 As Salaamu Alaykum, welcome to lecture number 31 of the course on statistics and probability. You will recall that in the last 15 lectures, we discussed basic probability theory. We started from the basic definitions of probability and we went on to discrete and continuous probability distributions. Also, we considered the bivariate situation and we considered the coefficient of correlation. Students, in today's lecture, we begin the third and last segment of this course and that is inferential statistics. That branch or that part of statistics which enables us to draw conclusions about real life phenomena on the basis of real data that we have collected on sample basis. Now, as you see on the slide, there are two branches of inferential statistics. Statistical inference can be divided into the two main areas of statistical inference are estimation and hypothesis testing. Estimation itself can be divided into two types, point estimation and interval estimation. I will be discussing with you, first of all, point estimation then we will go on to interval estimation and after that, we will discuss the other way of doing statistical inference and that is hypothesis testing. But students, before I talk about any of these, the most important point to be noted is that there is a concept called sampling distributions and this concept forms the basis for both estimation and hypothesis testing. So, in today's lecture and in the next lecture, we will be concentrating on this fundamental concept that of sampling distributions. As you now see on the screen, the probability distribution of any statistic such as the mean, the standard deviation, the proportion of successes in a sample, this is known as its sampling distribution. Students, now let me explain this point to you in some detail. I said the probability distribution of any statistic is called its sampling distribution and you will remember that in the beginning of this course, I shared these words with you. I said to you that there is a word called statistic, meaning statistics, not just statistics and what was that? Any quantity such as mean, median, standard deviation, any quantity that you compute from a sample that is called a statistic. We are talking about all the inferences that we have to draw, conclusions that we have to draw about the population. They are going to be based on that sample, that one sample that we will draw from the population. So, the sample we have drawn from this population now we have only this sample available and with this sample data we can do whatever we want and we will be computing the mean or the proportion of successes, whatever success may be and median and so many other quantities whatever we are interested in from this one sample. What we are doing is that the probability distribution of any statistic computed from the sample is called its sampling distribution and students, I will explain this concept to you with reference to an example in considerable detail, so that you do not have any confusion Bakinare. As you now see on the slide, let us examine the case of an annual Ministry of Transport test to which all cars irrespective of age have to be submitted. The test looks for faulty brakes, steering, lights and suspension and it is discovered after the first year that approximately the same number of cars have zero, one, two, three or four faults. Students should first try to understand this situation, that is the Ministry of Transport conducts a test and in that country it is low that cars have to be tested and they test these four things but they saw after about a year that all the things they tested, no fault or one fault or two or three or four. These five values of X, where X represents the number of faults in the car, they found that these five values are equi probable. The statement that I read was that they found that approximately the same number of cars have zero, one, two, three or four faults, but I presented it in this way, that with the same relative frequency definition of probability, which we have discussed a lot, that we can say that these five values of X, zero, one, two, three, four, five, they are equi probable and if you agree with this explanation, then students, we can talk about the probability distribution of the population. Sample ki baat to baat mein hoge, pehle we talk about the probability distribution of the population, the population of those many, many cars that are going to undergo this particular test. So, as you now see on the screen, the probability distribution of the population is, the column of X represents the number of faulty items in a car and this number goes from zero to four and the probabilities are 1 by 5, 1 by 5, 1 by 5, 1 by 5, 1 by 5 and 1 by 5. The reason being that as I just said, these five