 As Salaamu Alaykum, welcome to lecture number 29 of the course on statistics and probability. Students you will recall that in the last lecture I discussed with you in detail the binomial distribution. Also we discussed the fitting of the binomial distribution to real data. Towards the end of the last lecture I began the discussion of the hyper geometric distribution and today I will convey to you a detailed explanation of this particular distribution. As you will recall in the last lecture I mentioned to you that the four basic properties of a hyper geometric experiment are number one every trial results in a success or a failure. Number two the successive trials are not independent of each other and number three the probability of success changes from trial to trial last but not the least property number four that the number of trials is fixed in advance. Let us apply this concept to an example as you now see on the screen. Suppose that the names of five men and five women are written on slips of paper and placed in a hat. Four names are drawn. What is the probability that two are men and two are women? Students, let us tackle this problem step by step. As you have seen there is no mention in the problem that it is a hyper geometric experiment. It is obvious that this decision has to be taken by myself. It is a binomial experiment, a hyper geometric experiment, a Poisson process or something else. See in this problem we have five men and five women, a total of ten. So we can say that we are dealing with a finite population of size ten. Now we are drawing four names at random. Out of those four names which are drawn, two are men and two are women. Now students, it is our decision whether we would like to regard men as success or women as success. So let us suppose that we say that the occurrence of the name of a man is success and the occurrence of the name of a woman is failure. If I do that then I have to think of the three parameters of the hyper geometric distribution assuming that this is a hyper geometric experiment. If you pay attention to those three parameters, as I said in the last lecture, capital N is the population size at the beginning of the experiment. So in this case it is equal to ten because there are five women and five men and small n is the sample size, and what was k? The number of successes in the population in the beginning of the experiment. Since we said that occurrence of a man is success, it means that in the beginning of the experiment there are five men out of those ten, so k is equal to five. Students, if this is the case, then we have defined the parameters. But actually before defining these parameters, I wanted to think whether the four conditions are fulfilling. Let us look at it step by step. Condition number one, every trial results in a success or a failure. Yes, either the name that you draw, it will be the name of a man or that of a woman. Number two, the successive trials are not independent of each other. Students, this is a crucial point. Since your population size is finite, it is only ten. And because you are sampling without replacement, because if you are taking out four names, then it is implied that you are doing this work without replacement. It is not that you have taken out one name and put it back. And then again, obviously if you take out one, then you can keep it aside and take out the other. And this is called sampling without replacement. Because you are not replacing the pieces of the paper into that hat. So, if we do sample without replacement from a finite population, students, the trials are not independent. Why? For the first time, a woman's name came out and we took it out. So, how many elements remained in that population? Only nine. And how many women are there? Only four. So, what is the probability that this time you are going to have the name of a man? Students, it is five over nine. It was five by ten. Because there were ten in all and five men. So, what happens in the second row depends on what happened in the first. And now it is only four men and five women in that total of nine. So, everything falls in place. Both the conditions, second and third of the hyper geometric experiment, they are fulfilled. The successive trials are not independent of the previous ones. And the probability of success in this case, the probability of getting the name of a man changes from trial to trial. And the last condition is that the number of trials is fixed in advance. Our institution. So, now that everything is clear, I think we are in a position to apply the formula of the hyper geometric distribution. As you now see on the screen, the hyper geometric formula is the probability that my random variable capital X takes the value small x is equal to k c x n minus k c n minus x divided by n c n. Substituting the values of capital N, small n and k in this formula, I obtain 5 c x, 5 c 4 minus x and divide this product by 10 c 4. Students, that we should have two men and two women, because x represents the number of successes in small n trials. So, substituting the value x equal to 2 in the formula that I just presented to you, we obtain as you now see on the slide, 5 c 2 multiplied by 5 c 4 minus 2 divided by 10 c 4. And solving this expression, we have a number of successes in small n trials. So, substituting the value x equal to 2 in the formula that I just presented to you, we obtain as you now see on the slide, 5 c 2 multiplied by 5 c 4 minus 2 divided by 5 c 4 minus 2 divided by 10 c 4. And solving this expression, the required probability comes out to be 10 by 21. Students, I would like to encourage you to compute the other various probabilities that you