 Hello, welcome back. We have already completed one example set. In today's lecture, we will be looking at the second example set. As I had suggested, please read the problem statement and try to work out the problem on your own and compare the answers after the answers are discussed. Compare your answers with the answers that are discussed in the power point. The first example, we are going to take the normal distribution. We know that the parameters of the normal distribution are mu and sigma. Mu directly standing for the mean of the distribution and sigma standing for the standard deviation of the distribution. This is a unique case as in many other cases, the parameters of the distribution do not directly correspond to mu and sigma. In the log normal distribution, you have seen that the parameters were quite different from the mean and variance of the log normal distribution. So, let us take the normal distribution. The probability density function is given by f of x is equal to 1 by root 2 pi sigma squared exponential minus x minus mu whole squared by 2 sigma squared. The 2 parameters are mu and sigma. The range of this distribution is from minus infinity to plus infinity. The question is, mean of the normal distribution is indeed mu. We will go to the definition of the mean for continuous probability density functions. It is minus infinity to plus infinity x f of x dx. We plug in f of x here and then do the integration and see whether the answer on the right hand side matches with the answer on the left hand side. So, that is what we do here. We plug in 1 by root 2 pi sigma squared exponential minus x minus mu whole squared by 2 sigma squared here and this is the expression here. So how do we go about solving this integral? I hope all of you have exposure to integral calculus, how to do integration and how to handle the limits, how to do integration by parts. What we do here is, we make use of the substitution so that the integration gets simplified. We take p is equal to x minus mu by root 2 sigma. So, when you simplify this, it becomes x is equal to mu plus root 2 sigma p. So, I differentiate x with respect to p. I get dx by dp equals root 2 sigma. Elementary calculus, the important thing is to identify the correct form of the substitution for x in terms of p. So, that the integral will get considerably simplified. We said p is equal to x minus mu by root 2 sigma. So, when you square it, this expression inside the argument of the exponential term becomes minus p squared and x was replaced by mu plus root 2 sigma p. So, you are having x replacement for x here and this root 2 pi sigma is there in the original functional form and we substitute dx in terms of root 2 sigma dp. So, this can be integrated. I am just splitting the integral into 2 parts. So, you have the root 2 sigma cancelling out with this root 2 sigma. I am taking this pi, root of pi to the other side. That is why you have root pi into mu minus infinity to plus infinity mu into exponential minus p squared dp plus minus infinity to plus infinity root 2 sigma p exponential minus p squared dp. Please work out these steps on a paper so that you can make sure that you are doing the manipulations correctly. So the first term is mu into minus infinity to plus infinity e power minus p squared dp. This minus infinity to plus infinity e power minus p squared dp is a standard result. You will get root pi. So, you have mu root pi here and that is the same as the left hand side. Which means that the second term here, since the first term here became root pi mu, the second term here should become 0. That can be shown rather easily and so you have here root 2 sigma p exponential minus p squared dp and this is in a very convenient form already. You have e power minus p squared and you have minus half of 2p here okay and into minus 1. The reason why I am saying minus half of minus 2p is when you differentiate this particular term minus p squared you get minus 2p and so the integration becomes very simple and you will eventually get minus half into root 2 sigma e power minus p squared plus root 2 by 2 sigma e power minus p squared. You apply the upper limit here, you apply the lower limit here and you can see that the term vanishing. Hence we have proved that the mean of the normal distribution is indeed mu. That is what is given in the slide. The second term on the right hand side vanished and you have root pi mu is equal to root pi mu. Show that the variance of the normal distribution is indeed sigma squared. Again we write down the form f of x which by now you should know almost by heart 1 by root 2 pi sigma e power minus x minus mu whole squared by 2 sigma squared and again we do the same transformations p is equal to x minus mu by root 2 sigma x becomes mu plus root 2 sigma p dx is equal to root 2 sigma dp very straight forward. You know the variance is given by integral of x minus mu whole squared into f of x dx. So when that happens and you substitute for x minus mu, x minus mu whole squared will become p squared into 2 sigma squared and that is what is being written here. You have 2 sigma squared p squared 1 by root 2 pi sigma exponential minus p squared into root 2 sigma dp. Some steps have been omitted here. I would like you to do these steps and see whether you get this particular form of the integral. Alright. Now this integral is in fact very easy to evaluate. I am taking root pi by 2 to the left hand side and this sigma squared will cancel out. So essentially we have to show that root pi by 2 is equal to minus infinity to plus infinity p squared e power minus p squared dp. You may ask that since the x has been transformed into p how come the limits have not changed. So when x goes to plus infinity p also goes to plus infinity and when x goes to minus infinity p also goes to minus infinity. So the limits do not change even after