values are equi probable. Now, in order to find the mean and the standard deviation of this particular probability distribution, of course, we resort to the same method that we had before. We multiply the X column with the column of f of X, where f of X of course, represents the probabilities and multiplying the two and adding, we find that sigma X into f of X is equal to 2. Also, to find the variance of this distribution, we construct one more column and that is, X square into f of X and the sum of that particular column comes out to be 30 over 5 and that is 6. Now, the mean of any discrete probability distribution is sigma X into f of X and therefore, the mean is equal to 2, but the variance is given by E of X square minus E of X whole square as we have discussed earlier and hence, the variance of this particular probability distribution comes out to be 6 minus 4 and that is 2. Taking the square root of this quantity, the standard deviation of this particular probability distribution is equal to 1.414. Students, we know that it is a distribution that is discrete because number of faults in a car can be 1 or 2 and we know that the mean number of faults is 2 and we also know that the standard deviation of this distribution is 1.414. What about the shape of this distribution? Students, I hope that you have all answered my questions spontaneously and said that it is a discrete uniform distribution. After all, you do not remember that we had discussed the substantial distribution on a discrete uniform distribution and we noted that whenever the probabilities are equal, we get a uniform distribution. Now, a clear idea has been built in our mind about population. What are we wanting to do, students? We are wanting to talk about all possible samples of a particular size that I can draw from this population. Why? Because, as I said earlier, ye jo inference hona hai, ye jo tamam tar area hai of statistics, wo kya hai? To draw conclusions about this population that we have on the basis of a sample. To zahire ke jab hum ek real life research keringi, us waqt to hum sirf ek sample draw keringi. But sampling distribution ka concept samajne ke liye, we have to consider not just one but all possible samples that you can draw from any particular population. Sampling can be done in two ways. Sampling with replacement and sampling without replacement. Sampling with replacement kya hoti hai? You draw an element out of the population, you note down its main features and you replace it into the population. Or sampling without replacement kya hai? You draw an element out and you throw it away. Yaani mera matlab hai, ke you do not put it back into the population before drawing the next element. To students, sampling distribution ka jo concept hai, we will be discussing this with reference to both sampling with replacement and sampling without replacement. Jai jo main aapse abhi pehle kaha, ke we have to consider now all possible samples that could be drawn from this population of any particular size, ye baat bhi iss reference se ke ya to wo tamamthar samples jo draw ki ye jange, they will be with replacement or ya they will all be without replacement. So, in this example that we are considering, let us suppose that the Ministry of Transport decides that they will just be testing two cars at a time, yaani sample size jo hai that is two. Ye main itna chota sample iss liye liya hai students, ta ke iss concept ko samajne mein aapke liye asani ho. Now students, in this scenario that we are discussing, we are actually talking about sampling with replacement aur iss point ko mein kuch der ke baat zyada tabseel se explain karungi. Lekin phil hal aap ye note ke jaye that whenever you are drawing a sample of size small n from a population of size capital N and if you are doing sampling with replacement, the total number of possible samples that can be drawn in this manner that number is capital N raised to small n as you now see on the screen. In this example, capital N can be regarded as 5 and small n the sample size is 2 and 5 square capital N raised to small n gives us 25. The 25 possible situations are as you now see on the slide, they are 0 0 0 1 0 2 0 3 0 4. Similarly, 1 0 1 1 1 2 1 3 and 1 4. Acha aap yaha pe aap confuse hone ki yakinan koshish kare hain. Bilkul hi koi zeroorat nahi hain confuse hone ki students. Main hi ye jo 25 situations aapke saamne present ki hain. Ye wo hain ki jab do cars ko aap stop kare hain by the roadside at random aur unko test karne for these four things, faulty lights, suspension and the other two that I mentioned earlier. So aap ye 25 mumkina situations hain. 0 0 ka kya matlab hain? Ki wo jo pehli car hain usme bhi there is no fault aur jo dosri car hain usme bhi no fault. And what is 1 3? 