can also compute in this particular problem. I hope that you will realize that the random variable x denoting the number of successes in your sample, in other words the number of names of men that you might obtain, this variable goes from 0 to 4. Because it is possible that out of those four names, not even one is a man or one of them is a man or two or three or a maximum of four. Now, if you compute the probabilities of all these x values, students you are dealing with a proper probability distribution, because none of these answers will be negative and the sum of these probabilities will come out to be 1. And when you draw the line chart of this probability distribution, then of course you can start thinking about the center, the spread and the shape of your probability distribution. I would definitely like to encourage you to work on this problem on your own. Now, when you do that, then you will be able to verify the mathematical properties of the hypergeometric distribution that I am going to present to you now. As you see on the slide, the mean and the variance of the hypergeometric distribution are mu is equal to small n multiplied by k divided by capital N and sigma square is equal to small n into k by capital N into capital N minus k by capital N and this whole expression multiplied by capital N minus small n over capital N minus 1. Now, you will say again something very complicated, but believe you me it is not so complicated when you practice it. And as I have indicated earlier, we will not be going into the mathematical derivations of these formulae in this course, but rest assure every one of these formulae has its own mathematical background and a very solid mathematical derivation. And then students find the mean and variance of that probability distribution by the general formula that I indicated to you earlier. And what was that? That for any discrete probability distribution, mu is equal to sigma x into f of x, x column you have f of x. So, multiply the x column with the f of x column, add it up and you get the mean. But the property I have just conveyed to you, that will come out to be equal to small n minus k by capital N into small k over capital N or its problem may as you remember small n is equal to 4, k is equal to 5 and capital N is equal to 10. Hence the mean for this particular hyper geometric probability distribution is equal to 4 into 5 over 10, in other words 4 by 2 and that is equal to 2. So, this is the what we can expect on the average. If you do this process again and again, sometimes you will get no man, sometimes you will get all four men, sometimes one, sometimes three, but in the long run on the average you can expect to have two men and two women in your sample of size 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . start, draw k j, verify the formulae of the mean and the variance and decide for yourself about the shape of the distribution. What I would like to convey to you now is another very important property of the hyper geometric distribution. As you now see on the slide, if capital N becomes indefinitely large, the hyper geometric probability distribution tends to the binomial probability distribution. Students, to explain this point, I have a long discussion with you. But, you have to believe that if you pay attention to the different points of this, then the shape of the equation will become clear and, inshallah, whenever you tackle a problem like this, you will not have a problem deciding whether you should apply the hyper geometric formula or the binomial formula. First of all, you see that there are two types of sampling. Either you sample with replacement or you sample without replacement. The first type is that you have taken out, noted and put it back so that next time the same element can come again. And the second type is without replacement. The other thing is that you can sample either from a finite population or from an infinite population. So, we get a situation in which there are four possibilities. Either you sample without replacement from a finite population or you sample with replacement from a finite population or you sample without replacement from an infinite population or last but not the least, you sample with replacement from an infinite population. Now these four situations, students, in which the first situation is that you are sampling without replacement from a finite population, that is exactly the situation when you will apply the hypergeometric formula as I have explained earlier. But note that in both the situations in which you are sampling with replacement, you will not be needing to apply the hypergeometric formula, rather you will apply the binomial formula. What is the reason for this? This is a crucial point and you should pay attention to this. When you have taken out, noted what is the purpose of that element and replaced it into your population students, the constitution of your population remains undisturbed. That is, as it was before, it has been changed again. And now the second element you will take out, it will not have any relation to what happened before. So, the trials become independent of each other and the binomial experiment conditions are applicable now. Hence, in both these situations when you are sampling with replacement whether it is finite population or infinite population, you have to apply the binomial formula. I have discussed three situations with you. The fourth situation is that it is related to this property which I have just presented. The fourth and last situation is that you are sampling without replacement from an infinite population. This is the most interesting situation. When we say infinite