 As Salaamu Alaikum, welcome to lecture number 28 of the course on statistics and probability. Students, you will recall that in the last lecture, I discussed with you the discrete uniform distribution and after that we began the discussion of the binomial distribution. I will take up the discussion of the binomial in detail with reference to the same example that we started last time. As you now see on the screen, suppose that we toss a fair coin 5 times and we are interested in the number of heads. Thus, our random variable x which denotes the number of heads goes from 0 to 5 because we can have either no head or one head or 2 or 3 or 4 or a maximum of 5. Now, the question arises that in this situation, how do we obtain the various probabilities? Now, this is not a uniform distribution, because we say that all these x values probabilities are equal. In fact, in any such situation where the four basic properties of a binomial experiment are fulfilled, the ones that I shared with you in detail last time, students, the formula of the binomial distribution is probability that our random variable capital X takes the value small x is equal to n c x p raise to x q raise to n minus x. In this formula, n and p represent the two parameters of the binomial distribution, n is the number of trials that has been fixed in advance and p is the probability of success in each trial. q represents the probability of failure in each trial and therefore, q is equal to 1 minus p. Students, I will simply convey to you how to apply this formula in real life situations where we have the binomial experiment going on. In this example, n is equal to 5, because we are tossing the coin 5 times and p is equal to half, because we have said that the coin that we are tossing is a fair coin. So, substituting these two values in the formula that I just conveyed to you, we obtain the probability that my random variable capital X takes a value small x is equal to 5 c x half raise to x into half raise to 5 minus x. The point to be noted is that since p is equal to half, therefore, q which is 1 minus p is equal to 1 minus half and that is half. Now, that we have the formula of the binomial distribution, students, what we have to do is to substitute the various values of x for which we want to find the probabilities. As I said earlier, the x values are 0, 1, 2, 3, 4, 5 and so, substituting these values in this formula 1 by 1, we have the following situations. When x is equal to 0, the probability is 5 c 0 half raise to 0 into half raise to 5 minus 0 and solving this expression. The probability comes out to be 1 by 32. When x is equal to 1, the probability is given by 5 c 1 half raise to 1 into half raise to 5 minus 1 and therefore, the answer is 5 over 32. Proceeding in this manner, we obtain all six probabilities and in this example, they are 1 by 32, 5 by 32, 10 by 32 and 10 by 32, 5 by 32, 1 by 32. The sum of the probabilities is 32 by 32 and that is exactly what we wanted and of course, we note that none of these probabilities is negative. Hence, we are sure that we are dealing with a proper discrete probability distribution. Students, abhi jo distribution mein aapke saamne present ki, iss mein ek cheese, I hope ke aapne note karli hoge, it is an absolutely symmetrical distribution. Abha aap kahange ke line chart to amne drawing kia, to kaisi puta chel kia? I think it should be quite obvious. The probabilities were 1 by 32, 5 by 32, 10 by 32 and uske baad 10, 5 and 1 by 32. To aap table hi se andaza lagaasakte hain ke wo agar us table ke ter mein aap ek horizontal aina karakare yani litaye, to the top of the table is the mirror image of the bottom part of the table as far as the probabilities are concerned. Eka aur bade interesting baad note ki je, hamne dekhaha ke jab hamne x ki value 0 iss formulae mein rakhi to hamara ansar aaya 1 by 32. Students, ye ek fair coin tha jisko hamne paach dafa toss kia. To kia ham ye kaisakte hain ke because of the fairness of the coin, the chances of head occurring are exactly the same as the chances of tail. Of course, that is why we said that P is equal to half and Q is also equal to half, yani the probability of head is half and the probability of tail is half. The point to understand is that because of this fact that head and tail are equally likely to occur, therefore these probabilities can also be computed by the direct formula which is the valid in the case of the classical definition of probability. The simplest formula that we all know m over n where m is the number of outcomes favorable to what I want and n is the total number of possible outcomes. To ab dekhte hain ke ye jo 1 by 32 ansar aaya hain for x equal to 0 kia iss direct classical definition waale formulae se bhi hamne ye result milna tha students note ke ye ke we are tossing the coin five times. So, there are two ways of tossing the coin first time and two ways of tossing the coin second time and two ways of tossing the coin third time and so on aur agar ham ye kaam paach dafa karenge toh because of the multiplication theorem that we did sometime back what is the total number of ways in which we can toss the coin five