1 3 ka matlab hua ki jo pehli car roki gayi that has one fault out of those four things and the second car that you stopped that has three faults. To istra I am sure you will agree now that these are the 25 possible situations. Last situation me ne arz kya ke 4 hain, yani dono ki dono kare jo hain me charon ki charon cheeze karab. Now as I just said a short while ago ke n raise to n, capital N raise to small n ye number of samples hota hain when we are sampling with replacement. Yaha pe ye aap kahange ki what do you mean aap ek kar ko nikalke to bhi rse kar ko baapis draal to nahi reina. Yes you are right there students. Yaha aap iss point ko istra se understand ke je ke agar humare paas ek hat hain aur humne paach percheo pe ye paach number alag alag lik diye 0 1 2 3 and 4 aur un percheo ko fold kar ke us hat me daal diya. And then we shuffle or shake the hat and we draw first we draw one of those slips and suppose it says 0 humne usko baapis daal diya. And then I again shake it and then I draw another one and once again it is 0 students chunke ye with replacement hain hain ke mai ne baapis daal diya tha issi liye ye possible hua ke dosri mertaba bhi I could get a 0 and therefore I get the pair 0 0. Dekha aapne this is the way to understand the point ke agar aap without replacement karte tabto aap 0 0 aap ko nahi mil sakta tha. Agar pehli mertaba me 0 nikal aaya and you have thrown it out of that hat to aap to us hat ke andar sirf chaar par chi hain baak hi 1 2 3 4. To aapto next time you can either get a 1 or a 2 or a 3 or a 4. Now what is the total number of possible samples that can be drawn without replacement from a population? Students the answer is capital n c small n aap ko yaad nahi when we were discussing the rules of permutation and combination us svakth combinations rule ke ki discussion ke hawale se me ne aap se yehi kaha tha na that if you want to draw r things out of n things without replacement. Waha agar yeh lafs nahi bhi kaha students to it was implied. Kyuke aam zindagi me jab aap nikal te hain aap without replacement hi nikal te hain aap das students baith hain klaas me un mese 2 students ko lekar jaana hain haam ne seminar pe to haam yon to nahi karenge na ke ek nikal aap vapis daala aur dobara bhi wo hi nikal liya. So in real life of course most of the time we are dealing with situations when we are sampling without replacement. To us svakth jab combinations rule ki baat kahi to yehi aap se kaha ke the total number of ways of drawing r things out of n things is n c r. Exactly yehi formula aap yaha pe apply ke jhe the population size is capital n and the total number of ways of drawing samples without replacement of size small n from this population of size capital n that number is n c n. And as you now see on the slide in this example n c n is equal to 5 c 2 and 5 c 2 is equal to 5 factorial over 2 factorial into 3 factorial and that is equal to 10. So if the population values are 0 1 2 3 4 then the 10 possible samples of size 2 drawn without replacement from this population are 0 1 0 2 0 3 0 4 1 2 1 3 1 4 2 3 2 4 and 3 4. Students aap ne gaur kiya aap ne gaur kiya ke yeh jo 10 samples hain, in me first element joh hai that is not repeated. Kome se pehle joh element hain, kome ke baad usse mukhtali fe element hain and this is what you will obtain when you whenever you are sampling without replacement. Lekin is svakth hain hain, wapis ussi example pe chalthe hain joh ministry of transport wala hain jis me they are just stopping 2 cars at random by the road side and they are testing them. Yeh aap students sampling with replacement wala case valid hain, kyuke wo karein joh hain they are independent aur mukhen hain ke pehle karein me bhi 0 hi false ho aur 2sri jo roki jaye ussme bhi 0 hi false ho and so you get 0 0 as one possible situation. So as I said a few minutes earlier in this kind of a scenario the total number of possible samples is n raise to n 5 raise to 2 and that is 25 and once again if you see on the screen the 25 samples are 0 0, 0 1 and so on going up to 4 3 and 4 4. Yeh joh ministry of transport hain iska aim kya hain. Yeh isvak tham yeh assume kar rahe hain ke hain, yeh jahana chahat hain. Yeh ministry yeh jahana chahat hain. Ke what is the average number of false per car in this particular city? Yehani jist of the whole story ke hum muh main interested hain muh of course represents the mean of the population. Aap yeh joh samples of size 2 hum draw kar hain, inse to hain hum muh nahin mil sakta na. Inne se to kisi bhi sample ka agar aap mean compute karenge to that will be x bar to baathi yeh hi hai ke x bar joh hain that will be an estimate of muh aur yeh joh saari discussion mein ish waqt kar rahe hain aur aage agle lectures me bhi hogi that is in this area. X bar joh hum e sample se milega joh muh ki estimation karega ish