the transformation from x to p. So you have root pi by 2 integral minus infinity to plus infinity. I am writing p squared as p into p e power minus p squared dp. The integration is now possible. We had already discussed how to integrate p e power minus p squared in the previous example. We get root pi by 2 equals minus p by 2 e power minus p squared at infinity plus p by 2 e power minus p squared at minus infinity and we know that both these terms will vanish on the application of the limit and you are only left with half into minus infinity to plus infinity e power minus p squared dp. This result came after applying the integration by parts method and when you do this you also by now know that minus infinity to plus infinity e power minus p squared dp is equal to root pi. So you have root pi by 2 equals root pi by 2 and hence equating the integral x minus mu whole squared f of x dx where f of x is the normal distribution. We can equate that integral the second moment about the mean to the variance sigma squared and the result is proved because we get the same answer on both sides after the manipulations have been carried out. Now we are going to look at the use of the normal probability plot. The normal probability plot runs to 2 pages as far as the normal distribution is concerned. The first page involves negative values of z and that corresponds to the left portion of the standard normal curve. You know that the standard normal curve has mean 0 and variance 1. So when you have the standard normal curve you have the negative part and you have the positive part. Since the distribution is symmetrical you need only one chart. You do not need both the charts. If you have only the negative portion of the chart you can even use it to find the probabilities for the positive portion of the curve. Anyway I have provided data corresponding to both the negative as well as positive values of z. We will be referring to those charts to pick up the probabilities for different values of z. Reading the chart is pretty easy. Sometimes you may not get exactly the number you want. However you can do interpolation between 2 values if your z value falls between 2 tabulated values. Sometimes you may also have to do the reverse. You are given the probability and then you are asked to find what is the value of z that is going to give the desired probability. So this is the inverse problem. We will first do a few examples. What is the probability of z? The standard normal random variable less than or equal to 0. This corresponds to the cumulative distribution function of 0. You do not even need a chart because the mean of the distribution is 0 and the curve is symmetrical. The area under the curve below 0 will be equal to the area under the curve beyond the 0 or above 0 and so the probability will be 0.5. To verify this let us go to the actual probability chart. So here we are the z value is 0 and we know that the distribution is expressed in such a way. The cumulative distribution function is expressed in such a way that it refers to the area under the curve below the listed z value. So you go to z corresponding to 0. This is minus 0.09 and so it comes to 0 here and you can see the probability to be 0.5. Now let us look at the second part. What is the probability of z less than or equal to 0.5? We have to find the cumulative distribution value of 0.5 and how do we do that? So the z value is 0.5 and we go to 0.50 and this is 0.51, 0.52 up to 0.59. For the normal distribution at z equals 0.5, the area under the curve or the probability is 0.6915. So this is the value here and you can see here that is the answer we have also reported. Similarly you can show that probability of z less than or equal to minus 0.5 will be 0.309 approximately and that happens to be 1 minus 0.691. So you can use the symmetry property of the normal distribution and find probability of z less than or equal to minus 0.5. The probability of z less than or equal to minus 0.5 would have been the same as the probability of z greater than or equal to 0.5. We just now found the probability of z less than or equal to 0.5 and that came to around 0.691. Hence what is the probability that z greater than or equal to 0.5 will be? That will be 1 minus 0.691 and we get 0.309. We have to do the inverse problem now. Suppose it is desired to do the inverse problem to find z corresponding to the given probability. For example, find z such that probability of z less than or equal to small z. This is a random variable and this is the value is equal to 0.95. We have to see what is the value attained by the random variable z such that the probability becomes 0.95. So we go to the appropriate table and we search for the probability of 0.95 in this table and we see that the probability of 0.95 falls between 1.64 and 1.65. You come horizontally along this direction. Here you have 1.6. When you come to 1.64 the probability is 0.9495 and when you come to 1.65 the probability is 0.9505. So the appropriate value of z would be somewhere between 1.64 and 1.65. So we can take it as 1.645. 