properties of the normal distribution. So, if you want to find out the pdf of y then we start with the cumulative distribution function. So, probability y less than or equal to t is probability alpha x plus beta less than or equal to t which reduces to this and since alpha is positive the inequality remains intact. So, this is t minus beta upon alpha and so this is your cumulative distribution function y which is equal to the cumulative density function of x, but the parameter the t replace is gets replaced by t minus beta by alpha. So, now if you differentiate both sides that means differentiate with respect to t then this will become the pdf and this is d dt of f x t minus beta by alpha which will be a 1 by alpha into f x of t minus beta by alpha derivative of capital f x will be the pdf of the random variable x. And so when you substitute you write down the expression for this it will be 1 upon alpha 1 upon under root 2 pi into sigma e raise to minus 1 by 2 sigma square and your t gets replaced by t minus beta by alpha minus mu whole square. Just simplify the expression this gives you t minus beta minus alpha mu whole square here you have alpha sigma and this is 2 sorry the alpha in the denominator comes here. So, this will be there will be a into alpha square also I hope you can read it anyway I am speaking it out or may be let me just rewrite it whole thing here. This is 2 alpha square sigma square and so by our definition of the normal pdf this will be and the mean now becomes beta plus alpha mu and the variance becomes alpha square sigma square instead of sigma. So, therefore, you see that if you make this transformation where x is normal mu sigma square then for y the expectation will become alpha mu plus beta which anyway you can show from here also this is alpha expectation of x plus beta because beta is a constant. And so this is alpha mu plus beta and the variance will be just in because the constant will not matter since you will see when you write down the variance you will write down this minus I mean this minus this. So, beta will cancel alpha will come outside and so it will become square and so either way you can verify that the for the for this random variable the mean will be beta plus alpha mu and the variance will be alpha square sigma square. So, you can see that you can carry on the properties of normal variate quite easily. Now, an immediate consequence of this result is that if x is n mu sigma square then z which you write as x minus mu by sigma will be normal 0 1 because what is happening your here alpha is actually 1 by sigma if you compare it with the expression alpha x plus beta and your beta is minus mu by sigma. So, now if you substitute because we said that the mean will become alpha mu plus beta for that 1. So, here it will become mu by sigma minus mu by sigma. So, this is 0 and similarly you can show that the variance will be 1. So, the transformation x minus mu by sigma results in a standard normal variate and which we refer to as n 0 1. Now, we have tables for computing the various probabilities for normal 0 1 and you see that you can then compute the probabilities for any random normal random variate through this and that is what because of the transformation right and I will work out a few examples to show you how it goes. So, anyway this is the standard notation for a standard normal variate, this is your probability minus infinity to x. So, that means this is the cumulative density function. So, this will be 1 upon root 2 pi minus infinity to x e raise to minus half y square d y. So, this probability is given the notation. Now, the tables are given for x non negative for you know values of x going up to well we will also later on see that we do not need the values to be table for very large values of x then for x less than 0 we use the symmetry of the p d f around because this standard normal. So, this is this is symmetric about the origin right and this is your this is this right. Now, the symmetry means that if you have x here then the area to the right of this number is the same as the area to the left of minus x. This is what symmetry means because this area and this area are equal therefore, and since this is half area and this is 0.5. So, therefore, this shaded portion here is the same as the shaded portion here and so the formula is this. So, if you have table your values for x positive then for x negative you can get by this and we can verify this formula right away. If you want to compute phi of minus x then that is minus infinity to minus x of f x d x. Now, if you write y as minus x then x becomes minus y. So, the limits go from infinity to y right infinity to y f of minus y d y right and there is a minus sign here. So, if you interchange the limits this becomes y to infinity f of minus y d y which is 1 of minus now phi minus y right because this is now y to infinity. So, therefore, by the formula this is this, but phi of minus y is phi of x. So, therefore, I have shown you that this is 1 minus phi x. So, this formula has been verified. So, now you can get values of the cumulative density function for negative positive both so let us look at a few examples. Suppose x is normal 2 comma 4 that means the minus 2 and the variance is 4. So, then you want to find the probability that x is between 2 and 4. So, I will use the transformation. So, here I am sorry this should be z that means what I am doing is I should I will write in detail I am subtracting 2 and dividing by 2. So, this is less than x minus 2 by 2 less than 4 minus 2 by 