 As Salaamu Alaikum. Welcome to lecture number 24 of the course on statistics and probability. Students, you will recall that in the last lecture, we discussed in detail the concept of mathematical expectation. And we dealt with this concept with reference to discrete probability distributions. Today, I will begin with the Chebyshev's inequality and after that, we will proceed to the concept of continuous probability distributions. As you now see on the screen, the Chebyshev's inequality in the case of a probability distribution is stated as follows. If x is a random variable having mean mu and variance sigma square and if k is any positive constant, then the probability that a value of x falls within k standard deviations of the mean is at least 1 minus 1 over k square. Stated differently, the Chebyshev's inequality is given the probability distribution of the random variable x with mean mu and standard deviation sigma. The probability of observing a value of x that differs from mu by k or more standard deviations cannot exceed 1 over k square. Students, you will recall that this is a very confusing situation. See, its name is so sweet, Chebyshev, that it is very important to understand it. There is no such thing that you are confused. The simple thing is that the statement I gave earlier, that was one way of talking about it. And when I gave the grocery statement D, all I was doing was that I was stating the same thing, but I was doing it through the rule of complementation. You remember that if a is an event, its complement that is a bar, then if a is the probability 1 over k square, then its complement is its probability 1 minus 1 over k square. So, I would like to encourage you, as I said once earlier, please shed off your fear of probability theory and try to understand every concept step by step by a methodological approach. And believe me, if you do it patiently, you will inshallah understand it quite well. Let me now apply this Chebyshev's theorem to the example of the number of petals on the flowers of a particular species that we considered in the last lecture. As you now see on the screen, if a biologist is interested in the number of petals on a particular flower, this number ranges from 3 to 9 and each one of these numbers will have its own probability. The probability distribution is 0.05, 0.10, 0.20 and so on against the x values 3, 4, 5 and so on. Now, the mean of this distribution is 5.925 as you will recall from the last lecture and the standard deviation comes out to be 1.3. According to the Chebyshev's inequality, the probability is at least 1 minus 1 over 2 square that is equal to 0.75 that x will lie between mu minus 2 sigma and mu plus 2 sigma that is between 5.9 minus 2 times 1.3 and 5.9 plus 2 times 1.3. Now, these two values come out to be 3.3 and 8.5. And what we are saying is that this probability is at least 75 percent that if we randomly choose a flower, the number of petals that number will lie between 3.3 and 8.5. In simple words, we are saying that if there are a lot of flowers of this species, then at least 75 percent will be of this type in which the number of flowers will lie somewhere between 3.3 and 8.5. So, you have seen that it is not as difficult as one conceives when you read a mathematical equation. Actually, if you develop the ability to read mathematical equations as if they are verses of Milton and Shakespeare, you will be very well off. You should not be afraid of the equation, but try to read it as if it is a statement written in ordinary English. Let us apply this concept to yet another example. As you now see on the screen, suppose you invest a fixed sum of money in each of five business ventures. Assume that you know that 70 percent of such ventures are successful. The outcomes of the ventures are independent of one another and the probability distribution for the number of successful ventures out of five is given by x values 0, 1, 2, 3, 4, 5 where x of course represents the number of ventures that are successful and the probabilities are 0.002, 0.029, 0.132 and so on. Students, first of all, since we are talking about the number of ventures which are successful, it is obvious that if you have invested in all five ventures, then there are six possibilities, right? None of them might succeed, one might succeed, two, three, four or all five of them might succeed. So, the x values will obviously go from 0 to 5 and the probabilities available because of the past experience, because I am sure that you will recognize that these probabilities, they can only be found through the relative frequency definition of probability. Past experience, 0.002 percent of the time not even one of the five ventures were successful and so on. The sum of these probabilities will have to be one if we are in a position to say that we are dealing with a complete discrete probability distribution. Proceeding to the questions of this problem, we have number one, find mu that is the expected value of x and interpret the result. Number two, find the standard deviation. Number three, graph the distribution of this random variable x and its probabilities and number four, apply either the Shev's rule or the empirical rule to approximate the probability that x falls in the interval mu plus minus 2 sigma and after applying these rules also compare them with the actual probability of lying in this interval and also we are interested in determining should we expect to observe fewer than two successful ventures out of five. Now, students we have so many parts to this question, but of course when we do them step by step we are able to do all of them. As you now see on the screen, the first quantity that is the mean of this distribution comes out to be 3.50 and it is according to exactly the same formula that I have already discussed with you. And as far as the interpretation of this answer is concerned of course this can be said that on the average the number of successful ventures out of five will equal 3.5. Now, this type of discussion has been discussed. 