times 2 into 2 into 2 into 2 into 2 and students this number is equal to 32. What we will obtain by tossing the coin five times in any one go is not a pair not a triplet, but that thing which has five entities for example, tail tail tail tail tail or head tail tail tail tail or head head tail tail tail and so on and so forth. Now, the total number of such entities or such groups is 32 as I just mentioned, but we are interested in x equal to 0 that is the number of heads equal to 0 that is all tails and students you will agree with me that out of these 32 groups one and only one is favorable to what I want and which one is that the first one tail tail tail tail tail. Hence according to the classical definition one outcome favorable 32 total number of possible outcomes and the probability of x equal to 0 that is no head is 1 by 32. Dekhah aapne agar aap methodically problem ko approach karen toh everything falls in place aur kisi kisimki koi problem nahi aati. You see when you deal with things mathematically scientifically things fall in place and one feels so happy and delighted that you have obtained the correct result and you have an intuitive feeling that it is correct. Now as I said earlier I will not be dealing with the derivation of the binomial formula in this course, but we will be applying it and I hope students that by this explanation that I have just presented you are confident that the formula that I presented to you n c x p raise to x q raise to n minus x is indeed the correct one for this kind of a situation. Having computed all the probabilities of course we are interested in drawing the graph of this distribution and just as before we will be drawing a line chart in which we take the various x values along the x axis and the probabilities along the y axis. So as you see on the screen we obtain an absolutely symmetrical distribution. That is the next thing we are interested in as usual the mean and the spread of the distribution. Students for any discrete probability distribution the mean is obtained by the formula expected value of x is equal to sigma x into f of x and the variance is found by the formula expected value of x square minus expected value of x whole square. But you will be interested to know that in the case of the binomial distribution these two rather cumbersome formulae can be replaced by two extremely simple ones. And as you now see on the screen the mean of the binomial distribution is e of x is equal to n p and the variance of x is equal to n p q. Of course these formulae have their own derivations which I will not be doing in this course. We will simply apply them and if you are interested you can also find the mean and variance of a binomial distribution by the other formulas that we discussed earlier and you can compare the results of the two. So in this particular example since the number of tosses is 5 and since p is equal to half hence the expected value of x is equal to 5 into half and that is 2.5. Similarly since q is equal to half therefore the variance of x is equal to 5 into half into half and that is equal to 5 over 4. When we take the square root of this quantity the standard deviation of our binomial distribution comes out to be 1.12. Dividing the standard deviation by the mean and multiplying by 100 we obtain the coefficient of variation which in this particular example comes out to be 44.8 percent. The binomial distribution that we considered in this particular example came out to be absolutely symmetric but students this is not the case all the time. In fact, as you now see on the screen if p is equal to q in other words if both p and q are equal to half then we obtain an absolutely symmetric distribution. If p is less than q in other words p is less than half whereas q is greater than half then our distribution is positively skewed and if p is greater than q in other words if p is greater than half and q is less than half then our distribution is negatively skewed but the degree of skewness decreases as the number of trials n increases. Having understood the basic concept of the binomial experiment and the binomial distribution students I would now like to convey to you what is called the fitting of the binomial distribution to real data. Let me explain this concept with the help of an example. The following data has been obtained by tossing a loaded die 5 times and noting the number of times that we obtain a 6 fit a binomial distribution to this data. The data is as follows the random variable x takes the values 0, 1, 2, 3, 4, 5 and x represents the number of 6's that we have in 5 tosses of the die and the frequencies corresponding to the various x values are 12, 56, 74, 39, 18 and 1 so that the sum of the frequencies is 200. Let us try to understand this problem. You may be confused that we tossed 5 times so how did we get this 200? We will also approach this methodically step by step. The first thing is that this binomial experiment contains small n that is 5 because the die we are tossing it 5 times but the point to understand is that this particular binomial experiment of tossing a die 5 times is being repeated 200 times. We must