 Hello and welcome to the session. In this session, we learn about graphical location of medians and quartiles by Ojeef. Now we shall learn about the first method that is by drawing only one Ojeef by less than method. And the first step is to draw the cumulative frequency curve that is Ojeef by less than method. Second step is to find m by 2 and not n plus 1 by 2 and mark this point along the y-axis. This is the cumulative frequency curve and we mark the point m by 2 along the y-axis. Step 3 is at this point draw a line parallel to the x-axis meeting the Ojeef at the point the draw a line parallel to x-axis meeting the Ojeef at the point a. Now the next step is from the point a draw a line perpendicular to x-axis meeting it at say m then the x-coordinate of m gives the value of the medium that is the draw a line perpendicular to the x-axis meeting the x-axis at point m and the x-coordinate of point m is the required medium. Now to locate the value of your quartile q1 and at the quartile q3 we mark the point m upon 4 and not m plus 1 upon 4 3 into n upon 4 and not trice of n plus 1 by 4 on the y-axis respectively proceed with step 3 to locate the value of lower quartile that is q1. Then we mark the value of n by 4 along the y-axis then we draw a line parallel to the x-axis meeting the Ojeef at some point from that point draw a perpendicular on the x-axis and the point where this perpendicular meets the x-axis gives the value of the required lower quartile. Similarly to find the value of the upper quartile q3 we mark the point 3 into n by 4 along the y-axis from this point we draw a line parallel to the x-axis meeting the Ojeef at some point from that point draw a perpendicular on the x-axis and the point where this perpendicular meets the x-axis is the value of the required upper quartile. Let us take an example find the values of median q1 q3 and form an Ojeef for the following frequency distribution also verify the result by calculation. The marks obtained which is represented by x is given as 0 to 5, 5 to 10, 10 to 15, 15 to 20, 20 to 25 with the corresponding frequency represented by f that is 4, 6, 10, 9, 5. Now we shall rewrite the distribution as follows. Now we shall find out the cumulative frequency for the corresponding marks less than series for marks less than 5 that is the marks from 0 to 5 the cumulative frequency is 4 for marks less than 10 that is the marks obtained from 0 to 10 the cumulative frequency will be 4 plus 6 that is 10 for marks less than 15 that is the marks obtained from 0 to 15 the cumulative frequency will be 4 plus 6 plus 10 that is 20 for marks less than 20 that is the marks obtained from 0 to 20 the corresponding cumulative frequency will be 4 plus 6 plus 10 plus 9 that is 29 and for marks less than 25 that is the marks obtained from 0 to 25 the corresponding cumulative frequency will be 4 plus 6 plus 10 plus 9 plus 5 that is 34. Now taking the variable on the x axis and cumulative frequency on the y axis we shall plot the points and the points are 5, 4, 10, 10, 15, 20, 20, 29, 25, 24. For x and y axis we have taken 1 square is equal to 5 units the points are 5, 4, 10, 10, 15, 20, 20, 29, 25, 34. Now we join these points by a smooth curve and this is called the less than curve. Now the second step is to find the position of the median that is given by m by 2 and m is equal to the summation of the frequencies that is equal to 34. Therefore the position of the median that is given by m by 2 is 34 by 2 which is equal to 17. We mark this point along the y axis. Now from this point draw a line parallel to x axis and let it meet the less than curve at point a. From a draw a perpendicular on the x axis and the point where this perpendicular meets the x axis gives the value of the required median that is 13.5. Now we shall verify our result by calculating median with the help of the formula that is median is equal to l plus i upon f into n by 2 minus c where l is the lower limit of the median class, i is the width of the class, f is the frequency of the median class and c is the cumulative frequency of the class just lower than the median class. Here the position of the median is given by 17 and the cumulative frequency just greater than 17 is 20 with the corresponding class as 10 to 15. Therefore the median class is 10 to 15. Therefore l which is the lower limit of the median class is 10, i the class interval is given by 15 minus 10 that is 5, f is the frequency of the median class which is given by 10 and c is the cumulative frequency of the class just lower than the median class and is 10 and n is given by 34. So median m is equal to l that is 10 plus i upon f that is 5 upon 10 into n by 2 that is 34 by 2 minus c that is 10. So this is equal to 10 plus 1 by 2 into 17 minus 10 which is equal to 10 plus 1 by 2 into 7 which is equal to 10 plus 7 by 2 that is given by 27 by 2 which is equal to 13.5 which is the same value as we have got from the graph. So median m is equal to 13.5. Now we shall calculate lower quartile q1 and the position of q1 is given by n upon 4 and n is given by 34. So we have 34 by 4 which is equal to 17 by 2 that is 8.5. Now cumulative frequency just greater than 8.5 is 10 with the corresponding quartile class as 5 to 10. Therefore lower quartile q1 lies in the class 8 to 10. We plot the point 8.5 along the y axis from this point draw a line parallel to x axis meeting the less than curve at point b from b draw a perpendicular on the x axis and the point where this perpendicular meets the x axis gives the value of the lower quartile that is 8.7. Now we shall verify our result with the help of the formula that is lower quartile q1 is given by l plus i upon f into n by 4 minus c which is equal to l that is the lower limit of the quartile class that is 5 plus i which is the class integral given by 10 minus 5 that is 5 by f that is the frequency of the quartile class that is 6 into n upon 4 that is 34 upon 4 which is given by 8.5 minus c that is the cumulative frequency of the class just lower than the quartile class that is given by 4 which is equal to 5 plus 5 upon 6 into 8.5 minus 4 that is 4.5 which is equal to 5 plus 3.75 that is 8.75 which is the same value as we have got from the graph now we will find upper quartile position of upper quartile is given by 3 into l upon 4 and we know that l is 34 so we have 3 into 34 upon 4 which is equal to 102 upon 4 that is 25.5 now cumulative frequency just greater than 25.5 is 29 with the corresponding quartile class as 15 to 20 therefore upper quartile q3 lies in the class 15 to 20 now in the graphical method we plot the point 25.5 along the y axis now draw a line parallel to x axis meeting the left hand curve at point c and from c draw a perpendicular on the x axis and the point where this perpendicular meets the x axis gives the value of the upper quartile that is 18.05 now we shall verify our result with the help of the formula which is given by l plus i upon f into twice of n by 4 minus c which is equal to l is the lower limit of the quartile class that is 15 i is the class interval that is given by 20 minus 15 which is 5 by f that is the frequency of the quartile class which is 9 into twice of n by 4 which is given by 25.5 minus c that is the cumulative frequency of the class just lower than the quartile class which is given by 20 so we have 15 plus 5 upon 9 into 5.5 which is equal to 15 plus 27.5 by 9 that is equal to 15 plus 3.05 which is given by 18.05 value of q3 is same as we have got from the graph that is 18.05 this completes our session hope you enjoyed this session