3.5 ventures cannot be successful in any one time investment, but the first thing is that if this experiment is repeated again and again we expect that in any ten such ventures we will have 35 successful ventures out of 50 that we have invested in. The next thing is the standard deviation and as you now see on the screen the standard deviation is computed by the square root of the variance and the variance itself is expected value of x minus mu whole square. Hence, we need to construct a column of x minus mu mu and it is 3.5. So, 0 minus 3.5 gives us minus 3.5, 1 minus 3.5 is minus 2.5 and so on. Squaring this column we obtain x minus mu whole square and multiplying this particular column by the column of probabilities we obtain the column that you see at the end of the table and the sum of this column is 1.05. Now, since the variance has come out to be 1.05 as you just noticed. Therefore, the standard deviation is the positive square root of this quantity and that is equal to 1.02. You have noted that this particular problem may instead of applying the shortcut formula that I conveyed to you last time that is the variance of x is equal to e of x square minus e of x whole square. Today I applied the direct formula and the reason is that I wanted to convey to you that in many situations if your mu is a convenient number then you do not necessarily have to go for the shortcut formula. We have found the mean and the standard deviation. What was the next part of the questions that we had in this problem? As you now see on the screen, the next question was graph this particular distribution and as I have indicated earlier we can draw the line chart of this distribution because x is a discrete variable. The next point is to apply the Shebyshev's inequality and you remember that our question was that we should determine the equality of Shebyshev's inequality that mu minus 2 sigma and mu plus 2 sigma have so many values. Let us look at the screen once again and notice that mu plus minus 2 sigma is equal to 3.50 plus minus 2 times 1.02 and this comes out to be the interval from 1.46 and 5.54. Students, you be careful that the line chart that you just saw this is approximately like a mound or a hump. If we depict this through histogram, as you will see in some books that although they are dealing with a discrete variable, they are sometimes representing its visual picture in the form of a histogram and the reason for that is that we can develop a better idea about its overall shape by drawing a histogram. If you do this, then you will see that this particular distribution may, we do have something like a hump and it is not extremely skewed, it is only moderately skewed. Therefore, we are in addition to applying the Shebyshev's theorem, we are also in a position to apply the empirical rule. As you recall a few lectures ago, I also mentioned the empirical rule. If our distribution approximately symmetric here, then 68.26 percent approximately of the values will lie within the interval mu minus sigma and mu plus sigma, 95.44 percent approximately will lie between mu minus 2 sigma and mu plus 2 sigma and 99.7 percent approximately 100 percent will lie between mu minus 3 sigma and mu plus 3 sigma. I hope you recall all these points that I mentioned to you earlier. So, in this particular example, now that I have determined the two points mu minus 2 sigma and mu plus 2 sigma, I am now in a position to apply both the Shebyshev's rule and the empirical rule. As you now see on the screen, the value 1.46 is toward the left of the distribution and the value 5.54 is to the extreme right and according to the Shebyshev's rule, at least 75 percent of the values lie in this range, but according to the empirical rule, approximately 95 percent lie in this range. Now, let us compare the actual probability of lying within this interval. As you now see on the screen, in this probability distribution, the probability of 2 plus the probability of 3 plus the probability of 4 plus the probability of 5 is equal to 0.969 and we can say that 96.9 percent of the probability distribution lies within two standard deviations. According to this, you have seen that the actual probability of lying within that interval is consistent with both the Shebyshev's rule and the empirical rule. Since this distribution is hump shaped, the empirical rule was doing better than the Shebyshev's because Shebyshev could only tell us that at least 75 percent of the probability distribution lies in this range. The empirical rule told us more accurately that approximately 95 percent lies in this range and when we compute the actual probability for this range, we found that it is 96.9 percent quite close to what we had from the empirical rule. Students, the last part of this problem was would you expect to have fewer than two successful ventures in this situation? Let us try to solve this problem. As you see on the screen now, the exact probability that x is less than or equal to 1 that is x is less, fewer than 2 is 0.002 plus 0.029 and that is 0.031. Consequently, we can say that in a single such experiment, the probability is very small only 3 percent that we will have less than 2 successful ventures out of 5. So, if we have to make a decision in this regard, then we can be quite happy. Students, this is the way we try to interpret mathematical equations and we try to apply those results that we obtain mathematically to real world situations to our advantage. Mean or standard deviation ka aapas me cho li daman ka saath