 Assalamu alaikum. Welcome to lecture number 4 of the course on statistics and probability. Students, you will recall that in the last lecture, you did the frequency distribution of a discrete variable, the line chart, the relative frequency distribution and the percentage frequency distribution. Also, at the end of the lecture, we talked about the cumulative frequency distribution of a discrete variable. In today's lecture, I will discuss with you the frequency distribution of a continuous variable and we will be discussing the graphical representation of the continuous frequency distribution by way of three different graphs, the histogram, the frequency polygon and the frequency curve. Students, you will recall that in lecture number 1, I explained to you that the continuous variable takes values over a continuous interval. For example, a normal Pakistani adult males could be anywhere between 5.25 feet and 6.5 feet. Hence, in such a situation, the method of constructing the frequency distribution is somewhat different from the method that we did last time. Let me explain this point with the help of an example. Suppose that the environmental protection agency of a developed country performs extensive tests on all new car models in order to determine their mileage rating and suppose that we obtain the following 30 measurements by conducting such tests on one particular car model. So, ye hamare figures aagay, ye environmental protection agency ko ham EPA se denote kar rahe hain aur 36.3, 30.1, 40.5 ye EPA ki mileage ratings hain un 30 cars ke liye jin ko ono ne examine kia hain. So, ab dekthe hain ke iss khism ke data ki situation mein, jo ke ye a continuous variable ka data hain, hum kis tara se proceed karenge? The first step is to determine the largest and the smallest value in the data set. In our example, the smallest value is 30.1 and we will denote it by x naught. The largest value is 44.9. It is the maximum value and so we denote it by x m. The next step is to find the range of our data set. Ab range se kia muradhe? It is extremely simple. Range is defined as the difference between the maximum value and the smallest value. So, as you can see on the screen, the range in this example is x m minus x naught that is 44.9 minus 30.1. In other words, the range is equal to 14.8. As you can now see on the screen, if we plot the x values along the x axis, which is also known as the real line, then our value our smallest value 30.1 is to the left side and the largest value 44.9 is to the right and the distance between the two is 14.8. Lehaaza range ka concept, geometrically speaking, ek distance ka concept hain. Ke hamara poora jo data hain, wo yani kaha se kaha tak range karta hain. Uski jo maximum width hain, geometrically us data ki jo width hain, uska jo spread hain, wo kaha se kaha tak hain. The next step is to decide on the number of classes that you want to have in your frequency distribution. Ab classes se kia muradhe? Classes se matlab hain ki wo jo complete range thi usko ham divide karna chaate hain, chote-chote intervals mein. These small intervals, the sub intervals of that big interval, they are called classes. Students, they are no hard and fast rules for the determination of classes. Ke exactly how many classes you want to have. It is somewhat subjective. It also somewhat depends on the size of your dataset. Agar aapki goh zyada values, thousands mein values agar nahi hain. If it is a small dataset having 50, 130, 70 observations, you would generally have 5, 6, 8, 10, 12 classes. So, in this example, let us suppose that we decide to form 5 classes. Yani wo jo hamari big range jyut hai usko ham 5 intervals mein goya divide karna chaate hain. So, what we will do is to divide the range by the number of classes that we want in order to get the approximate value of the class interval. Class interval se muradhe hain ki wo jo classes, wo jo chote-chote intervals amni form karne hain, unki width kya hoge? The class interval is denoted by h, small h and as I just said, when we divide the range by the number of classes that we want, we will get the class interval. So, now as you can see on the screen, we have h is equal to 14.8 divided by 5 and that is 2.96. But obviously, 2.96 is not a convenient number to work with and hence we will be rounding it upward in order to obtain h equal to 3. Agla step ye hai ke ham ye decide karein ke what is going to be the lower class limit of our first class. Yani ye to ham ne decide kariya ke classes