 Hi and welcome to the session. Let us discuss the following question. Question says, during the medical checkup of 35 students of a class, their weights were recorded as follows. This is the given data. Draw a less than type ojive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula. First of all, let us understand what is an ojive. Graphical representation of cumulative frequency distribution is called ojive or we can say it is called cumulative frequency curve. This is the key idea to solve the given question. Let us now start with the solution. First of all, let us rewrite the data given in the question. This is the given data to us. We know this table represents a cumulative frequency distribution of less than type. And here, these values represent upper limits of the respective class intervals. Now to represent the data given in the table graphically, we mark the upper limits of the class intervals on the horizontal axis, that is, x-axis and their corresponding cumulative frequencies on vertical axis, that is, y-axis. We know number of students represent cumulative frequency here. Now here we have shown upper limits of the class intervals on x-axis, that is, x-axis represents weight in kg and y-axis represents the cumulative frequency. Now to draw an ojive, we will plot points 38, 0, 43, 42, 5, 44, 9, 46, 14, 48, 28, 50, 32, 52, 35, on the graph. Here we have shown all the points on the graph. This point represents 38, 0, this point represents 43, this point represents 42, 5, this point represents 44, 9, this point represents 46, 14, this point represents 48, 28, this point represents 50, 32 and this point represents 5235. Now joining these points by a free hand smooth curve we get a cumulative frequency curve or an ojive. This is a ojive of less than type. Now let us find out median of the given data graphically. Now we know total number of students is equal to 35. So here n is equal to 35. Now n upon 2 is equal to 35 upon 2 that is 17.5. Now locate this point 17.5 on y-axis of the graph. Now from this point draw a line parallel to x-axis cutting the curve at this point. Now from this point draw a perpendicular to x-axis. Now this is the required perpendicular. Now the point of intersection of this perpendicular with the x-axis determines the median of the data. Clearly we can see this point is 46.5. So the required median weight is equal to 46.5 kg. Now here we can write graphically median weight is equal to 46.5 kg. Now we will find the median by using the formula. Now first of all we will change this less than type cumulative frequency distribution to continuous class intervals and normal frequency. So we can write here 36 to 38, 38 to 40, 40 to 42, 42 to 44, 44 to 46, 46 to 48, 48 to 50 and here we can write 50 to 52. We know these are all upper class limits. So here we have mentioned upper class limits and we also know that these are continuous intervals. So upper limit of this interval is lower limit of the next interval. So we can write all these intervals this way. Now we will write corresponding cumulative frequencies of every interval. So they are 0, 3, 5, 9, 14, 28, 32, 35. Now let us find out individual frequencies of each interval. Now clearly we can see number of students having weight in the interval 36 to 38 is equal to 0. Number of students having weight in the interval 38 to 40 is equal to 3 minus 0 that is 3. Similarly number of students having weight in the interval 40 to 42 is equal to 5 minus 3 that is 2. Here number of students is equal to 9 minus 5 that is 4. Similarly for this interval number of students is equal to 14 minus 9 that is 5. For this interval number of students is equal to 28 minus 14 that is 14. Here number of students is equal to 32 minus 28 that is 4 and for this interval number of students is equal to 35 minus 32 that is 3. Now for finding the median we have to find the value inside a class which divides the whole distribution into two halves. We have shown above that n upon 2 is equal to 17.5. Now we will locate the class whose cumulative frequency is greater than or nearest to 17.5. Now clearly we can see this value of cumulative frequency is greater than 17.5 and this is the median class. So we can write median class is equal to 46 to 48. Clearly we can see 46 to 48 is the class whose cumulative frequency 28 is greater than 17.5. So 46 to 48 is a median class. Now let us write the formula for median. Median is equal to L plus n upon 2 minus CF upon F multiplied by H where L is the lower limit of the median class, n is the number of observations, CF is cumulative frequency of class preceding the median class, F is the frequency of median class and H is the class size. Now we know number of students represent the frequency. Now clearly we can see this is the median class and lower limit of the median class is 46 and CF that is cumulative frequency of class preceding the median class is equal to 14 and frequency of the median class is also equal to 14. So here we can write L is equal to 46, n upon 2 is equal to 17.5, CF is equal to 14, F is equal to 14 and H is equal to 2. We know class size is equal to 2, class size of our interval is equal to upper class limit minus lower class limit. So by subtracting upper class limit and lower class limit of any of the interval we can find class size. Now substituting all these values in formula for median we get median is equal to 46 plus 17.5 minus 14 upon 14 multiplied by 2. Now this is further equal to 46 plus 3.5 upon 14 multiplied by 2. Now it can be further written as 46 plus 7 upon 14. We know 3.5 multiplied by 2 is equal to 7. Now we will cancel common factor 7 from numerator and denominator both. Now this is further equal to 46.5. So we get median weight is equal to 46.5 kg. So we get median weight obtained from the graph is equal to median weight obtained from formula that is 46.5 kg. Now this verifies our answer. This completes the session. Hope you understood the solution. Take care and have a nice day.