silsle mein kuch intahai ehem mathematical points jinn ke under sampling distribution ka bahut zyada important concept bhi aata hain. Yeh sab kuch hum discuss ish waqt kar rahe. Alright, we have 25 samples and we are interested in the mean. So, students let us compute the mean of each one of these 25 possible samples. And as you now see on the screen, the 25 sample means are 0.0, 0.5, 1.0, 1.5, 2.5, 1.5, 2.5, 1.0 and so on. And the mean of the last possible sample is 4 plus 4 divided by 2 and that is 4.0. Aap ne note kiya, ke in means me se kuch means zyada mertaba akhar kar rahe hain aur kuch gam mertaba. So, naturally zehne mein ye baat aati yeh ke agar hum in mean values ki frequency distribution form kar lein. So, we will have a better idea of the situation. So, we would like to do exactly that and as you now see on the slide, the frequency distribution of x bar in this particular example is the column of x bar starts from 0.0 and goes up to 4.0 and the number of samples having these values of x bar starts from 0.0 to 4.0 or 1, 2, 3, 4, 5, 4, 3, 2 and 1. So, that the sum of the column of frequencies is 25 and this is exactly the number of samples that we had. Dividing these frequencies by the total frequency 25 students, we obtain 1 by 25, 2 by 25, 3 by 25 and so on and these numbers represent the probabilities of the various values of x bar. The probability that x bar will be equal to 0 is 1 over 25, the probability that x bar will be 0.5 is 2 by 25 and so on. So, students, they represent the probabilities. If you have some confusion, then you go to the same old thing that if all items are equally likely, then the classical definition is valid and here in reality, the probabilities are occurring due to the classical definition. These samples, all possible samples, they are equally likely to be drawn. In a real life situation or equal chance. And we can apply the formula favorable over the total and so, when we see that there is only one sample, which means 0.0. This means that only one of those 25 is favorable to what we want here and so, the probability that x bar will be equal to 0.0 comes out to be 1 by 25. And in a similar way, you can interpret all those probabilities. Students, this particular probability distribution that we have just constructed, this is called the sampling distribution of x bar, meaning a very simple concept. A column of x bar along with a column of probabilities such that the sum of the probabilities column is 1, that is the probability distribution of any statistic is called its sampling distribution. If we are talking about x bar, x tilde, that is the median, and these 25 samples are the median's compute. And due to the same procedure, if we compute the probabilities of the various values of median, then students that would have been called the sampling distribution of x tilde. So, I hope that you have no confusion whatsoever. The sampling distribution is a fundamental concept for statistical inference. Estimation or hypothesis testing, dono me iska kalidhi kirdar hai. Or, ye chees hum discuss karenge as we go on. At this point in time, I would like you to think of the shape of this particular distribution and also its center and its spread. After all, aapko patato hai, har waqt hamein teen cheezho mein interest hota hai. The center or the location of our distribution on the x axis, the spread which is a measure of variability or variability wo cheez hai, jo Allah toala ne iss kainat mein, har taraf rakhi hui hai. After all, what is statistics? Students, statistics is the science of dealing with variability. Shape mein hum bara interested rahehte hain. Is it symmetric? Is it skewed? Let us look at the shape of this particular distribution. As you now see on the screen, if we plot x bar along the x axis and the probabilities along the y axis, we obtain the vertical line segments which yield an absolutely symmetric triangular distribution. Now, aapne dekha ke bilkul hi symmetric hai aur triangular hai. What will be the mean of this distribution? I think you will agree that I do not have to apply any formula. I just simply locate the center of this distribution that we have on the slide and we find that the mean is equal to 2. But then what about the spread? Uske liye aap thora sa problem hai. Spread ke liye to shahid hame actual formula compute kar nahi pare. So, what is the formula by which we can compute the spread, the standard deviation or even the mean of a distribution of this type? As you now see on the slide, students, we have already learnt that in the case of the probability distribution of x, mu is given by sigma x into f of x and the variance is given by sigma square equal to e of x square minus e of x whole square which is further equal to sigma x square into f of x minus sigma x into