not mix small n with capital N. In this problem small n is 5 but capital N is 200. That one binomial experiment consisting of 5 tosses of the die that experiment is being repeated 200 times. Now that you have understood this basic point the other thing is what is the value of P? Now if this is a fair die then we say that 6 that is 1 over 6 according to the classical definition but here we are saying that it is a loaded die. The one which is not fair, it is not a fair die. Some of the faces have greater probability of occurring than some others. So now in this situation students, how will we compute probability of getting a 6? I hope you remember the relative frequency definition of probability that if an experiment is repeated again and again and again a very very large number of times then the proportion of successes is taken as the probability of success. This example getting a 6 may be regarded as success or if we toss this die too much then the proportion of 6 that would be the probability. Now the problem is that according to this definition you should toss this die infinitely and the proportion of that will be the probability distribution. That is equal to NP. That the mean of any binomial distribution is equal to NP. But you agree that in any real life scenario you cannot repeat the experiment infinitely. I may be able to repeat this particular experiment of tossing a die 5 times 200 times or I might repeat it 2000 times or even 2 million times but students I cannot do this infinite number of times. So what should we do now? This means that if we compute the mean value of the data we have, then we can only call it X bar. So that is only a sample of tosses out of the infinite number of tosses that we could have done. Since this is only a sample, this means that it is eligible to be called X bar and not mu. Since X bar itself can be available, students the best we can do is to replace the theoretical value mu by its estimate X bar. After all sample mean is always an estimate of the corresponding population parameter. So this how I say our equation mu is equal to NP is replaced by the equation X bar is equal to NP and if we bring N to the other side, students we obtain P is equal to X bar over N and this is the formula that we will apply in order to find P. As you now see on the screen in this particular example the column of X is 0, 1, 2, 3, 4, 5 and the column of frequencies 12, 56 and so on. Hence multiplying the X column with the column of frequencies, we obtain the column of f x the sum of which is equal to 398 dividing that by the sum of the f column we obtain 398 over 200 and that is equal to 1.99. Substituting this value in the equation X bar equal to NP, we obtain 1.99 is equal to 5P or P is equal to 0.398. Students, you have seen that the two parameters N or P, we knew that we were tossing the die five times. So N is equal to 5, N is equal to 5, P we did not know but now from this method which I have conveyed, we have found the value of P which is actually an estimate of the true value of P. The value that we have obtained is 0.398 and that is equal to 0.167. That is very different from 0.167. 0.398 or 0.167 is very different. So it is a clear indication that the die is actually not fair at all. Having found the value of P and already having the value of N students, we will now find all the probabilities for this particular example. As I conveyed earlier, the formula for the binomial distribution is N c x, P raise to x, Q raise to N minus x and so in this example as you now see on the screen, the formula is 5 c x 0.398 raise to x into 0.602 raise to 5 minus x. The value 0.602 is obtained by subtracting 0.398 from 1 because Q is always equal to 1 minus P. Now this formula has to be applied to the various values of x which in this example are 0, 1, 2, 3, 4 and 5 and substituting these values of x in this formula 1 by 1, we obtain the following probability distribution. The probability that x is equal to 0 is equal to 0.07907. The probability that x is equal to 1 is 0.26136 and so on. Once again, none of the probabilities is negative. It is just not possible that a probability is negative and the sum of these 6 probabilities is equal to 1. Students, probabilities we have not removed and now we want to do one more step and that is the computation of the expected frequencies. The expected frequencies say, see the data I presented you earlier, the column of x and the column of f which had a sum of 200, those were what we call the observed frequencies. So, these were the frequencies which were obtained through actual experimentation. As I said, if 200 students experimented this repeatedly, that the loaded die was dosed 5 times, then when they actually performed this experiment, then these were the frequencies. In other words, this is what we observed in reality. Expected frequencies that we expect to have on the basis of the binomial distribution. That means, the probabilities that we have just computed, in the form of the validity of these probabilities, what were the frequencies? What could we expect if this binomial distribution was actually valid? Students, how do we compute them? Very simple. All you have to do is to multiply each one of these probabilities by 200, the total of the frequencies that we earlier had and as you now see on the screen, if we do so, we obtain the following expected frequencies. 