ki jo lambai hai that is going to be 3 units. Lekin where do we start from? Yani ham kaha se start karein ye kaam karna? The answer is that we should start from a value which is either equal to or slightly smaller than the smallest value in our data set. So, as you can now see on the screen, is example me hamari smallest value thi 30.1. So, students we can in this case start our first class from the number 30.0. Yani 30.1 se thoda peechhe ek convenient round number se ham start karein and that is 30.0. Ab jab ham ne ye number decide kariya, yani ye dono batin decide kariya ke the lower class limit of our first class is going to be 30.0 and the class interval of each class is going to be 3 then it becomes very simple. Now, all we have to do is to successively add the number 3 to the number 30.0. Yani, now as you can see on the screen the lower limit of the first class is 30.0 then 30.0 plus 3 gives us the lower limit of the second class as 33.0. The lower limit of the third one is 33.0 plus 3 and that is 36. Similarly, we have 39 and 42. Lower limits though ab determine ho gayin. Now we want to determine the upper limits. Ab yaha pe ek isme point hai which is very important. Lekin sometimes students do not pay so much attention to it. Isliye mai aapko hi toh aaj jo khas thor pe ab isse agle point pe dilwana chahti hoon. Now as you can see on the screen the upper limit of the first class has been written as 32.9. Apparently, it seems that the interval of the first class is 2.9. 32.9 minus 30 is 2.9. So, it appears that the class interval of 3 has been changed. Ke bo toh abhi toh hum ne kaha tha ke 3 hai aur ab hum ne usko 2.9 kar diye hai. Now this is exactly the point that I wanted to draw your attention to. Pehle aap stable ko puri tara se study kalli je. Jab ye pehle value humne 32.9 rakhli toh uske baaki jo values hai upper class limit ki bo bilkul usi tara jistra se lower class limit ke liye kiya tha. Ke aap successively 3 add karte jaye and you get all the upper class limits. Hence, you get the classes that you now see on the screen, 30.0 to 32.9, 33.0 to 35.9 and so on. So, aap iss point ko dekhte hain ki ye humne 32.33, 33 to 36, 36 to 39. Is tara se kyu ne likha tha? Iska jawab ye hai ke abhi to humne classes form ki hain. Masla hume tap pesh aayega jab hum apne data ki values ko in classes ki under tally karne ki koshish karenge. Agar me ne likha hota, 30 to 33, 33 to 36 aur mere data set ke under ek value hoti 33. Toh mujhe samaj me na aata ki me usko pehle class me enter karun ya second class me enter karun. You see, 30 to 33, 33 to 36. Iska matlab hai ke wo jo ek value hai 33, wo pehle class me bhi shamil ki jaya sakti hai aur second class me bhi shamil ki jaya sakti hai. And in this way we get the problem of duplication and confusion. This was exactly the reason why we take the upper class limits in a different way as you just saw. When we write 30 to 32.9, 33 to 35.9, there is no problem. Agar mere data, where day take enter, if the value is 33, I will put it in the second class and if the value is 32.9, I will put it in the first class. So agar yeh point clear ho gaya, toh uske baad, now let us go to the next step. Of course the next step is to distribute all our data values into these classes that we have formed. So as you can now see on the screen, we construct another column and that is the column of tally marks and then another column and that is the column for frequency. Me aapko frequency ka matlab toh thodi der ke baad baataungi. Pehle tally marks ke upar hum baat, tally marks ke baare main baat karthe hain. You know the tally marks are used in order to have convenience in sorting the data and putting the data in the proper, in its proper place. So in this example, if I go back to the original data that we collected, as you now see on the screen, our data values were, the values in our data set were 36.3, 30.1 and so on. Isle haas se sabse pehle value 36.3 jo hain isko hum tally karenge third class me, kyunke third class jo hain that is going from 36.0 to 38.9 to saab zahir hain ke 36.3 jo hain goi isi class me fall karega. Next jo value hain 30.1, obviously that will fall in the first class, which starts from 30.0 and goes up to 32.9. To iss tarike se hum apne data set ki har value ko apne frequency table ke andar tally karte chale jayenge. And we will get the resulting table, as you now see on the screen. Istra se humne saara data jo hain usko apne table ke andar tally kar liya. Lekin aap ke zahin me ek confusion