f of x whole square. But now, since we are dealing with the probability distribution of x bar, therefore we replace x by x bar in these formulas. And hence we obtain mu x bar that is the mean of the sampling distribution of x bar is equal to expected value of x bar which is equal to sigma x bar into f of x bar. Similarly, sigma square x bar that is the variance of the sampling distribution of x bar is equal to expected value of x bar square minus expected value of x bar whole square and this is further equal to sigma x bar square into f of x bar minus sigma x bar into f of x bar whole square. Students, you have seen that this is very easy. All you have to do is to replace x by x bar and therefore you just replace x by x bar. Now, let us try to find the mean and the variance of this particular sampling distribution. As you now see on the screen, when I multiply the x bar column by the column of f of x bar which represents the probabilities of the various values of x bar. I obtain x bar into f of x bar is equal to 0, 1 by 25, 3 by 25, 6 by 25 and so on. And upon adding this particular column, the expected value of x bar comes out to be 50 over 25 that is equal to 2. Similarly, in order to find the variance, I first need to find expected value of x bar square and that is equal to the sum of the column x bar square into f of x bar. Therefore, I multiply the first column that of x bar by the third column that is x bar into f of x bar to obtain the fourth column which is x bar square into f of x bar. And the values in this particular column are 0, 1 by 50, 6 by 50, 18 by 50 and so on. Adding these values, the expected value of x bar square comes out to be 250 over 50 and that is equal to 5. Now, substituting these values in the formula of the variance, I obtain the variance of x bar is equal to 5 minus 2 square and that is equal to 1. When I take the square root of this particular quantity, it is obvious that the standard deviation of the sampling distribution of x bar is also equal to 1. Students, now I am going to tell you two or three very important points. The first thing is that the standard deviation of the sampling distribution of x bar is called the standard error of x bar or this term is a standard error. This is widely known worldwide and it is a very, very important term. Standard deviation of our distribution is called the standard error. Now, I am sure that you are wondering where the word error comes from. So, let's do this by pending this particular question and I will pick it up later. Now, the other very important point that I would like to convey to you is that there are certain relationships between the mean of the sampling distribution and the mean of the population and between the standard deviation of the sampling distribution and the standard deviation of the population. As you now see on the slide, the first property is that mu x bar is equal to mu. This may on the left hand side we have mu x bar which represents the mean of the sampling distribution of x bar and as you noticed a short while ago in this example, mu x bar is equal to 2. Now, in this equation on the right hand side we have mu and we are talking about the mean of the population and that also was equal to 2. Hence, this equation is verified that mu x bar is equal to mu. Students, this formula I have given you, this is valid in the case of sampling with replacement and also in the case of sampling without replacement. If you do sampling with replacement, then 25 samples were made, 25 means you have computed, 25 values of x bar and we have seen that mu x bar, the mean of the means that is equal to the population mean. If we do sampling without replacement, then 5C2 means only 10 samples or only 10 other samples mean, but you would have found that the mean of those 10 means would also have been equal to 2. Now, the other extremely important relationship is between the standard deviation or the standard error of the sampling distribution of x bar and the standard deviation of the population. And as you now see on the slide, in case of sampling with replacement, sigma x bar is equal to sigma over square root of n. On the other hand, in case of sampling without replacement from a finite population of size capital N, sigma x bar is equal to sigma over square root of n and this whole term is multiplied by the square root of capital N minus small n over capital N minus 1. The factor square root of capital N minus n over n minus 1 can be referred to as the finite population correction factor and it is abbreviated to be called FPC. All right. Now, it is a complicated situation, right? It is not necessary to get so confused. As I have told you many times, just relax and approach the problem in a methodical manner. What was our equation with replacement in the case of sigma x bar is equal to sigma over square root of n and I hope you do remember it was the square root of 2 and it was 1.414. Now, sample size is equal to sigma over square root of n minus 1.414 over square root of 2, i.e. 1.414 over 1.414 and that is equal to 1. But what is the standard deviation of the sampling distribution students? We have seen that sigma x bar is equal to 1. So, this equation is verified. 