15.8 is the expected frequency for x equal to 0 and for x equal to 1, the expected frequency is 52.5. Next, we have 69.1, 45.7, 15.1 and 2.0. It is interesting and important to note that the sum of the expected frequencies is 200 exactly the same as the sum of the observed frequencies. Students, aaye ye jo abhi abhi humne nikaali hai expected frequencies, in ko interpret karnegi koshish karte hain. Jaise maine pehle kaha, ke sabse pehli jo expected frequency hain that is 15.8 aur ye x equal to 0 ke agaist fall kari. Ab iss ka kya matlab hai? Iss ka matlab ye hai, ke ye jo experiment hain, if it is a proper binomial experiment in which the four conditions that I conveyed earlier are fulfilled and whose parameters are n equal to 5 and p equal to 0.398, then if we do this binomial experiment 200 times, we should obtain x equal to 0 15.8 times. Ab aap kahenge ke ye to baat confusing hoge hain. Once again, note ke je ke panic karnegi ko isorurat nahi hai. All you have to do is to think ke 15.8 relative to 200 means 158 relative to 2000, yani aap uska ek 0 barhadi je, to 15.8 ka decimal bhi to ek place ke his aap se shift ho jayega. Ke agar ham ye experiment of tossing the loaded die 5 times, 2000 defa repeat karte to 118 defa we would have obtained x equal to 0, yani no 6 out of the 5 tosses, yani 5 defa toss kia aur 8 p defa 6 nahi aaya. Ye wala jo outcome hain, ye jo result hain, this will be obtained 158 times in 2000 repetitions of this experiment. Bilkul ishi tara se we can interpret all the other expected frequencies. And students, what is the point in doing this, dekhhe main aap se kaha tha, ke iss vaakth ham discuss kar rahi hain fitting of a binomial distribution to real data. Real data jo tha usme jo observed frequencies thi, ab jo expected frequencies aai hain. Ham dekhna chahate hain, ke aaya ye expected frequencies jo hain, are they telling well with the observed ones or is there a discrepancy between them. Agar observed aur expected mein zyada fark nahi hain, then we can say that what we have fitted it is a good fit. Lekin agar ham dekhhe ke hamari fitted distribution or observed distribution mein bahot fark hain, then of course, we think that there is something wrong and may be this particular binomial distribution is not the proper one to be fitted to this particular data set. So, savali peyda hota hai ke phir hamne shuru mein fit hi kyun kya tha. Agar iss ne baad mein sahi nahi sabith hona, to phir to hame shuru se hi iss maamli pe gaur kar na chahiye tha. Dekhye students, shuru ham reasonable assumptions pehi kar rahi hain. Ham iss liye binomial fit kar rahi hain is particular situation mein, ke hamne dekhah ke 200 students jo classroom mein bethi we hain, baari-baari un sabne ek loaded die ko 5-5 defa toss kya. To hamne dekhah ke chaaro konditions puri ho rahe hain. The first one that every trial or in other words, every toss results in either 6 or not a 6, yani, either a success or a failure. Secondly, we noted that every toss was independent of every other toss. After all, jab aap ek defa phayke, uske baar aap usko utha hain aur dobaara phayke, to zahir hai, ke dosri mertu baar jo result aar hai uska pehle result ke saath koi tal lokh nahi hain. Pehla result, dosre result ko effect nahi kar rahe hain. Number three, the probability of success remains constant from trial to trial. Aap point ye hai, ke agar bo fair die hota, to ham kate ke probability of getting a 6 is 1 over 6 aur ham ye kate ke khaa ham iss ko 5-5 defa toss karne ka amal 200 defa bhi repeat kare yani in all ek 1000 defa bhi toss karne to it is still a fair die. Yani, our assumption is that the symmetry of the die is not distorted. Ishi tara ki assumption, ham loaded die mibhi kareinge, ke even if it is loaded, jis tara se bhi wo load kiya gaya hain, uski wohi haale trahti hain throughout the experiment aur us wo ghista nahi hain, jis tara se bhi hain, ushi tara se hain. And what is the fourth and last condition of a binomial experiment? The number of trials is fixed in advance. In this example, the number of trials is 5, it is fixed in advance aur ham ne tamamthar students po jo dosa student bhethe hoi hain, ham ne sab se kaha ke aap ne paanchi defa iss ko toss karne hai. So, you see students, ham ne binomial distribution fit hi is liye ki thi, ke we were reasonably confident that this situation is a binomial experiment situation. Ab ye jo main pehle baat kahi, ke ham compare karna chahti hain expected frequencies ko with the observed frequencies. The observed frequencies are 12, 56, 74, 39, 18 and 1, whereas the expected frequencies are 15.8, 52.5, 69.1 and so on. Now, if we compare each expected frequency with the corresponding observed frequency, you will probably agree that there is not a very big difference between them. 15.8 is not extremely different from 12, nor is 52.5 from 56, issi tara 74 or 69.1 mein bohat sadeed fark nahi hai and so on. Aap me se kuch students, shahid