1.645 would be the close enough answer. That is the answer which is given in this particular slide. What is the value of z such that the probability is 0.95. From the table you find the value of z to be 1.645. So you should be able to use the table for both direct calculation of the probability and the inverse calculation of z for a given probability. Now let us go to the next example. Example number 5. Here we have the question what is the probability of the random variable less than or equal to 2 z1 so that the probability is 0.93. We have to identify z1 for this situation. First we will identify what is 2 z1. So from the normal probability chart we can find that if 2 z1 is 1.476 the probability of z less than or equal to 1.476 is 0.93. Please look at the standard probability chart and verify whether this is indeed so. And then you can find out what is the value of z, z1 and that is coming to 0.738. Sometimes what may happen is you may think that for whatever reason probability of z less than or equal to 2 z1 may be taken as 2 into probability of z less than or equal to z1 okay. Then you may find the value of z1 but that value would be completely wrong. You cannot take the constant 2 or any constant here outside the outside. If you had taken it like that if you had equated probability of z less than or equal to 2 z1 as 2 times probability of z less than or equal to z1 the z1 value would have been minus 0.088 and this is completely different from the correct answer of 0.738. In my earlier lecture on normal distributions I was telling that the distribution of students marks is assumed to be normal or Gaussian and you can make the appropriate calculations for cutoff. For example mu plus 2 sigma and above may be taken to be s grade and so on. We will take a simple example and see how the cutoffs are fixed in this particular case. So the problem statement is you are having a large class of 200 students and the distribution of total semester marks is assumed or it appears to be normally distributed. The class average is 45% and the standard deviation is 10%. The instructors want to decide or identify the cutoff grades according to the following table. So our aim is to find the cutoffs. So you have the s grade, a grade, b grade so on to e grade. The instructors want to put the cutoffs such that 5% of the class gets s grade, 15% of the class gets a grade, 30% of the class gets b grade. You expect the majority of the students in the class to get a grade somewhere in the vicinity of b and c. So the majority of the class will have grades b and c. In this case 30% of the students should be having b grade and 25% of the students in the class should be having c grade and the d grade is 15% of the class and e grade is 5%. Here you also have the column of cumulative percentages. What I do here is simply add the total percentage of students. Here we have 5, 20, 50, 75, 90, 95 and if you include the students who are getting u grade then it comes to another 5% and that comes to 100. We are not going to look at the u grade okay. So we will be looking at the other grades. So the mark distribution is assumed to be normal with mean of 50 and standard deviation of 10 okay. So we have to first convert them into the standard form so that the probabilities can be obtained. We have to convert the random variable x into z. The random variable x is having its own mean mu and the variance sigma squared. So that normal distribution had a mean of 50 and standard deviation of 10 or variance of 100. If you want to convert it into the standard normal then it should be converted into a normal distribution of mean 0 and variance 1. For doing that we take z is equal to x-mu by sigma, mu is 50, sigma is 10. So we convert x into z by applying the transformation, z is equal to x-50 by 10. Now we have to find the cumulative distribution functions for different values of z. Let us look at the cutoff for grade s. So the grade s is defined such that the probability of the random variable z, the standard normal random variable capital Z is less than or equal to z subscript s. This is the z value corresponding to the s grade and that is equal to 0.95. We have seen from one of the earlier examples that the zs will correspond to 1.645 and you substitute 1.645 here and that is equal to x-50 by 10 and when you do the calculations 16.45 plus 50 is 66.45. So the s grade cutoff is 66.45. Any student getting value higher than 66.45 or keep it at 66.5 will be awarded the s grade. Now you want to see what is the probability of z less than or equal to zA. You have to identify zA such that the probability of z less than or equal to zA equals 0.8. How did you get 0.8? We want to put the cutoffs in such a manner that the percentage of students having the cutoff mark or above should constitute the 20% of the class. The A cutoff should be so fixed that 20% of the class should have A grade or s grade. So where is the cutoff for A? Even though the cutoff for A is 15% we have to add with the percentage of students we have to add the 15% to the percentage of students having the s grade also. What would be a nice idea here is to draw the normal distribution and on the normal distributions right hand side put some numbers as cutoff marks and you can put the s grade and A grade there okay. So you have to add the 5 to 15 so that the area beyond the cutoff for A will be 0.2. So to find that since the probability chart gives the area below the z the area below z will be 0.8 if the area above z A is 0.2. So the cutoff is decided based on the probability below the zA being 0.8 and in such a case we can see that zA is 0.842. I will show this so when you come to this particular table you locate z value of 0.84 but the probability is only 0.788. So you have to increase the value of z some more and so you see here corresponding to 0.84 the probability is 0.7996 and corresponding to probability of 0.802 you have a z value of 0.85. So the required z value would be somewhere between 0.84 and 0.85 for all practical purposes you may take the z value to be 0.84. I have used software so I have a more accurate value here and I plug in 0.84 or 0.842 here x-mu by sigma and then this becomes 8.42 and so 58.42 would be the cutoff for A. So the cutoff for A is 58.4. Now we have an interesting case the cutoff for grade B should be placed such that 50% of the students should have a grade B or above and 50% of the students should have the grade lower than B. You can see that even though the cutoff for B should be such that it encompasses 30% of the class we have to look at the cumulative distribution and see that 50% of the class will have either grade B or higher