population, it is obvious that practically speaking, it means that it is a very large population. In real life, there will be no infinite population, but if it is very large, we can say that it is an infinite population. Now from this, you are sampling without replacement. Now technically speaking, you should apply a hypergeometric formula because if you are doing without replacement, that element is taken out, thrown out, then now one element is reduced in that population. And as I gave you earlier, your probability of success in the second trial changes, but the point is that if it is a very large population, students, even if you draw one element and throw it away, the population is almost undisturbed. It is an example of this that as you are an island of sand, and if you pick up one element and throw it away, then that does not disturb that population. If you pay attention to this, then you will realize that we can say that the various trials are independent and we can apply the binomial formula. Of course, there is a mathematical derivation which shows that the hypergeometric formula tends to the binomial formula whenever capital N tends to infinity, I hope that you will think about it and appreciate the point. Now as a rule of thumb, whenever small n is less than or equal to 5 percent of capital N, then we say that capital N is much larger as compared with small n, so that capital N can be regarded as infinite compared with small n and in that situation we do apply the binomial formula. We can apply it if we want. On the other hand, as you now see on the slide, if the sample size small n is greater than 5 percent of the population size capital N, in other words small n is not extremely small as compared with capital N, then we apply the hypergeometric formula. Students, the next distribution that I am going to discuss with you is the Poisson distribution. This distribution was discovered by the French mathematician Poisson in the year 1837 and you will be interested to know that it has two main applications as you see on the screen. Number one, the Poisson distribution is a limiting approximation to the binomial distribution when small p the probability of success is very small, but n the number of trials is so large that the product N p equal to mu is of a moderate size. And number two, the Poisson distribution is a distribution in its own right by considering what is called a Poisson process, a process where events occur randomly over a specified interval of time or space or length. First point is that basically we are dealing with a binomial experiment and as such we should apply the binomial formula, but it has been mathematically proved that if the two parameters N and p of the binomial distribution are such that p is very small, but n is large then the binomial formula can be replaced by the formula of the Poisson distribution. Number two, the problem is that there is a certain kind of process that is called a Poisson process. What is this process? It is a process in which you first determine a time interval or a space interval and you see that the event of interest in this interval that is occurring in a random manner. For example, an interesting example is that a typist, he is typing page upon page upon page, if you understand a unit of space, then you will agree that typing error that can occur anywhere on that page, it will happen by mistake unintentionally and in that manner we can say that it is a random mechanism or the ninth line. This is one example and another example that of interval of time. For example, number of traffic accidents in a particular city in a 24 hour time interval. It will be unintentional or randomly, it can occur at any time. Such situations are called Poisson process and in this situation it has been mathematically proved that the Poisson formula is the appropriate formula. Alright, I have mentioned the Poisson formula twice or three times. What is the Poisson formula students? As you now see on the slide, the Poisson distribution is defined as the probability that capital X is equal to small x is equal to e raised to minus mu mu raised to x over x factorial, where x takes the values 0, 1, 2, so on up to infinity and e is equal to 2.71828. Students, you should realize that the Poisson distribution has only one parameter and that is mu. Formula kya tha? e raised to minus mu, mu raised to x over x factorial, to mu he wo wahid number hai jisko hum kahenge parameter of the Poisson distribution and if you do the mathematics, you find that mu actually is the same thing which is the mean of our Poisson distribution. Now jo doh main aapko applicatione bataiin, pehli kya thi that basically we are dealing with a binomial situation, but n is large and p is very small, then we can apply the Poisson formula instead of the binomial. Ab practically speaking kya values honi chahinye, generally it is acceptable that if n is greater than or equal to 20 and p is less than or equal to 0.05, then students we can apply the Poisson formula instead of the binomial formula. So, let us apply this now to an example, 200 passengers have made reservations for an airplane flight. If the probability that a passenger who has a reservation will not show up is 0.01, what is the probability that exactly three of these 200 will not show up? Aapne dekhah ki yeh kaafi interesting situation, pehli baat to yeh ke kya hum binomial experiment kasah deal kar rahe, dekhahe doh possibilities hain for any passenger who has made a reservation, either he will show up or he will not show up. And what we are interested in is what is the probability that out of these 200 people three will not show up? So, jis tiz me hum