1 is equal to 1 and therefore, we have verified for this example, this very important relation. Now, what we are doing at the moment is without replacement. This is equivalent to the situation when we are sampling with replacement. But if you do a sample without replacement, then you will find that sigma x bar will be equal to sigma over square root of n multiplied by the square root of capital N minus small n over capital N minus 1, i.e. the FPC. Students, ye jo term hai capital N minus small n over capital N minus 1, jaisake main do tin bar kaha, this is required whenever we are sampling without replacement from a finite population. Magar ek point bohat important yaha pe bhi aap note ki jai. If the sample size is much, much smaller than the population size, for example, if capital N is 1000 and sample size small n is only 2. Iis situation mein ye jo term hai ye kis tis ke berabar hogi. Capital N minus small n over capital N minus 1 is equal to 1000 minus 2 over 1000 minus 1, i.e. 998 divided by 999 and students, you will all agree that this is equal to 0.99 something and that is approximately equal to 1. So, this means that whenever the population size is much larger than the sample size, this term is almost equal to 1 and we do not need to apply it. If the sample size is less than 5 percent of the population size, then you do not need to apply this term, but if the sample size is greater than 5 percent of the population size, then you should attach this term to the other part sigma over square root of N. Alright, what are we wanting to discuss in this lecture? We are talking about the sampling distribution of X bar. Abhi hum ne sample size 2 ke liye dekhah, ke hamari sampling distribution triangular hai uska mean bilkul middle mein hai obviously aur uska standard error jo hai uski bhi ek relationship hai with the population standard deviation. Mean ki to nahayat interesting relationship hai, jaisa mein hai pehle kaha, ke sample means ka mean, population mean ke barabar hota hai. Students, jo property mein aap ke saath abh discuss karne wali hoon that is of crucial importance in statistical theory. Aur wo ye hai ke what happens to the shape of our sampling distribution if the sample size is increased. Abhi cho example tha usme toh it was an extremely small sample size, it was only 2. aaye dekhte hai ke hamari sampling distribution ki shape, kistra se transform hoti hai as we increase the sample size. Suppose that the ministry of transport decides to test at any particular occasion not just 2 cars, but 3 cars. Dusre lafzome abh hum ara sample size 3 hai. And according to the formula that I presented earlier, the total number of samples of size 3 that can be drawn with replacement from a population of size 5 is equal to capital N raised to small n that is 5 raised to 3 and that is 125. Students, agar in 125 samples ke liye har sample ka aap mean compute karein or frequency distribution banayi or probabilities nikalne, exactly the same way as we did earlier then you obtain the sampling distribution of x bar that we now have on the screen. This time the values of x bar are 0.00, 0.33, 0.67, 1.00 and so on. And the frequencies are 1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3 and 1 dividing each of these numbers by 125 we obtain the probabilities. And when we plot these on a graph paper, we find that even now our distribution is absolutely symmetrical, but it is very important to note that unlike the earlier one which was triangular, this particular distribution is more like a normal distribution. Students, I would like to encourage you to work on this problem on your own. Yani aap size 5 ke jo population hain a 0, 1, 2, 3, 4, usme se aap all possible samples of size 3 khut nikalye with replacement. To kya ainge samples? The first possible sample will be 0, 0, 0. Yani un kareon ke hawale se hum you samjhe ke bo tino karein jo stop ke gayin by the roadside un me se kisi bhi kaar me koi ek fault bhi nahi paya gaya. And the second possible sample will be 0, 0, 1. Yani pehli do kaarein toh sahi nikli, but in the third one there was one fault. And students you have to go very systematically. Kyuke aap systematically nahi chelenge toh there is a probability that you might get a bit confused. So, I would like to help you a little 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 3, 0, 0, 4. Aap iss ke baat kya