2. So, this probability goes here right and now. So, this is 2 minus 2 0 and this is your standard normal variant right x minus mu by sigma is normal 0 1 for which we always have this notation of z right and this is 4 minus 2 by 2 which is 1. So, this probability you can compute in terms of the standard normal variant in this way and by our notation this is phi of 1 minus phi of 0 and from tables I get that phi of 1 is 0.8413 and phi 0 will be 0.5 because we have said that the standard normal is symmetric the pdf is symmetric about the origin. So, this portion of the. So, this area under the curve will be equal to 0.5 and this will be also 0.5. So, phi 0 will always be 0.5 because this is the area for phi 0 less than or equal to x less than or equal to 0. So, that will be half of the area which is 0.5. So, therefore, this is the probability. So, therefore, very conveniently once you have the tables available for the standard normal you can compute it for any normal variant right and again probability x less than 0 for this variable. So, here again I make the transformation x minus 2 by 2 less than. So, this will be minus 2 by 2. So, this is probability z less than minus 1. So, which is phi of minus 1 and then I use this formula this formula. So, phi of minus 1 is 1 of minus phi 1 phi 1 I already know is 0.8413. So, this is 1 minus 1 0.8413 which is 0.1587 right. Now, when you want to compute absolute x minus 1 greater than 3. So, what are you saying here you are saying that your x minus 1 should be greater than 3 and x minus 1 should be less than minus. That means x minus 1 should be less than minus 3 and x minus 1 should be greater than 3. So, from here what do you get that x should be greater than 4. So, from 4 to infinity and from here you get that x should be less than minus 2. So, that means it should be from minus infinity to minus 2. So, that is what we have done I have written that this event is equal to the that means x must be either here minus infinity to minus 2 or it must be in the set 4 to infinity. Well I have used a different notation here then here does not matter. Now, since these two sets are disjoint you can see minus infinity to minus 2 and this is 4 to infinity. So, I can write the probability as the sum of the individual probabilities and so I get this and again I transform my variable x minus 2 upon 2. So, this remains minus infinity this is minus 2. So, minus 2 by 2 and here again I use the same transformation to reduce it to a standard normal variate and therefore, this becomes phi of minus 2 because phi of minus infinity is 0 plus 1 minus this will be what this is 1 and this is infinity. So, here phi infinity is 1 and this is minus phi of 1 because yes you all agree that this is phi of infinity I mean this phi of infinity will be 1. So, therefore, and phi of minus 2 you can write as 1 minus phi 2 again by our formula for computing negative probabilities in terms of positive this thing and therefore, this is it. So, I just substitute the values phi 2 from the tables is 0.9772 this is 0.51 and so this is the answer. So, one can go on and therefore, the whole idea would be that you sit down calculate a few of such probabilities by yourself to become familiar. Now, let us just take an example here. The annual rainfall in inches in a certain region is normally distributed with mean 40 inches and sigma that is the standard we call by the way I did not name it, but sigma this is under root sigma square is referred to as standard deviation this is also known as standard deviation. So, standard deviation is 4 that means variance is 16 what is the probability that starting with this year it will take over 10 years before a year occurs having rainfall of over 50 inches. So, let us understand the problem first they are saying that the annual rainfall is normally distributed that means if every year you add up the total rainfall in that particular region then those numbers will be fitting normal distribution which has mean 40 inches and standard deviation 4. So, the numbers that you get as the total rainfall in a year for different years. So, that has a normal distribution. Now, they are asking for a probability that starting with this year you go on for 10 years and within those 10 years no year will have rainfall more than 50 inches. So, it is you know actually compounded event it is not a straight forward. So, let us see how we go about computing this probability. So, to compute the probability x greater than or equal to 50 I will do the same trick reduce this probability in terms of a standard normal variant. So, since mean rainfall was 40 inches. So, x minus 40 standard deviation was 4. So, this now reduces to a standard normal variant. So, therefore, the required probability is the same as probability z greater than or equal to 5 by 2 which is this is 1 minus. So, this has to be z greater than 5 by 2 would be 1 minus 5 pi by 2 and so this will be 1 minus of this. So, this 1 minus 0.9938 and so this will not be the required probability we will have to write the number anyway. So, what we will do is I will continue to say that p is 1 minus 0.9938 or maybe I will just do the. So, this is 1 minus 0.9938. So, this is the probability of is the probability of rainfall. So, the probability would be standard normal variant greater than or equal to 5 by 2 which will be 1 minus 5 of pi by 2 5 by 2 and that is 1 minus 0.9938 because 5 pi by 2 is 0.9938. So, this is probabilities. Now, this we can look upon as the probability of rainfall being more than 50 inches in a