hain. Ke klas interval to 3 tha aur klas interval humne 2.9 kar diya. Me chahungi ke aap apne stable ko ek tofa pheer se gaur se deke. If you look at the upper class limit, limit of the second class is 35.9 and the upper limit of the first class is 32.9 and if you subtract 32.9 from 35.9, students you get exactly 3. Similarly, 38.9 minus 35.9 is exactly 3 and 41.9 minus 38.9 is also 3. Iss se ye aapko mein indicate karna jaar hain hoon ke klas interval jo hain ho abhi bhi 3 hain. Lekin mujhe maalum hain ke abhi bhi aapko thori si confusion hain. To abhi chan minute ke baad jaha hum klas boundaries pe hain ge, to inshallah taala aap ki ye confusion ussakth totally remove ho jayi ki. Now, let me explain to you the concept of frequency. Ye jo third column humne banaya iska title humne diya frequency. Ab aap soch rahe honge ke wo jo metric mein hum example wo jo experiments keya karthe the physics mein tuning fork hilaate the aur uski frequency determine karthe the shayi diistra ki koi cheez hai. Ye asa to bilkul nahin ye ji. It simply means how frequently something happens. So, as you see in the on the screen now, we had two tally marks in the first class and hence the frequency of the first class is 2. Yani iska mafum very simply sirf ye hain ke 2 values aisi hain jo ke pehli class ko bilong kar rahe hain. 4 values aisi hain jo ke second class ko bilong kar rahe hain and so on and so forth. Students, let us now consider the concept of class boundaries. Abhi jo pehli jitni baate hui un me maine class limits ka zikar kiya tha basically ke lower limit of the first class is 30 and the upper limit is 32.9 and so on. To class boundaries kya cheez hai. Dekhi aapko yaad hoga ke sbse pehli lecture mein main aapko errors of measurement ka concept diya tha. You will recall that I said that if a value is recorded as 30.0, actually it lies somewhere between 29.95 and 30.05. Similarly, if the value is recorded as measured as 32.9, actually it lies somewhere between 32.85 and 32.95. If you recall all that discussion, you will appreciate that we can utilize that concept in this situation and we can construct a column of class boundaries which will be actually the true limits of those classes. So, as you now see on the screen, we construct another column and that is the column of class boundaries, 30.0 ki jaga pe ham likhenge 29.95 or 32.9 ki jaga pe ham likhenge 32.95. Students, ye ham ne kya step liya hai? You know what we have done? We have stretched the class to the left and to the right, yani 30.0 to 32.9 jo hai usko ham ne stretch kar diya hai dono taraf in order to get the true boundaries, the true limits of that class. 32.9 is replaced by a value which is slightly larger and that is 32.95 and 30.0 has been stretched to a value slightly smaller and that is 29.95. Now, this process will of course continue for all the classes and thus in this way, we will get all the class boundaries. But I think some of you might be a bit confused ki ye stretching to chale hui, but how do we do it? I mean what is the formula? Well, the formula is very simple. All you have to do as you now see on the screen is to add the upper limit of the first class with the lower limit of the second class and divide by 2, yani 32.9 plus 33.0 divided by 2 gives you 32.95. And we write this new number that we have just obtained as the upper boundary of the first class as well as the lower boundary of the second class. Bilkul ishi tara se ham throughout proceed karenge jo next 2 values hai jinn ka average hume lena hai, they are 35.9 and 36.0. Ye aap diagonally ishtra ek diagonal tari ke se chalte jayenge aur dono values jinn ko aap ne add karke 2 se divide kiya hoga yani unka average liya hoga wo aap un dono values ko us average values se replace kar denge. Aur aap us pehli baat pe bhi baapis aajaye, wo aap ko confusion thi na ke wo class hai uska interval to 2.95 hochuka hai. Aap zara screen pe dekhye aur note ki jiye ke jo aap ki pehli class hai. The interval from the lower boundary to the upper boundary that is the difference between 32.95 and 29.95 it is exactly equal to 3. Exactly that which we wanted. Similarly, for the second class the interval from 32.95 to 35.95 is exactly 3 and I hope that this sorts out that confusion. Wo jo apparently lag raha tha ke humne class ki, class ke interval ko tabdeel kar diya hai. In reality we had not, we had not done so. The true class limits of these classes are such that the interval is exactly what we wanted. Students iss tamam interesting