karein? Should we talk about 0, 1, 0, 0, 1, 1, 0, 1, 2, 0, 1, 3 and 0, 1, 4? Yes, I think we are going in the right direction. Aap iss ko zaroor ek martaba khud ki jaye. It will take quite a bit of time, but I assure you if you go through this exercise once you will acquire a sense of confidence and you will feel at home with this idea. How do we obtain the various X bar values that you just saw on the screen? The first one was 0, 0, 0. Oto khaar bohat obvious hai. Ojo pehla sample hain a 0, 0, 0, 0, zahire ki agar aap in 3 values ko add karein or 3 se divide karein you will get a 0. But what about the second one? We have 0, 0, 1 or agar in values ko mean lena chahin to kya aega 0 plus 0 plus 1 over 3 and how much is that? 1 by 3 and that is 0.33 and that is exactly the value we had on the screen as the second possible value of X bar. Now, that we have looked at the distribution of X bar for the case sample size equal to 3. Let us also consider the sampling distribution for that case when sample size is equal to 4. Any 8 number or add kalee j is case may how many possible samples? N raise to N, 5 raise to 4 and how much is that? Students that is 625. So, if you want to draw samples of size 4 with replacement from a population of size 5, you have as many as 625 possible samples. Aap me aap se yeh nahi kumgi kya aap isko khud karein because I know that now you will say that it is too much 625 samples. The first of which is 0, 0, 0, 0 and the last one is also very simple 4, 4, 4 and 4. Lekin jo 623 mein mein that is a long story. But if you do not want to do it yourself then please accept what I am conveying to you as you now see on the slide. In this situation we have a sampling distribution of X bar as follows. The values of X bar are 0.00, 0.25, 0.50, 0.75, 1.00 and so on and the probabilities are such that when we plot this particular distribution it is again absolutely symmetrical. As you now see on the slide, students this particular one looks like a normal distribution even more than the one which we had in the case of sample size equal to 3. And this is exactly the point that leads us to one of the most important theorems in statistical theory. As you now see on the screen the theorem known as the central limit theorem states that if a variable X from a population has mean mu and finite variance sigma square then the sampling distribution of X bar approaches a normal distribution with mean mu and standard deviation sigma over square root of n as the sample size n approaches infinity. Now you have noticed that according to the central limit theorem, if sample size infinity tends then our sampling distribution approaches a normal distribution. And now the example I have presented to you, you have seen that infinity is a normal distribution of X bar and X bar. Now you have seen that infinity is very far away. Now we have only considered size 2, size 3 and size 4 and we saw that the distribution looked much more like a normal than what we had in the case of size 2. Now as you see on the slide as n tends to infinity the sampling distribution of X bar approaches normality with mean equal to mu and standard deviation equal to sigma over square root of n. This is a theorem of fundamental importance. Most of the time we can say that our population that will be something like a normal because for example, if you are talking about heights or temperature or blood pressure, as I discussed earlier, most of these phenomena are normally distributed. So in that case there is no need for this theorem. But it is obvious that there will be many situations where we cannot say that the population is normal and in that case how do we know what our sampling distribution will look like? This theorem comes to our rescue. We note that irrespective of the shape of the population as long as the sample size small n is adequately large, the sampling distribution of X bar is going to be approximately normal with mean mu and standard deviation sigma over square root of n. And what happens in the case when the population itself is normal, then we are better off. We have the situation that if the population sampled is normally distributed, then the sampling distribution of X bar will also be normally distributed regardless of the size of the sample. In other words, X bar will be normal with mean mu and standard deviation sigma over square root of n even if sample size n is not very large. Students in today's lecture, I discussed with you the fundamentally important concept of sampling distributions. We will be discussing this topic further in the next lecture and in the meantime I would like to encourage you to study various aspects of this concept in your textbook and in as many other works as you conveniently can. Until next time rest of luck and Allah Hafiz.