year and this we will treat as a success. That means, now you can say that in a year the experiment is to measure the rain and if the rainfall is more than 50 inches we treat that as a success. So, that is what I said that the problem requires because the event is complex. So, first we computed the probability of the rainfall being more than 50 inches in a year which came out to be this. Now, I want to find out that in 10 years rainfall not more than 50 inches in any year. So, that means, if I treat those 10 every year as a trial that means, the trial is you add up the total number of rainfall in that year and then you are saying that in 10 years there should be in none of the years that you count from beginning from this year the rainfall is more than 50 inches. That means, in other words there is no success in these 10 trials and so the probability of no success in 10 trials would be 1 minus p raise to 10 which was 0.9938 raise to 10. So, again please compute this number because I had made a mistake. So, here this is whatever this number you can now use your calculators to compute this probability to say that that will be the probability that in 10 years there will be the rainfall will be less than 50. So, this is so I am just trying to show you that how you first use the normal approximation and then I mean the standardization and then you use the binomial random variable to compute the actual probability. So, median and mode of a normal distribution this again see that is well I mean when you look at all these properties of this distribution you realize that it is a very interesting one and this is suppose what we mean by the median of distribution is the number for which the for which your cumulative density function has the value 0.5. That means, the area of the if you have this thing here then this is the this is the point where this is x naught. So, this area is 0.5 and this area is 0.5 and since we have already said that the normal distribution is symmetric about x equal to mu. So, that immediately clear that the area to the left of mu is 0.5 and area to the right of mu is 0.5. So, immediately we know that the median is also at x equal to mu. Now, to define the to obtain the mode point this is the point at which f x the probability density function attains its maximum value. We did this for the binomial also remember and what was the number at which the random variable attains its maximum probability. So, now in this case we will have to find out the maximum the value the value of x at which this function attains its maximum value. So, this will be done by finding out the differentiating it finding out the critical points. So, and putting this equal to 0 the derivative now you see that here this is the derivative. So, this portion is not 0. So, therefore, this is the only portion which will be 0 and therefore, that gives you x equal to mu. So, x equal to mu is a point of maximum or minima essentially or it could be a point of inflection, but. So, now we have to look at the further the second order derivative to determine the nature of this critical point. And I have written down the expression for second derivative that means you differentiate this expression again and compute it at evaluate it at x equal to mu. So, then you see this portion goes 0 because x equal to mu and here also when you put x equal to mu you essentially get this. So, this is again e raise to 0 which is 1. So, you simply get minus 1 upon 2 pi sigma square which is less than 0. So, if the second derivative has negative sign at a critical point that point must be a point of maxima. So, that much from calculus we can obtain and. So, therefore, it is really interesting that x equal to mu is the mean median and mode. So, it is all the things combine into 1. Now, another important as I told you I have been I have mentioned that demover was used to you initially defined this normal distribution to approximate binomial probabilities. So, let us formalize the procedure here. So, this is the demover Laplace limit theorem which is that if S n is the number of successes in n trials of a binomial random variable n comma p. Then for any a less than b. So, here of course, for the binomial the mean is n p and the variance is n p q. So, for any a less than b if you want to look at the probability of a less than S n minus n p upon root n p q less than or equal to b. So, you see what we have done is we have standardized this binomial random variable of successes in n trials number of successes in n trials. So, it will be S n minus n p upon under root n p q. So, then it is the demover Laplace theorem says that this probability can be approximated by phi b minus phi a where phi is your cumulative distribution function for the standard normal distribution for the standard normal variate. So, this will be phi b minus phi a as n goes to infinity. So, that means for large n you can approximate this probability by phi b minus phi a. And later on we will show that this actually is a very general I mean you can talk about a more general result where you do not really need to have the distribution as binomial. And for any distribution this kind of thing which we call as the central limit theorem we will talk about it later, but right now demovers Laplace limit theorem simply says that if you have a binomial random variable then if you want to compute the probability that S n minus n p upon under root n p q. So, this lies between a and b a strictly less than this then this can be approximated by phi b minus phi a. This is the