discussion ke andar ek aur bhi ek key point hai, jisko aap ko convey karna chaati hoon. In fact iss sari discussion ka ek tisam ki ek revision hai, aap baat ki jo aap bhi abhi hum kar rahe hain. Aap jo bhi data set ho, hamesha issko iss tra se aap proceed karein. Ke pehli initially aap jo class limits lein unka number of decimal places utna hi ho na chahiye jitna aapke data ka hai. Hamare iss data mein, as you remember all the values were up to one decimal place and hence initially all my classes were constructed in such a way that the values were of one decimal. As you remember 30.0 to 32.9, 33.0 to 35.9 abhi aap number jo hain, it is obvious ke ye ek hi decimal ke hain. Uske baad jab aap class, limits ko class boundaries mein convert kareinge to automatically aap ke jo numbers hain, wo ab 2 decimal pe ho jayenge, yani ek decimal baad jayega. So as you saw the class boundaries were 29.95 to 32.95, 32.95 to 35.95. Ab ye 2 decimal ho gaye, hala ke hamara jo original data hai wo sref ek decimal katha. This creates a lot of convenience. Ab koi imkaan nahi hai, koi possibility nahi hai ke haam apne data ki kisi bhi value ko kisi ghalat class ke andar enter karte. Yani, saway uske ke hain waise hi carelessly kaam kar rahe ho, otherwise there is no possibility of any confusion. Mera data ek decimal point k bala hai, lehaaza wo pehle jo class limits bana hi hain uske tahit wo kisi ek class me bati aasani ke saa telli ho jayega. Agar wo number, mera number 32.9 hai to pehle class mein chala jayega. Agar 33.0 hai to 2nd class mein chala jayega. Aur baad mein jab hum boundaries bana denge to wo jo gap tha, wo jo gap tha between 32.9 and 33.0, that gap is avoided. And we get a continuous distribution which this is exactly meant to be. After all, this is the frequency distribution of a continuous variable. Students, we have accomplished the construction of the frequency distribution of a continuous variable. Aap yakeen ki jaye ki ek intahai basic, lekin ek intahai important concept hai. Kyuke iske baad aap isse boh zyada advanced jo cheezein seekhenge, un mese boh jesee issi concept ke upar base karthi hi. So, I would like to encourage you ke aap apne jo text book hai, usme jo solved examples hai, unko bhi aap zaroot study kee jaye. Aur khas thorpe jo unsolved exercises hain, unko aap zaroot attempt karein. The more you practice, the better off you will be. Let us now consider the concept of the relative frequency distribution and the percentage frequency distribution. You will recall that in the last lecture, when we were doing the frequency distribution of a discrete variable, we have already done these concepts. Aur aap ko shahad yad ho kee, they were very simple concepts. All you have to do is to divide the frequency of any class by the total frequency and that gives you the relative frequency. Ab yaha pe aap kahin kee, yeh relatives ka kya chak karein, ki aunties aur uncles hain kdusre kee, aasa bilkul nahin yeh ji. Obviously, relative frequency means the frequency of any class relative to the total number of observations that you have. Yani agar kul observations, these hain, to these mese kitni hain, jo ke is particular class ko belong karein. Yani us khali us number ka alayada se itna fayda nahin hain. Jab hain usko dekhthe hain, in context where the total number of observations that we have, we get a better idea of the situation. Multiplying the relative frequency by 100 gives you the percentage frequency distribution and as I said in the last lecture, percentages to lemayan ke liye particularly, percentages ko understand karna toh sabse jada aasaan hota hain. So, as you can now see on the screen, the percentage frequency distribution for this example is 6.7 percent of the values belong to the first class, 13.3 percent belong to the second class, 46.7 percent belong to the third class and so on and so forth. Of course, the when you add up all these percentages, you get a total of 100. Students, relative frequency distribution banane ka ek bohat bada fayda hi bhi hain, ke aap comparison kar sakte hain between two frequency distributions having similar classes. Let me explain this point with the help of an example. Let us go back to that same example of the environmental protection agency and let us suppose that they have conducted this particular test not on just one particular car model, but on two different car models A and B as you now see on the screen. The frequency distribution for the model B is 7, 10, 16, 9 and 8 and the total number of