interested hote hain usi ko hum generally success kahite hain. So, let us regard no show as success, having done that what is the probability of this event abhi hum ne pada, ke the probability is 0.01 that the person will not show up aur ye probability aap ke khyaal me kise ye figure hasil hua hoga students, it is from the past records of that airplane company, ke hazaar yeh 10,000 passengers jis se pehle un ke pass reservation karane ke li aaye un me se 1 percent the who did not show up. So according to the relative frequency definition of probability, we can say that this probability is 0.01, agar hum ne p bhi determine kalliya 0.01 aur n bhi waze hain 200, tukke is bakth 200 logon ki baat ho rahe hain. So the number of trials is 200, then students, we can find the answer to our question. We have two options for that, either we apply directly the binomial formula n c x p raise to x q raise to n minus x. So is case may 200 c x 0.01 raise to x 0.99 raise to 200 minus x and then because we want to know what is the probability that exactly 3 will not show up therefore we put x equal to 3 in this formula and we obtain the exact answer. But the other option is that instead of applying the actual binomial formula students, we can apply the Poisson formula, why? Because n is large and p is very small, yeh hi dosha aaye tha na jis ke tha hai the binomial distribution can be approximated by the Poisson distribution. If we wish to do so, then how do I proceed? The first thing is that to apply the Poisson formula, I need the value of mu, the mean of my Poisson distribution. Lekin tukke bunyadi torpe bo binomial situation hai jis ke saath hum isko equate karne wale Hence we take mu equal to n p, the mean of the binomial distribution. Putting n equal to 200 and p equal to 0.01, what do we obtain? As you now see on the screen, we have mu is equal to 200 into 0.01 and that is equal to 2. Now applying this in the Poisson formula, we have the probability that capital X is equal to small x is equal to e raised to minus 2, 2 raised to x over x factorial. And since we are interested in x equal to 3, because we want the probability that exactly 3 persons will not show up, hence substituting x equal to 3, we obtain e raised to minus 2, 2 raised to 3 divided by 3 factorial. And solving this expression, the probability comes out to be 0.1804. In other words, we see that there is 18 percent chance that out of these 200 passengers, 3 will not show up. To abhi ye us airplane company ki discretion pe hai ki wo is 18 percent ko substantial probability samajti hai ya wo ye samajti hai ke this is quite small. And then they decide that maybe we can overbook by 3 persons ke agar hum 200 seat hai aur 203 logon ki reservation kar lein to biaasi fi sath chance hai ke wo jo 3 extra logon ki hum ne reservation ki hai, we will be able to put them on the plane. Dekhaha apne, kitna interesting hai real life application of probability distributions. This is the most fascinating thing about statistics. The ability to apply mathematical complicated formulas in real life day to day situations alright. Poisson distribution ki jo pehli application hai as a limiting approximation to the binomial wo to hum ne discuss kar li. Let us now get back to the second application that of the Poisson process. As you now see on the screen, a Poisson process may be defined as a physical process governed at least in part by some random mechanism. Stated differently, a Poisson process represents a situation where events occur randomly over a specified interval of time or space or length. Such random events might be the number of taxi cab arrivals at an intersection per day, the number of traffic deaths per month in a city, the number of radioactive particles emitted in a given period, the number of floors per unit length of some material and the number of typing errors per page in a book and many other similar examples. Ye sare example jo main aapke saam ne present ke aap agree karenge ke wo jo event hai that will be occurring in a random manner as I explained earlier. Now as you see on the screen, it can be said that under these properties it can be shown mathematically that the probability for the number of occurrences of a random event, the event of interest in an interval of stated length t is given by the Poisson distribution with the parameter lambda t. That is the Poisson process formula is the probability that capital X is equal to small capital X is equal to e raised to minus lambda t, lambda t raised to X over x factorial. Students, aap to aap jakinan karenge ke ye to bohat hi zyada complicated hoge, lekin meri bohi purani advice ke aap panic nakane and just consider every point step by step. Soce pheli baat ye ke pehele jab thodhideh pehele me ne Poisson distribution define ki to me ne aapko formula kya aapke saam ne page kiya the? e raised to minus mu, mu raised to X over x factorial. Aap note ke je ke jo formula me ne abhi aabi present kia that is exactly the same pattern, We are writing lambda t. So, let us see what exactly do we mean by lambda and what exactly is t. As you now see on the slide, lambda represents the average number of occurrences of the outcome of interest per unit of time. It may be time or it may even be space. t represents the number of time units under consideration and once again I want to remark that it can also be units of space. Also x represents the number of occurrences of the outcome of interest in t units of time or space. Let me explain this now to you with the help of a very interesting example. Telephone calls are being placed through a certain exchange at random times on the average of 4 per minute. Assuming a Poisson process, determine