Laplace theorem. So, now we will use this and so we say that now suppose you want to compute the probability that C is strictly less than S n is less than or equal to d. And here of course it is understood C is also strictly less than d. Then we will standardize the whole inequality in the sense that we will transform this to C minus n p upon under root n p q and this will be less than S n minus n p upon under root n p q which will be less than or equal to d minus n p under root n p q. And now you see this becomes standard normal variant and so your a gets replaced by this in the theorem and your b gets replaced by this. And so I can say that by the demover Laplace theorem that this probability which is the same as this can be approximated by phi of d minus n p upon under root n p q and phi of C minus n p upon under root n p q. So, this is the whole idea. So, therefore, and these values are tabulated. So, given C d and n p you can look up the tables and compute this probability. And of course, if you want to compute the actual probability then we will see through examples that it can be very cumbersome. So, in fact, this is a very useful theorem and it in a very simple way allows you to compute approximate these probability. And of course, then we will continue with the computations for approximating these probabilities. What they said is that this is good enough as long as your n p into 1 minus p is greater than or equal to 10. That means, you do not require too many very large values of n, but as long as of course, if this number is larger than 10 you get a better approximation. And that you can also for yourself experiment with the problems where you try to increase the value of n and then see that your the estimate will improve. So, anyway, but this gives good approximation as long as n p into 1 minus p is greater than or equal to 10. Now, there is a important aspect to this approximation and that is the continuity correction factor which I will discuss here. So, you see look at this curve. So, I have the binomial curve here bar chart and then the red line curve is the normal approximation and you see what we are saying here is that if you are asking for the probability x greater than see this is the shaded portion that you are looking for. So, here you want to say that probability x greater than 7 if you are asking for this probability x greater than 7. So, then if x greater than 7 means that x is greater than or equal to 8. So, to get that event to get that probability if you want a good approximation see your the rectangle the bar that represents the probability at 0.8. So, that is starting from 7.5 you want to cover that whole area because you are wanting probability x greater than or equal to 8. So, therefore, you will need to say that your the approximate approximating standard normal variate should be greater than or equal to 7.5 and not 8. So, because the so you will want to say the for the continuity factor you will say that you will do it from 7.5 because you want to use up the approximate you include the area which is at the point a the probability which is represented by at the point a by this bar should get added to the to your estimate right. And similarly, if you are wanting let us say so I have that table here just look at the example. So, if you have x equal to 6 that means a single value if you have x equal to 6 then you would want it between 5.5 and 6.5 because that is the rectangle the bar that you constructed at 6 that stands at from 5.5 to 6.5 and the length the height is the probability of that x equal to 6 right. So, therefore, the continuity factor that you require would be for probability x equal to 6 it will be 5.5 and 6.5 your x must vary then because see the discrete situation you are now approximating by a continuous situation then as I said x greater than 6 would be x greater than 6.5 and if x is greater than or equal to 6 then you are including 6 then you will have to go little further down right. That means you will have to start from 5.5 if it is x greater than or equal to 6 then you will begin from 5.5. So, x would be greater than 5.5 if x is less than 6 that means x is less than or equal to 5 then again you will say that x is less than 5.5. So, that is very clear from this graph and x less than or equal to 6 would be x again see the moment you have equal then you have to go a little ahead of it 6.5 and if it is strict inequality then. So, the rule is very clear and by looking at this figure you can always. So, if you keep this in mind then you will not go wrong because you will realize that you have to include the area under the bar or the rectangle which is for that particular value the limiting end value and so you will have to accordingly to give the correction factor right. So, this is called the continuity correction factor because you are estimating a discrete situation by this thing. So, if you use the continuity correction factor now let us give this thing let us give let me give you show you the calculations through an example. So, let me show you an example through an example how we what I mean by the continuity correction factor. So, let x be the number of times a fair coin. So, fair coin means that the probability of showing a head and a tail are the same and that equals half right. So, lands heads when it is flipped 49 times x is the number of times head shows up when I have flipped a fair coin 49 times. Now, find the probability that x is equal to 25 use the normal approximation and then compare it with the exact