hai. Mean compute ki jai, standard deviation compute ki jai and draw the graph of the distribution if you can and you know you have a certain idea regarding the two limits within which the bulk of your data lies. Alright, we have completed the elementary discussion of discrete random variables and their probability distributions. Students, the next very exciting and important area is continuous probability distributions and the most important distribution that you will study in this regard after a few lectures is the normal distribution. And accept it as a challenge rather than being intimidated and pessimistic. It is so much fun if you are able to try to get involved in the philosophy of any subject and then you are able to learn better. As you now see on the screen, a random variable x is defined to be continuous if it can assume every possible value in an interval a, b where a and b may either be finite quantities or they can be minus infinity and plus infinity respectively. The function f of x is the probability density function and it is abbreviated to pdf. Sometimes we simply say density function of the random variable x. Now, these are mathematical points that I would not have seen in the previous lecture like to go into the detail of in this particular course and if you are interested, you are most welcome to study the mathematics courses in this regard. Students, now what is the difference between the discrete random variable and the continuous? For example, the number of sisters you might have or the number of books in your bag or continuous variable when we are measuring. So, the height of a person can be any number between for example, 5 feet 4 inches and 6 feet. So, it can be 5.7 feet or it can be 5.77192 feet. Now, because of the continuity, the graph of the probability density function of a continuous probability distribution is going to be a continuous curve as you now see on the screen. Students, continuous probability distribution case probability is compute. Now, this is a very important point and I would like you to concentrate really well. In this situation, we have to compute areas under the curve within certain ranges in order to find the probability that we are interested in. For example, suppose that I want to find the probability that the next person who walks into this particular room is going to be somewhere between 5 feet 10 inches and 5 feet 11 inches. If I want to compute this probability, then I will be finding the area between the two values, 5 feet 10 inches and 5 feet 11 inches. This again relates to pure mathematics and it relates to integral calculus. We will not go into details in this course, but we will apply it and make sure that this is the method by which we will become continuous computing all the probabilities. So, the crux of the matter is that in case of a continuous random variable, we will be finding the areas to represent probabilities and since the total probability is always 1, therefore the total area under the curve of any continuous probability distribution is also always going to be 1. In Keebunyadpe, we can say that a continuous probability distribution has the following two very important properties. Number 1, f of x is greater than or equal to 0 for all values of x and f of x is number 2, the integral from minus infinity to infinity of f of x with respect to x is equal to 1. Also, the probability that x takes on a value in the interval c d, where c is less than d is given by the integral from c to d of f of x. And as you see on the screen, the area that we will be computing will be of the form that is shaded in the figure that is in front of you. Pure maths, a definite integral gives you the areas of the form that you just saw in the figure. As you see on the screen, if we find the integral of f of x with respect to x and we find it from K to K, the answer is going to be 0. In other words, if we say that if I am interested in finding the probability that x is equal to K, then I should integrate in this manner that I go start from K and I also end at K, my answer is going to be 0. It is an impossible event. You know that 0 is the only probability when we are dealing with an impossible event. So, what did I say? I said that probability that x is equal to K is equal to 0. Probability that a person's height is exactly 5 feet 11 inches, probability of this is 0. Now you will say that this is not possible that a person's height is 5 feet 11 inches. Students, just recall what I said to you in the very beginning of this course. You remember that I said that if you are measuring that this number is this height for example is 11 inches. Actually, it is somewhere between 10 and a half inches and 11 and a half inches. So, mathematically speaking, we cannot identify, we cannot pinpoint that a person's height is exactly this particular value. But we can always specify and interval that a person's height is or is. And if we apply this logic students, then of course we are able to find the integral because the lower limit will be say K 1 and the upper limit will be K 2 and we find the integral and the answer is not 0. If we narrow that interval too much, then K 1 and K 2 will have a gap in the two end points and against that base you will have a strip under your curve and between that strip you have a certain region whose area is found by finding this integral and this area represents the probability that the height of this person lies in that range. So, this is a very, very important and interesting point that the probability of any particular point in case of a continuous probability distribution is 0. For example, variable X equal to 1, 2, 3, you get probabilities on those points and the regions between them, the probabilities were 0. Now, here it can be said that somewhat