but when there are students who are crossing a certain cutoff like 58.4 in the present case they will automatically get the A grade and that would constitute about 15% of the class and if they cross a certain value corresponding to the cutoff for S they will automatically get the S grade and that will encompass about 5% of the class. So you can do the calculations and you will find that since the cumulative percentage is 50 or the cumulative probability is 0.5 that will definitely correspond to a z value of 0 and the standard normal random variable taking a value of z equals 0 means the cutoff for the B grade is 50. Since zB is equal to 0 we put 0 here and so the cutoff for the B grade is around 50. Similarly you can find out the cutoffs for the C grade and D grade and E grade the same exercise is followed the zC values have been identified using the same procedure as I outlined earlier. Now you can find out the cutoff for grade C, cutoff for grade D and the cutoff for grade E. You can summarize the results. You can see that the students who are getting 66.5 or above get a S grade and that encompasses about 5% of the class so in the class of 200 students 10 students get S grade. So the number of students who get the A grade is 30 and the cutoff for the A grade is 58.4. So any student getting grades or marks rather between 58.4 to 66.5 will be awarded an A grade and 30 such students meet this criteria. The cutoff for the grade B as we discussed earlier is 50 for grade C is 43.3, grade D is 37.2, grade E is 33.6 and any student getting marks below 33.6 would be automatically assigned a value of U. He may have to write the supplementary or the repetition of the course. If you count the total number of students you can see that the number comes to 10 plus 30, 40, 100, 150, 180, 190. So 10 students have failed the course unfortunately. Let us now go to another problem. So far we have been discussing about the normal distribution. Now we have a case where the distribution is described by a funny kind of expression f of x is equal to A power x. x can take values 1, 2, 1, 3. We had discussed this problem earlier in the case of discrete probability distribution. x was taking values 1, 2, 1, 3. But now when you discuss the same problem for the continuous distribution case then x will be ranging from 1 to 3 or x can take any value between 1 to 3. Here x can take values 1 or 2 or 3 if the probability distribution function is discrete. We are talking about a random variable which is discrete in nature. In that case x can take values only 1, 2 and 3. If the random variable x capital X is continuous then the range for x will be from 1 to 3, from 1 up to a value of 3. So the interval is now 1, 3. What is now the permitted value of A and what is the mean and variance of the distribution in the stated interval? To find the value of A we use the criterion that between the lower limit to the upper limit the area under the curve should be equal to 1. So using this criterion we integrate A power x between the permitted limits 1 and 3 and c for what value of A we will get this integral to be unity. So we carry out the integration. The integration is quite interesting. You put A power x is equal to p then take natural log on both sides x ln A is equal to ln p and then you differentiate p with respect to x and you will get 1 by p dp by dx is equal to ln A. Hence dx becomes dp by p ln A and the integral attains eventually this following form the lower limit becomes A and the upper limit becomes A cube dp by ln A is equal to 1 and that is nothing but p by ln A. So you get this expression A cube minus A is equal to ln A. By trial and error which can be quite easily done with the help of a spreadsheet you can find that A takes the value of 0.7. To find the mean value we evaluate mu is equal to 1 to 3 x A power x dx and this would require integration by parts recall that mu for a continuous probability distribution function is given by mu is equal to lower limit to upper limit integration x f of x dx here the lower and upper limits are 1 and 3 respectively. We have x A power x dx we can do integration by parts and integral of A power x dx is equal to A power x by ln A we get the value of mu to be 1.8838 this integration I am not showing all the steps I am expecting you to do it on your own and see whether you get the answer. Sometimes you will carry out the integration sometimes the integration may be quite lengthy and after doing it you may want to check whether you have done it correctly. There are a few softwares which can do the integration between limits and give you the numerical value. One such software is the MATLAB okay. If you have access to MATLAB you may want to use the integral option to see whether the answer is correct. In the next case we want to find the variance. Variance by definition is x minus mu whole squared f of x dx. So f of x is given by A power x one important thing you have to remember is the value of x is varying between 1 to 3 but you have fixed the value of A to be 0.7 that was the first subdivision of the present exercise. What was the permitted value of A so that the criterion for the probability distribution is satisfied. The criterion for the probability distribution was integral of 1 to 3 A power x dx is equal to 1 and after carrying out the integrations and putting in the limits you found that A was 0.7. Now you have to put the value of 0.7 wherever you see A. What I have done is kept A as it is so that the integration is first done and the substitution is finally done. So in this case A actually should be 0.7 here but I have left it as A because it will help me in my integration. Finally I will substitute the value of 0.7 into A. So once you do that you can again do the integration by parts and find out that sigma squared is 0.3253 or sigma is equal to 0.5704. I am taking the square