cars for that model is 50. Whereas, the distribution for the first model is exactly the same as what we were discussing earlier. Now, in order to compare the situation of the two car models, it is very simple if we do the relative frequency distribution or the percentage frequency distribution. So, as you now see on the screen, we have 6.7 percent of the cars of model A have mileage 30 to 32.9. Lekin agar aap model B ko dekhe to 14 percent of the cars of this model have this particular mileage. To aap isska kya matlab hai? isska matlab hi hai ke model A seems to be better than model B issliye ke ye mileage to bahut kam mileage hai. This is not a very good mileage. Or model B jo hai uski 14 fee sathkare hain aisi hai, which is more than double of the percentage that we have for model A aisi jin ka mileage itna kam hai. Ishi tara agar aap aap agar aap second class ko dekhe to model A main 13.3 percent are falling in that mileage range, but for model B 20 percent, which is again substantially higher than 13 percent, 20 percent is falling in that range. Coming to a better mileage, which is 36 to 38.9, we see that almost 47 percent cars of model A are having this much mileage, but only 32 percent of the cars of model B have this particular mileage. To mera khyaal hai students aap mera point samajge hain ke iss tari ke se hum step by step hum compare kar sakte hain. Aur agar hame zyada iss me experience ho jaaye to aap a glance, almost at a glance we are able to have a proper comparison between the two sets of data. Let us consolidate all these ideas with the help of another example. The following table contains the ages of 50 managers of childcare centers in five cities of a developed country. As you can see the ages are 42, 26, 32 and so on. Convert this data into a frequency distribution. Now students in this connection, the first point to be noted is that just as we had in the last example, the data in this example, the data of the ages of the managers is a pile of numbers and we would like to convert it into a compact and comprehensible frequency table. So, in order to construct this table, of course the first step is to find the range of the data set just as we did in the last example. As you can see the smallest value in this data set is 23 and the largest value is 74. Therefore, the range is equal to 74 minus 23 equal to 51. The second step in constructing the frequency distribution is to determine how many classes we would like to have and the rule of thumb is to select something between 5 and 15 classes. In this example, suppose that we would like to have 6 classes, the next step is to determine the width of the class interval and an approximation of the class width can be calculated by dividing the range by the number of classes. So, dividing 51 by 6, we obtain 8.5 and if we round it, it is equal to 9. But students, the point to note is that generally rather than having a number like 9 or 7 or 8, we would like to have a rounded number like 10. You know, 10 year age intervals are easily recognized and understood by common people rather than a 9 year interval. So, the next step is the determination of the lower and upper limits of the various classes and we must keep in mind that the frequency distribution must begin at a value which is equal to or less than the lowest number in our raw dataset and also it should end at a value which is equal to or greater than the highest number in our dataset. In this example, the youngest age is 23 and the oldest is 74. So, students, we can start our frequency distribution from the number 20. Just as in the case of the class interval, we would like to have a convenient number. Similarly, for class limits as well, we would like to have a rounded number like 20 or in some other case, it might be 15. That would be regarded perhaps as a bit more convenient than perhaps 23. Now, that we are clear that the class interval is 10 and the lower limit of the lowest class is 20. Students, then we can very conveniently construct the classes as 20 to 29, 30 to 39, 40 to 49 and so on. The next step is to construct the column of frequencies and applying the tally method, we obtain as you now see on the screen the frequencies of the various classes as 6, 18, 11, 11, 3 and 1. Of course, the sum of the frequencies is 50 exactly the number of managers that we had in our data set. So, this is how we have consolidated a pile of 50 numbers into a compact and comprehensible tabular form. Today we have discussed the frequency