it is an opposite kind of a situation that you will not get any particular point probability. You can never say that the height of this person was 11.1 to 9, 1 because you can counter argue that whatever you have measured as 11.91 or whatever I just said, it is actually 11.905 or 11.915. So, it is always an interval. And because of this fact, students can in a continuous situation you can express the probability of any particular interval C to D in four different ways that I will now explain as you now see on the screen. You can write probability that X lies between C and D such that you write C is less than or equal to X less than or equal to D or you write C is less than X less than equal to D or you write C is less than or equal to X less than D or you may write C is less than X less than D. t. That means, if you put equal sign or not, it does not matter. Because if you put the equal sign, then you say that c is less than or equal to x. So, you are saying that x can be equal to c. But if you are not putting it in, then you are saying that c is not included. So, the same thing is that when probability of x equal to c is going to be 0, then that 0 should be included in it or not. The answer is going to be the same. If you are talking about discrete probability distribution, then equal or not, that makes a lot of difference. Students, let me now explain to you the concept of the distribution function of a continuous random variable. You remember that in the case of a discrete situation, we were simply accumulating the probabilities the way we did when we did the cumulative frequency distribution. The first value as it is that plus the next gave us the next one and so on. Because we have to deal with the concept of distribution function with reference to integration. And that integral will always start from minus infinity and will go up to a particular value a. As you now see on the screen, capital F of a is equal to the probability that x is less than or equal to a. Exactly the same definition that we did in the case of the discrete random variable and this probability that x is less than or equal to a is equal to the integral from minus infinity to a of the function f of x with respect to x. If we draw the graph of the distribution function of a continuous random variable, students you will find that the graph starts from the level 0 and it reaches the level 1 and this happens in a continuous fashion and there is no gap and no jump the way you had in the case of the discrete random variable. We were bound to have all those jumps, starting from level 0 and reaching the level 1. Students there is a very interesting and important relationship between small f of x which is the probability density function of our continuous random variable and capital F of x which is the distribution function. And as you now see on the screen the relationship is that the derivative of capital F of x with respect to x equals small f of x. Let me now illustrate all these concepts, a lot many concepts that I have conveyed to you with the help of a simple example. Suppose that we are interested in finding the value of k so that the function f of x defined as follows may be a density function. The function is f of x is equal to k times x for the range x lying between 0 and 2, but the function f of x is equal to 0 elsewhere. Students, what you just read that f of x is equal to something in a certain range, but it is equal to 0 elsewhere is this kind of expression in many books and the reason is that if you just write that f of x is equal to kx for x lying between 0 and 2, then the function will be the same, but mathematically speaking it is better that you write the rest as well. kx equal to 2 ke aage jo sara ilaqa hai real line ka ya x axis ka that is the range from 2 to plus infinity ya x equal to 0 se pehle ka jo sara region hai minus infinity to 0, yani bo jo hamara x axis hai uske bo jo baqti portions hai, waha pe kya ho raha hai. Waha pe to hamara function 0 hai, is liye we write that f of x is equal to 0 elsewhere, elsewhere se matlab ke us range ke ilawa. Now, if we look at the questions that we have in this particular problem, we are required to compute the probability that x is equal to 1, also the probability that x is greater than 1 and in addition we are also interested in finding the distribution function capital F of x of this particular pdf. Last but not the least students we are interested in finding the conditional probability that x is less than half given the information that x lies between 1 by 3 and 2 by 3. In order to solve this problem, we will have to apply a methodological approach and if we commence in a step by step manner, the very first thing to note is that in order to determine the value of the constant k, we will have to apply the two basic properties of probability density function. The first one is f of x is greater than or equal to 0 for every x value and the other one as stated earlier that the total area under the curve has to be 1. In other words, the integral from minus infinity to infinity of this function has to be 1. Now, students ye to pehli kondition henna k f of x must be greater than or equal to 0. This problem mein f of x kis tis ke baraabar tha kx aur x values ki range kya diy hui hain hain hain 0 to 2. To saaf zahir hain k k jo hain that will have to be a positive number if we want this condition to be fulfilled. Agar ke negative ho aur x ki value 0 aur 2 ke dar mein koi value aap leh hain. For example, x equal to 1 to ke ki value ho minus 3 to ke x kya ho jayega minus 3 into 1 and that is minus 3 to apki pehli kondition to fulfill mein hui na that your f of x must be greater than or equal to 0 for that entire range 0 to 2. As you now see on the screen, the second condition is satisfied if