the probability that in a 15 second interval there are three or more calls. Students, what is the first thing? Are we actually dealing with a Poisson process? Yes, we are because you will agree that at any telephone exchange in any specified interval of time, may it be one hour or one minute, the telephone calls will come in a random manner. So, in that in this way it is a Poisson process and now let us concentrate on what we are interested in finding. We are saying that in a 15 second interval, three or more calls will be obtained. What is the probability of this? Maybe we can interpret it in a way that it will be appropriate. We want to say that in a 15 second interval, it is very difficult to handle three or more calls for a telephone operator. So, what is the information that we had in this example? The statement was that on the average, how many calls are we getting per minute at this particular exchange? Four calls per minute. So, students, I hope you realize that if we regard one minute as one unit of time, then lambda is equal to 4. After all, what was lambda? The average number of occurrences of the outcome of interest per unit of time. So, if you agree with that, then the only other thing we have to think about now is t, the number of time units that we are interested in. We were saying that in a 15 second interval, three or more calls are obtained. So, how many minutes are there in a 15 second interval? Obviously, one fourth of a minute and therefore, we can say that t is equal to 1 over 4. When you have specified lambda or t, half the problem is solved. The parameter of our Poisson distribution then is lambda t which is equal to 4 multiplied by 1 by 4 and that is equal to 1 and hence as you now see on the screen, the Poisson formula is e raised to minus 1, 1 raised to x over x factorial. Having determined the formula, ab aakhri baat kya hai, ke what is x? Yani, kitne telephone calls me ham interested hain students. Ham ne kaha tha, what is the probability that we receive three or more calls in a 15 second interval? Aagar ham isko directly tackle karna chahin, toh we will be in a bit of trouble, kyunke ham determine nahin kar satte exact number. Yani, aap tekhin hain ke three or more ka matlab hai, aayidhar three calls or four or five or seven or twelve to ye toh ek indefinite sa ho gaya. But, if we apply the rule of complementation which we have already dealt with, probability of a bar is equal to 1 minus the probability of a, then how do I tackle this problem? The probability that capital X is greater than or equal to 3, where capital X of course represents the number of telephone calls that I receive. This probability is equal to 1 minus the probability that X is less than 3 and as you now see on the screen, this means that we are trying to compute the probability 1 minus the probability that X is equal to 0 or X is equal to 1 or X is equal to 2. And according to the addition theorem, this means that we are computing the probability 1 minus the sum of the probabilities of X being 0, 1 and 2. Hence, our formula becomes the probability that X is greater than or equal to 3 is equal to 1 minus the sum of e raise to minus 1, 1 raise to X over X factorial, where X goes from 0 to 2 and substituting the value of e raise to minus 1, which is 1 over e, it is 0.3679. The formula is 1 minus summation 0.3679 into 1 raise to X over X factorial such that X goes from 0 to 2. Now, substituting the values of X in this formula, putting X equal to 0, 1 and 2 respectively, we finally obtain the probability that X is greater than or equal to 3 is equal to 0.08025. This means that there is only 8 percent probability that 3 or more than 3 will be received at this particular telephone exchange in a 15 second interval. So, it is a very fairly small probability and hence the management can decide that the workload is not extra ordinary for the telephone operator and therefore, they do not need to make a policy decision to change the way the telephone exchange is operated. Students, the next thing that I would like to convey to you are some important properties of the Poisson distribution. As you now see on the slide, the first property is that if the random variable X has a Poisson distribution with parameter mu, then its mean and variance are given by E of X is equal to mu and variance of X is also equal to mu. Yes, this is one striking property of the Poisson distribution that its mean and variance are numerically equal. Of course, is the center of the distribution and the variance and its square root are measuring the spread of the distribution, but it is an interesting mathematical relationship between the measure of central tendency and the measure of spread that the mean and the variance are numerically equal if we are dealing with the Poisson distribution. Of course, this property has again its own mathematical derivation, but for the purposes of this course, I would like you to simply verify this property for any particular Poisson distribution that you might deal with rather than going into the mathematical rigorous derivation. Now, the second property of the Poisson distribution is that the shape of the Poisson distribution is positively skewed, but the distribution tends to be symmetrical as mu becomes larger and larger. It is positively skewed if P is less than half, but it is negatively skewed if P is greater than half, but here we see that the Poisson distribution will never be negatively skewed. It is basically a positively skewed distribution, but if mu is large then it will appear like an approximately symmetrical distribution. I would like to convey to you that just as we can fit a binomial distribution