distribution of a continuous variable. Let us now turn our attention to the visual representation of this distribution. As I said in the beginning of today's lecture, I will discuss in this regard the histogram, the frequency polygon and the frequency curve. The first step is to mark off adjacent rectangles whose bases are marked off by class boundaries along the x axis and whose heights are proportional to the frequencies associated with the respective classes. Students, the first step is to mark off the class boundaries along the x axis and construct a scale for the frequencies along the y axis as you now see on the screen. I will be considering the cars of model A that we were considering in the first instance. As you remember the lowest class boundary was 29.95, then we had 32.95, 35.95 and so on until we covered the largest value in our data set. As far as the y axis is concerned, as I said a short while ago, we should construct a scale in order to represent our frequencies. As you remember the highest frequency was 14 and as such we have constructed a scale so that we can cover the height equal to 14. We have to draw rectangles with the bases along the x axis and heights according to our frequencies for the various classes that we have. So, as you now see on the screen, Isitara for the second class 32.95 to 35.95 you will recall that the number of cars falling in that class was 4 and so we draw a rectangle of height equal to 4 units adjacent to the first rectangle. The class from 35.95 to 38.95 contained 14 observations and so the height of the third rectangle is much taller than the first two and we have this very interesting diagram coming up. Proceeding in the same manner we obtain the histogram that you now see on the screen. As I said earlier, this kind of a diagram is called the Histogram and it gives you the indication of the overall pattern of your frequency distribution. The second graph that I will discuss with you is called the frequency polygon. A frequency polygon is obtained by plotting the class frequencies against the midpoints of the classes and connecting the points so obtained by straight line segments. Now, the important thing you must have noted is that what we have just made before is that we plotted the class boundaries against the x axis but now we have to take midpoints along the x axis. Midpoints extremely simple. All you have to do is to add the lower boundary of any class with the upper boundary of that class and divide by 2 and that gives you the midpoint. So, as you now see we have the midpoints for this example as 31.45, 34.45, 37.45, 40.45 and 43.45. As you must have realized it is simply the average of the lower boundary and the upper boundary of any class. Now, plotting these along the x axis and constructing the scale for the frequencies along the y axis as you now see on the screen we are ready to draw our graph which is called the frequency polygon. Now, note that our first midpoint was 31.45 and our last midpoint was 43.45 but on the x axis the values I have taken you are noting that before 31.45 I have taken one more value on the same interval as the rest of the values. That is the same interval h is equal to 3 and because of this my first value is on the x axis that is 28.45. Similarly, on the right side of the graph my last midpoint was 43.45 and that is 46.45. Now, why did I do this? The reason for this is that I want my frequency polygon to be a closed figure. As you will recall from your school days, in geometry polygon is called a many sided closed figure. So, the one class that I have increased in the beginning and the last class that I have added is 0. So, it is appropriate to write frequency equal to 0 for this first class and this last class that I have added. It is appropriate to write frequency equal to 0 for this first class and this last class that I have added. As I will just now explain, all you have to do is to plot the frequencies that you have against the x values which are the midpoints. So, as you now see on the screen I have our midpoints are 28.45, 31.45 and so on and the frequencies are 0, 2, 4 and so on. So, plotting these points and joining them by straight line segments, we get the graph that is called the frequency polygon. Now, I will stress once again that this class that we have formed in the beginning with 0 frequency and the last one that we formed with 0 frequency, because of that graph it looks like a closed figure. If we do not form these two classes then