bohat haraan ho rahe hoon, ke main kaha ke 15.8 aur 12 mein zyada fark nahi hai, ya baaki jo comparisons kiye. Baat asal mein ye hai, ke agar ham iss maamle mein intuitively so chahi, to some of us will think that these differences are not very big, some others might think that they are quite big. So, this is not the way to actually deal with this problem. In fact, there is a procedure known as the chi-square test of goodness of fit and that is the proper statistical procedure by which we can determine whether or not we should say that these differences are big. The chi-square test of goodness of fit comes under the realm of statistical inference and that is the third and last segment of this course, which I will inshallah begin in the 31st lecture. I would like to present to you the line charts that you now see on the screen. The line chart of the observed frequency distribution has been superimposed on the line chart of the expected frequency distribution. Aap in line charts ko compare kiye aur decide kiye ke aapke apne khyal mein, kya ye fark zyada hai, ya gam hai. Alright, having conveyed the basic concepts of the binomial distribution, I would now like to apply it to a real life example, an example from industry. As you now see on the screen, suppose that the past record indicates that the proportion of defective articles produced by a certain factory is 7 percent and suppose that a law newly instituted in this particular country states that there should not be more than 5 percent defective. Suppose that the factory owner makes the statement that his machinery has been overhauled so that the number of defectives has decreased. In order to examine this claim, the relevant government department decides to send an inspector to examine a sample of 20 items produced in that factory. What is the probability that the inspector will find two or more defective items in his sample so that a fine will be imposed on the factory? Students, aap ne dekhaha ke kitna interesting problem hai, and it is a very pertinent real life problem. Aap dekhte hai ke iss situation mein, why is it that we can say that we are dealing with a binomial experiment? Subse pali baat ye, ke in a factory thousands and thousands of items are produced. So we can say that the experiment of the production of the item is being repeated a very large number of times. Or iss wajah se, relative frequency definition hai, that is valid. Jo relative frequency definition hai us kitahethe, the probability of success which in this particular case is defective item, how much will that be? According to the past record of this factory 7 percent of the items are defective. And so we can say that p is equal to 0.07. Now what about n? The number of trials, students aap ne dekhaha ke wo jo inspector hai us ne 20 items ko inspect karna hai. Iska matlab hai that the number of trials is 20. So we have identified both the parameters of our binomial distribution. n is equal to 20 and p is equal to 0.07. Now what is the x value whose probability we want to find? Hamara question tha ke what is the probability that this inspector will find two or more defective items in his sample. So that a fine will be imposed on this factory. To us hawale se as you now see on the screen, we are interested in finding the probability that x is greater than or equal to 2. And x of course represents the number of defective items in the sample of 20. According to the law of complementation, the probability of a bar is equal to 1 minus the probability of a and applying this formula in this example, we have probability that x is greater than or equal to 2 is equal to 1 minus the probability that x is less than 2. And this is further equal to 1 minus the probability that x is equal to 0 or x is equal to 1. And this can be written as 1 minus the probability that x is equal to 0 minus the probability that x is equal to 1. Students basic equation hamay mil gaye or ab ham ye jo probabilities hain of x equal to 0 and x equal to 1 ye ham compute karna chahti hain. So what is the formula of the binomial distribution n c x p raise to x q raise to n minus x. And in this particular example repeating myself n is equal to 20 and p is equal to 0.07. So as you now see on the screen, the required probability is equal to 1 minus 20 c 0 0.07 raise to 0 into 0.93 raise to 20 minus 0 minus 20 c 1 into 0.07 raise to 1 into 0.93 raise to 20 minus 1. Solving these expressions the probability that x is greater than or equal to 2 comes out to be 41.3 percent. Students ye jo result hain i a s ko interpret karne ki koshish karthe hain. Haam ne dekhah ke agar us factory owner ne ye jo claim kia tha ke us ki machinery overhaul hoge hi hain or number of defective items decrease hoge hain. Agar aisa actually nahi hain or actually the machinery is in the same situation as before so that the probability of defective item is still 0.07 percent. Then the probability is 41 percent that this inspector will find two or more defective items in his sample. Ab agar 20 ke sample me usko do defective item milte hain to iska matlab hoa 10 percent aur jo nahi law hain uske to hai to 5 percent si zyada nahi honi