to real data, we can also fit the Poisson distribution to real data. Mu is equal to n p, mu was replaced by x bar and thus we estimated the value of the parameter p. We will equate mu to x bar and the x bar, the mean of the observed frequency distribution will be used as the parameter instead of mu and hence our formula will become e raise to minus x bar, x bar raise to x over x factorial. So, the procedure is absolutely the same and I would like to encourage you to work on this problem on your own. You can find such problems in your text book and in many other books and I would like to also repeat that after we have fitted any distribution to real data, we also want to determine is it a good fit or is it not a very good fit and as indicated earlier there is a proper procedure called the chi square test of goodness of fit which you can apply after you have fitted a distribution to real data, but we will be doing the chi square test and many other tests in the third part of this course. At this point in time students before I close the segment on discrete probability distributions, I would like to convey to you another very interesting point. You have noted earlier that in certain situations the Poisson formula approximates the binomial formula. And as you now see on the slide we can say that the Poisson distribution can be used to approximate the hyper geometric distribution when small n is less than or equal to 0.05 times capital N, small n is greater than or equal to 20 and small p is less than or equal to 0.05 and in this expression that you have just read small p itself is equal to k over capital N. Once again no need to be confused just put the various points together and you will arrive at the result. Baharhal by and large we have discussed in detail various important discrete probability distributions and the next very important segment of this course my dear students is continuous probability distributions most importantly the normal distribution. Prior to talking about the normal distribution I would also like to discuss with you at some length the continuous uniform distribution. And let us now define it in a formal manner as you now see on the slide. A continuous random variable x is said to be uniformly distributed if its probability density function is defined as f of x is equal to 1 over b minus a such that x lies between a and b. The graphical representation of this particular density function explains why it is called a uniform distribution. As you see on the slide if f of x is equal to 1 over b minus a this means that the height of my curve which in this case is a straight line that height is equal to 1 over b minus a for all the x values that are under consideration that is all the x values from a to b. And because of this fact that it is a horizontal line we can say that it is a uniform distribution. Also it is interesting to note that we obtain something like a rectangle and therefore this is also called a rectangular distribution. Students what are the basic properties of any continuous distribution f of x is greater than or equal to 0 and the total area under the curve should be 1. It is a uniform distribution it is like a rectangle so you do not even have to apply the integration formula to find the area under this distribution. Now, you see its height that is equal to 1 over b minus a as you just saw and its base how much is that a to b how much is the distance obviously b minus a because b is greater than a and therefore multiplying the height 1 over b minus a by the width of the base b minus a of course we get 1. The total area under the uniform distribution is exactly equal to 1 and what are the other properties of any distribution that we are interested in the mean the spread. So, in this case students I hope you realize that because it is an absolutely symmetric distribution in the sense that if we place a mirror in the middle the left hand side is the mirror image of the right hand side therefore the mean will be at the exact middle of the distribution the left edge is a the right edge is b and therefore a plus b over 2 gives us the center therefore it is very easy to find the mean you do not have to do integration and what about the variance as you now see on the screen the variance of the uniform distribution over the interval a b is equal to b minus a whole square over 12. Once again you can verify it by first finding the variance by the general formula e of x square minus e of x whole square and later finding it by this simple formula that I presented and you will see that they are one and the same. Now, students when do we apply the uniform distribution? Jab kabhi bhi hame is khasam ka koi phenomenon hamaare saam ne ho ki jaan hama ye kais sakein ke the various possible values of x are evenly spread out over the entire range of the values or there is no clustering at any particular point then we are dealing with the uniform distribution. As you now see on the screen examples of such phenomena are number one if a short exists in a 5 meter stretch of electrical wire it may have an equal probability of being in any particular 1 centimeter segment along the line. Example 2 if a safety inspector plans to choose a time at random during the 4 afternoon work hours to pay a surprise visit to a certain area of a plant then each 1 minute time interval in this 4 work hour period may have an equally likely chance to being selected for the visit. Also the uniform distribution arises in the study of rounding of errors etcetera. Students this brings us to the end of today's lecture and in the next lecture inshallah we will deal with the normal distribution best of luck and until next time Allah