root of this value to get 0.57. The next example is involving the log normal distribution. We were discussing the log normal distribution in one of the earlier lectures and we saw that when the random variable x was subject to a transformation okay and you took the natural logarithm of this random variable x it became a new random variable and this new random variable started to behave or obey a normal distribution okay. So this was a very useful result and what we can do is do the transformation for our data and use the properties of the normal curve in the analysis of the data. So the problem statement is particle sizes of represented by D from a crushing equipment may cover several orders of magnitude and their size distributions are often described by a log normal distribution. For one such distribution the parameters of the log normal distribution are alpha is equal to minus 1.1515 and beta is equal to 1.919. So these are the values of the parameters. The log normal distributions parameters are minus 1.1515 and 1.919. So the question is how can the parameter alpha be negative? Is it making physical sense? When we are converting D to ln D okay we are making the transformation from the original variable to the transformed variable. In this particular case we are taking the natural logarithm and so we have ln D. So if the value of D is less than 1, ln D can become negative. So the D value here was 0.162. The D value here was 0.3162 microns. So ln D became negative. The next question is quite simple. What is the form of the actual log normal distribution? The form of the actual log normal distribution was f of x is equal to 1 by x into 1 by root 2 pi beta squared e power minus ln x minus alpha whole squared by 2 beta squared. The difference from the normal distribution was you have ln x here instead of x and then you also have an additional 1 by x term here. So whenever you want to find the cumulative distribution using this function you have to be careful. You have to integrate the function in terms of x and not ln x okay. When you want to find the probability of a less than x less than b it should be integral of a to b f of x dx and so this will be integral of a to b and then this particular function dx okay. When you are using this form directly the question is what is the normal form of the log normal distribution okay. So you have changed the random variable x to ln x. So you have to consider ln x as a random variable a new random variable and think always in terms of ln of x okay. So to do that you can put ln x as some variable p and you can write dx by x as d of ln x okay and so dx by x is d of ln x. So you are going to have dx by x replaced by d of ln x as shown here and then you are also having ln x here and then 0 to m in fact becomes minus infinity to ln of m. So you take ln x as the random variable and if you start thinking in terms of ln x this is nothing but a normal distribution with mean alpha and standard deviation beta. I would suggest to you to carry out this derivation or this kind of transformation and integration on your own to convince yourself and become familiar with the use of the log normal distribution otherwise things can become slightly confusing. So what is the mean and variance of the log normal distribution? They are not directly alpha and beta. You have to use the formula mu of the log normal distribution is e power minus alpha plus beta squared by 2 and in this case it comes to 1.9933 microns. The variance is e power 2 alpha plus beta squared into e power beta squared minus 1 and that comes as 153.95 microns squared and the standard deviation comes to 12.407 micron. So there is another subdivision or a couple of subdivisions to this example 8 which we will see shortly. Coming to part t of this example we have really come to the part t. What is the probability of particles having sizes beyond mu of the log normal distribution plus sigma of the log normal distribution? Remember that both the mean and standard deviation have the same units. So in this case it is microns. What we have to do is to find the probability we have to convert to z the standard normal form by the transformation ln of d minus alpha by beta. So this d will be based on what is the expression given here. So ln of d1 minus alpha would be ln of mu plus sigma comes to 14.401. You may want to verify that minus alpha is minus 1.1515 and beta is 1.919 and you get z value of 1.99 and what is the probability? z greater than 1.99 is 1 minus probability of z less than 1.99 and we get the answer as 0.0233. So the probability of finding particles beyond mu plus sigma of the log normal distribution is pretty small at 0.0233. Part f what is the probability of particles having sizes below mu log normal minus 0.1 sigma log normal and let d2 be mu log normal minus 0.1 into sigma log normal that comes to 0.7525 microns and this we have to convert to ln. So ln of 0.7525 is minus 0.2844 and then we have to convert this into z ln of d2 minus alpha by beta which is ln of 0.7525 minus of minus 1.1515 divided by 1.919 and that z value comes to 0.4519 and the probability of z less than 0.4519 is equal to 0.6743. So I have given a few typical examples for the log normal distribution as well as the normal distribution. There are many variants and versions of the examples you can do. I would suggest to you to take up any standard statistics and probability textbook and work through some of the problems to see not only whether you are using the probability charts correctly but whether you are able to understand the problem statement correctly and then do the necessary calculations. Sometimes you may do the calculations correctly but the answer is not what the problem statement was looking for. So it is important to understand the question properly and then do the mathematical calculations also correctly. Thank you.