probability. So, I want to compute probability x equal to 25 and as I told I just discussed with you that since this is a binomial random wearable you are trying to approximate it by a normal distribution. So, the bar is actually when x equal to 25 the bar starts from 24.5 and ends at 25.5. So, this is the bar and this the height is the probability that you associate with x equal to 25. So, 24.5 and 25.5. So, therefore, I have to change the to approximate again I should say this that is the approximate this thing will be 24.5 less than or equal to x less than or equal to 25.5. So, I am now approximating this event by this event right because of the continuity factor. And then because the mean yeah I am sorry this should not be 49 I have correctly here. So, now for this binomial random wearable your number of times you flipping the coin is 49 probability that your p is half. So, therefore, n p is 49 into 1 by 2 which is 24.5. So, it should be 24.5 and your variance is n p q. So, that is 49 into 4 divided by 4 and so under root of that would be and that is why I chose this 49 to make it perfect square. So, that is 7 by 2. So, this is now standardized. So, x minus 24.5 divided by 7 by 2 would make it standard normal and so this is the event that you want to you want to now find out the probability. So, this is this right and therefore, this is equal to 0 and this is 1 upon 2 by 7. So, therefore, this is equal to phi of 2 by 7 minus phi 0. Now, this is phi of 0.28 minus phi 0 phi of 0.28 from the normal tables is equal to 0.6103 minus 0.5 right and so this comes out to be 0.1103. So, this is our probability that we have obtained for x equal to 25 through a normal approximation and remember the condition was that your n p q must be greater than or equal to 10. So, in our case our n p q is how much n p q remember is 49 by 4. So, which is more than 10 right that is our n p q is 49 by 4 which is equal to 42 sorry 12. something 12 holes are 48. So, 0.25 in fact. So, this is more than 10 right and so we applied the approximation and this is the result that we got right. Now, exact probability if you want to compute using your binomial computations then you see it is actually 49 choose 25 and 1 by 2 raise to 49, 49 trials and your p and q are the same. If you write out this number 49 into 48 up to 25 because that will be 49 minus 25 plus 1 and then 25 factorial 1 upon 2 raise to 49 it took me almost half an hour to compute from sitting at the computers and this number comes out to be 0.11275. So, if you just look at it this here and that is what I am saying now you compare it with. So, third decimal place the number differs the value differs and therefore, I would say that this is a good approximation and here you see our n p q was only 12.25 and if you take larger n that means if you had flipped the coin more than 49 then this number would have improved. And just see the phenomenal calculations you had to do even for n equal to 49 I mean multiplying these numbers 24 numbers getting 25 factorial multiplying 2 49 times and then dividing. So, this the amount of calculations that you would have to do for the exact probability certainly is not required if you can approximate this probability by this simple method because these table these values are already available to you. And so this is what I am trying to say that as we go on you will see the numerous applications uses of this concept of normal distribution and its computations. So, I will continue with the examples of you know binomial the normal approximation of binomial probabilities. And these are two good examples from the book by Rothschilden as I told you that this is the book on probability theory and at the end of the course I will give you all these references. So, I just thought I will you know give you a feeling of you know some more examples to reinforce the idea that the approximations of discrete probabilities can be done by the continuous random variables and in a very effective way. So, let us look at this example the ideal size of a first year class at a particular college is 150 students. From past experience the college knows that on the average 30 percent of those accepted for admission will actually attend the college. So, therefore you know people tend to apply to more than one college and then they of course decide which is the best one for them. So, the college has the experience that only 30 percent of the people who have been accepted for admission will actually attend the college. So, therefore the college adopts the policy of giving admission to 450 450 students. So, that finally when the you know draw out after the draw out rate they will still be left with the class of size 150 this is the idea. So, they decide to offer admission to 450 students. Now, compute the probability that more than 150 first year students attend the class. So, given that 30 percent is the so it means the probability of a person attending who has been given admission will attend college is 0.3. So, therefore we want to find out. So, here of course we will say x is the number of students who attend college and so the value of x will be equal to the we can treat the person who is been given admission and attends college as a success. So, out of 450 people x will be the number of students who attend college and so this will be a binomial random variable with n as your 450 and your p as 0.3. So, this is binomial 450 comma 0.3 because we are treating that the experiment has been performed that means admissions have been offered to 450 students and