of course our graph would have looked as what you now see on the screen. Now, actually the point is that from the statistical point of view it really does not matter, but if you consider the very basic mathematical definition of a polygon this graph which is not touching the x axis from either side you know it is not a closed figure. So, this is exactly the reason why we do what we just did. The next concept is that of the frequency curve. The frequency curve is also a very important concept and a very easy concept. All you have to do is to smooth your frequency polygon by the freehand method. So, as you now see on the screen the dotted line represents the frequency curve whereas of course the continuous straight line segments represent the frequency polygon. What is important in this is that this frequency curve does not mean that this smooth curve must touch all the plotted points. Many of our students are confused and I want to stress that you keep this in mind that this is a smoothing by the freehand method in order to indicate the overall pattern of your frequency distribution. It is not necessary that your curve must pass through all the points that you plotted. Let us consolidate all these ideas by considering the frequency distribution of the ages of the managers of child care centers. The class intervals were 20 to 29, 30 to 39, 40 to 49 and so on and the frequencies were 6, 18, 11 and so on. Now, in order to construct the histogram the first step is to form the column of class boundaries and as explained earlier the class boundaries are obtained by adding the upper limit of any class to the lower limit of the next class and dividing by 2. For example, when we add 29 to 30 and divide by 2 we obtain 29.5 or students note ke 29 ke jaga pe bhi hum 29.5 likhenge or 30 ke jaga pe bhi 29.5. Similarly, we can find the average of 39 and 40 to obtain 39.5 the average of 49 and 50 to obtain 49.5 and so on. Now, in order to construct the histogram we will take the class boundaries along the x axis and the frequencies along the y axis. The histogram is a set of adjacent rectangles and the heights of the rectangles will be according to the frequencies. The first frequency is 6 and so the first rectangle has a height of 6 units. The second frequency is 18 and therefore the second rectangle is much taller than the first one as you can see on the slide. Similarly, the third frequency is 11 and we have the third rectangle with height equal to 11 units. Proceeding in this manner students we obtain the histogram that you now have on the screen. Next we consider the construction of the frequency polygon. As you will remember for this purpose we first need to find the midpoints of all the classes which in this example are 24.5, 34.5, 44.5 and so on. Of course, they are obtained very easily by adding the lower class boundary of any class to the upper class boundary of the same class and dividing by 2. Now, since we want the polygon to touch the x axis on the left side as well as the right hand side. Therefore, we add a class in the beginning and in the end of our table and these classes are 9.5 to 19.5 and 79.5 to 89.5. Now, since our data set does not fall in age groups therefore the frequencies of these classes are 0. The midpoint of the first class that I just added is 14.5 and that of the other one is 84.5. Now, taking all the midpoints along the x axis and the frequencies along the y axis we are ready to construct our frequency polygon and by plotting the frequencies against the midpoints and joining the points so obtained by straight line segments students we obtain the frequency polygon that you now see on the screen. Last but not the least we would also like to draw the frequency curve of this particular data set and as you now see on the screen when we smooth the frequency polygon by a free hand curve we obtain a moderately positively skewed frequency curve. So, we see that the visual representation of the frequency distribution gives us a fairly good idea about the main properties of our data set. Next time we will take this discussion further and I will discuss with you the various types of frequency curves that we encounter in practice. Also we will go to the concept of the cumulative frequency distribution and the cumulative frequency polygon. In the meantime as I said earlier I would like you to practice with all that you have learned today and also to attempt the assignment that you will find on the website. Best of luck and Allah Hafiz.