the integral is equal to 1 and in this problem if we try to compute this integral students, we can do it in the manner that is in front of you. The integral from minus infinity to infinity can be split into three parts minus infinity to 0 plus the integral from 0 to 2 plus the integral from 2 to infinity and if this entire quantity is equal to 1 which it must if it is a proper density function then we obtain 1 is equal to the integral from minus infinity to 0 of 0 plus the integral from 0 to 2 of k x plus the integral from 2 to infinity of 0 and this comes out to be equal to 1 is equal to 0 plus k times x square over 2 and the limits 0 to 2 applied on that plus 0 and this is equal to after applying the limits 2 k. Hence, k comes out to be equal to half and we have determined the solution to the first part of this question. Now, since k has been determined, we can say that our probability density function is f of x is equal to half x or x by 2 for the range x lying between 0 and 2 and f of x is equal to 0 elsewhere. Students, you remember that the second question is what is the probability that x is equal to 1 or third part is what is the probability that x is greater than 1. You know for the first question out of these two you do not have to compute any integral for the same reason that I have discussed earlier as you now see on the screen. Since f of x is a continuous probability function, therefore the probability that x is equal to 1 is 0. As stated earlier, the probability of x being equal to any particular value is always 0 in the case of a continuous probability function. But in order to compute the probability that x is greater than 1, of course we will find the area under the curve which in this case happens to be a straight line. The area under this line between the values x is equal to 1 and x is equal to 2. Why did I say that we find the area between 1 and 2? It is obvious that you are saying that what is the probability that x is greater than 1. So, it is obvious that we can go from 1 to 2 and from 2 to infinity you remember that f of x is equal to 0. So, even if we try to compute the area from 2 to infinity, that answer is going to be 0. So, in essence all we have to do is to compute the area between 1 and 2. Now in order to compute this area of course we will find the integral from 1 to 2 of f of x. In other words the integral from 1 to 2 of x over 2 and that comes out to be x square over 4 with the limits 1 to 2 and applying those limits the probability is equal to 3 by 4. Students, agar aap iss distribution ke graaf ko ek madhapa phir gaur se dekhe to aap recognize kar lenge ke jo area shaded 1 or 2 ke darmean hai that is exactly 3 by 4 of the entire area under that line between 0 and 2. Yeh is liye ke yeh jo example hum kar rahein that is an extremely simple example or humari jo distribution hai that is giving us a straight line. Is liye elementary geometry ke ru se we can even visually compute the area. The situation is not that simple in case of distributions that we encounter in practice. The normal distribution followed by so many real life phenomena is totally different from a straight line and similarly the exponential distribution and other distributions. So, la muhala hume integration he ka sahara le na padta hai aksar okaat me we will be finding integrals. As you will recall in this problem we also want to find the distribution function of this probability distribution. As you now see on the screen capital F of x which is the probability that capital X takes a value small x is given by the integral from minus infinity to small x of capital F of x and in this particular example we will have to do it step by step the way we will do in any such example. First of all we compute capital F of x for the range x lying between minus infinity and 0 and the integral of 0 from minus infinity to x comes out to be 0. Next for the range x lying between 0 and 2 our capital F of x is given by the integral from minus infinity to 0 of 0 plus the integral from 0 to x of x over 2 which is our pdf for the range 0 to 2 and students computing this quantity capital F of x comes out to be x square over 4. Finally for the range x lying between 2 and plus infinity capital F of x comes out to be the integral from minus infinity to 0 of 0 plus the integral from 0 to 2 of x over 2 plus the integral from 2 to x of 0 and computing all these integrals the answer comes out to be 1. Students, this is the last answer why is it 1? Because we are computing the area capital F of x the area from minus infinity up to a certain point x if we go ahead from 0 to 2 from minus infinity up to a certain x value beyond x equal to 2 this puri range 0 to 2 puri area 1 that is covered. As you now see on the screen the combination of all these results gives us the following picture capital F of x is equal to 0 for all x values that are less than 0 capital F of x is equal to x square over 4 for any x value that lies between 0 and 2 and capital F of x is equal to 1 for any x value that is between 2 and plus infinity. Students, that is perhaps the most exciting one that we also wanted to find a conditional probability. I will take up this particular point in the next lecture and in the meantime I would like to advise you to practice with all the concepts that you have done today. You have realized that in lectures you have the text book and I hope that you are also making the effort to acquire material from other many books. So, I would like to advise you to practice and practice and practice because as you know practice makes a man perfect and a woman too. So, you practice Kijie inshallah you will find that you are comfortable with the concept of continuous probability distributions. My very best wishes to you and until next time, Allah Hafiz.