chahiye the. Iska matlab hain ke chances 41 percent hain that the factory owner will be fined for this kind of a production. Right having discussed the binomial distribution in considerable detail students the next distribution that I would like to discuss with you is the hyper geometric distribution and whenever we conduct what is called a hyper geometric experiment then we have the distribution which is called the hyper geometric distribution. Ab sabali payada hota hai ke hyper geometric experiment se kya murad hain? Iski bhi chaar hi properties hain jinn me se pehli aur chaathi bilkul identical hain with the binomial experiment. Lekin jo second or third properties hain they are very different. In case of the binomial experiment students we said that the trials are independent and the probability of success remains constant from trial to trial. Is case me hum kahenge that the trials are not independent of each other and the probability of success changes from trial to trial. Iskasam ki situation kab payada hoti whenever we draw a sample without replacement from a finite population we are encountering this kind of a situation. Fars ki jaye ke taash ka jo ek deck hai jis me bhaavan patte hain uss me se hum ek patta draw karein and suppose it comes out to be a black card. Peheli martabha jab hum ne draw kiya tha, what was the probability of getting a black card? 26 over 52 isliye ke bhaavan me se 26 patte kale hote. Now that one card has been drawn and if we put it away and we do not replace it in the deck so that we can say that we are sampling without replacement students ab uss deck me ek aavan patte baaki regein aur 26 me se ek kala patta nikal chuka hai. So now we only have 25 black ones. Ab agar hum ek aur patta uss me se draw karein what do we have? The probability of getting a black card this time is equal to 25 over 51 and this is not exactly the same as 26 over 52. Hence you can see that in this kind of a situation the successive trials are not independent of each other and the probability of success changes from trial to trial. Dhusri martabha mujhe kya hasil hua? Depend krta hai is baat pe peheli martabha red patta nikla tha ya black nikla tha. Agar red nikla hota toh dhusri martabha me the probability of getting a black card would have been 26 over 51 aur agar black nikla toh phir a probability is case me ab kya hai? 25 over 51. Toh second draw ki jo outcome hai it depends on what happened in the first draw aur dekhah apne saathi ke the probability of success, the probability of obtaining a black card is changing from trial to trial. Is scenario me the distribution that we have is called the hyper geometric distribution and students as you now see on the screen the formula for the probabilities in this particular situation is the probability that my random variable capital X takes a value small x is equal to k c x into n minus k c n minus x over n c n. Students iss formulae me aapne dekhah ke 3 quantities exist kar rahi hai capital N, small n and k ab jo example me aabhi present kya tha usawaale se capital N is equal to 52 the total number of cards that we have in the deck in the beginning of the experiment and k represents the number of successes in this population if I may call it a population in the beginning of the experiment. Toh 26 patte chu ke kale hute hain therefore we say that k is equal to 26. Last but not the least what is small n? The number of cards that I would like to draw from this deck without replacement. So agar hamne yeh faisla kiya waha hai ke ham 5 patte draw karengi, then we say that small n is equal to 5. So capital N, small n and small k are the 3 parameters of the hyper geometric distribution. Students yeh jo loves meh istimal kar rahi hu this is very very important aapne dekhah ke meh ne aapse kaha tha ke binomial distribution ke 2 parameter hote hain small n and small p aur hyper geometric ke liye meh aapse aabhi kaha ke capital N, small n aur small k yeh 3 parameter hain for the hyper geometric distribution. What do we mean by parameters students? These are those quantities which determine our distribution. Yani agar aap in quantities ki values badal dein you will obtain every time a new hyper geometric distribution. Wo hyper geometric distribution jis meh capital N 10 hai, small n 5 hai aur k yeh jo hai that is equal to 8. Uska line chart uski shakal it will be different from that hyper geometric distribution jis ka capital N 20 hai, small n 11 hai and k is equal to 16. So this is a very important point any distribution that you will take up you should be very clear about the parameters of that distribution. Ab savali peyda hote hain ke iss tara ki situation kab arise hogi students I will discuss with you this problem in detail in the next lecture. In the meantime I would like you to practice with the various concepts concerning the binomial distribution that we have dealt with today. In particular I would like you to fit binomial distribution to real data wherever you think that you are dealing with what can be called a binomial experiment. I wish you the very best in your study of probability distributions and students until next time. Allah Hafiz.