out of this those who attend who actually come attend the college is a success. So, the number of successes we are saying is a binomial random variable. So, therefore you have to compute the probability. So, actually you want that the class size should be. So, you have to compute the probability of more than 150. That means it should be 151 for 152 and so on. So, therefore when you write probability x greater than or equal to 151 then your continuity factor when you apply then this will become 150.5 because you are actually asking for the probability that x is greater than or equal to 151 because more than 150. So, therefore we just standardize this you know this variate and therefore subtract x minus n p, n p is 450 into 0.3 and then divided by n p q. So, 450 into 0.3 into 0.7 and do this to the right hand side also. So, you get this and then of course the advantage of taking an solved example is that you have the calculations done for you. So, here this is actually the phi of 1.59. So, this number reduces to 1.5. So, therefore this is your standard normal probability from minus infinity to sorry this is 2 1.59. So, we are writing this as 1 minus phi of. So, this number is 1.59. So, this required probability is 1 minus phi of 1.59 which comes out to be this 0.0559. So, hence this reduces to 6 percent of the time more than 150 of the 450 accepted students will attend college. So, the college is in good situation because it is only 6 percent chance that more than 150 people will actually come and attend the college those who have been offered admission. So, this was one situation. Now, another interesting problem, this is see to determine the effectiveness of a certain diet in reducing the amount of cholesterol in the blood stream 100 people are put on a diet. After they have been on the diet for a sufficient length of time their cholesterol count will be taken. The nutritionist running the experiment has decided to endorse the diet if at least 65 percent of the people have a lower cholesterol count. So, after the trial period at the end of the trial period you will again take test their cholesterol in the blood stream and if the count is lower for 65 percent of the people then after going on the diet. So, the nutritionist will have I made the sentence complete the new running the experiment has decided to endorse. So, the nutritionist will endorse the diet and say yes it is just proven to lower the cholesterol in the blood. So, now what we want to find out is what is the probability that the nutritionist will endorse the diet that the nutritionist endorses the diet will endorse. So, I write well will endorse the diet if in fact it has no effect on the cholesterol level. So, what we are saying is that suppose the diet has no effect on the cholesterol, but what is the probability that the nutritionist will still endorse it. So, therefore, now the way we are arguing out and so we have to now decide how to model the situation and the idea here is that you know it is either way. See people may on their own have their cholesterol count come down and so the chance of that happening is you know equally likely either your cholesterol count goes up or it comes down. So, that is a equally likely situation and therefore, it is being said that you can just take p equal to half. So, therefore, we are assuming that the diet has had no effect on the cholesterol level. So, but the since the count is going to be taken after the trial period and then if it turns out that 65 percent of people have lowered their count then they will be then the diet will be endorsed. So, therefore, we will work with p equal to half. So, I hope it is clear. So, what we are assuming is that chance of the count being going up and down is equally likely. So, therefore, we will take p to be half and x is the number of people whose cholesterol level is lower. Then see what we are looking for is. So, the binomial random variable and therefore, this is the probability sigma 65 to 100, 100 i half raise to 100 because r and n minus r both are will add up to n since p is half. So, this is the thing and of course, you can see that this is the stupendous task you know trying to compute it this way. So, therefore, we will say that essentially we are looking for probability x greater than or equal to 65 and again add the continuity correction factor. So, it will be 64.5 and then n p would be 100 into half and n p q will be 100 into half into half which becomes 25. So, under root 5 and this is 50. So, this is what you are looking for and this number comes out to be 2.9. So, therefore, it is the required probability is 1 minus phi of the normal standard normal probability 2.9 from minus infinity to 2.9 and this comes out to be 0.0019. So, which is very small and therefore, the chance that you know with p as half the chance that the 65 people out of this 100 will have their cholesterol lowered is very low and therefore, the diet will not get endosed. So, anyway because the diet was not having as we said that probably the diet has no effect on the cholesterol level. So, therefore, it does not get endosed. So, no loss nobody is lost. So, this is how you know one can go on and look at different situations and then try to see. So, I just thought that these two will also add to your you know experience of handling you know these problems and also reinforce the idea of you know computing you know computing these what shall I say messy probabilities by you know approximating through continuous random variables and making your task easy.