 Hello and welcome to the session. In this session we are going to discuss the following question which says that obtain the value of median md lower quartile q1 and upper quartile q3 by drawing two oj's. Here marks are represented by x and x represents the number of students and there are 10 students who got marks from 0 to 10, 12 students who got marks from 10 to 20, 9 students got marks in between 20 to 30, 7 students got marks from 30 to 40, 6 students got marks from 40 to 50 and 18 students got marks from 50 to 60 to draw a less than oj's take cumulative frequency on the y axis and variable that is the upper limit on the x axis and to draw more than oj's take cumulative frequency on the y axis and lower limit on the x axis the intersection of both these oj's gives the value of the median with this key idea we shall proceed with the solution we are given the following distribution to draw less than oj's now we shall find the number of students that is the cumulative frequency for the x1 series that is marks less than series. Now the number of students who got marks less than 10 will be the number of students who got marks from 0 to 10 that is 10. Now the number of students who got marks less than 20 is the number of students who got marks in between 0 and 20 that is the sum of the number of students who got marks from 0 to 10 and 10 to 20 that is 10 plus 12 which is equal to 22. Next we have the number of students who got marks less than 30 and this is equal to the number of students who got marks less than 20 plus the number of students who got marks from 20 to 30 that is 22 plus 9 that is equal to 31. Next is number of students who got marks less than 40 and this is equal to the number of students who got marks less than 30 plus the number of students who got marks in between 30 and 40 and this is equal to 31 plus 7 that is 38. Next the number of students who got marks less than 50 that is the sum of the number of students who got marks less than 40 and the number of students who got marks in between 40 and 50 which is equal to 38 plus 6 that is 44 and the number of students who got marks less than 60 will be equal to the number of students who got marks less than 50 plus the number of students who got marks in between 50 and 60 that is 44 plus 18 which is equal to 62. Now for less than 12 the points are 10, 10, 20, 22, 30, 31, 30, 38, 50, 44 and 60, 62 in the graph we have taken marks along the x axis and number of students along the y axis. Now we shall plot these points for less than curve on the graph. Our first point is 10, 10, next point is 20, 22, the next point is 30, 31, then we have 30, 38, next is 50, 34 and then we have 60, 62. Now we shall join these points by a smooth curve and this is called the less than curve. Now we shall find the number of students that is the cumulative frequency for marks more than series. The number of students for marks more than 0 will be the sum of all the number of students who got marks from 0 to 10, 10 to 20, 20 to 30, 30 to 30, 30 to 50 and 50 to 60 that is 10 plus 12 plus 9 plus 7 plus 6 plus 18 that is 62. The number of students who got marks more than 10 will be the number of students who got marks more than 0 minus the number of students who got marks from 0 to 10. So we have 62 minus 10 that is 52. Now the number of students who got marks more than 20 will be the number of students who got marks more than 10 minus the number of students who got marks from 10 to 20 that is 52 minus 12. So we have the number of students as 40. Number of students who got marks more than 30 will be the number of students who got marks more than 20 minus the number of students who got marks from 20 to 30 that is 40 minus 9 which is equal to 31. Number of students who got marks more than 40 will be the number of students who got marks more than 30 minus the number of students who got marks in between 30 and 40. So we have 31 minus 7 which is equal to 24 and the number of students who got marks more than 50 will be the number of students who got marks more than 40 minus the number of students who got marks in between 40 and 50. So we have 24 minus 6 that is 18. Therefore for more than curves the points are 0, 62, 10, 52, 20, 40, 30, 31, 40, 24, 50, 18. To draw the more than curve now we shall plot these points on the graph. The first point is 0, 62, next we have 10, 52, then there is 20, 30, then we have 30, 31, next is 30, 24 and then 50, 18. Now we shall join these points by a smooth curve. This is the required more than curve and from the key idea we know that the intersection of both less than oj and more than oj gives the value of the median. p is the point of the intersection of the two curves and from p when we draw a perpendicular on the x axis the point where this perpendicular meets the x axis is the value of the required median which is equal to 30. Therefore we get the value of median as 30. Now we know that m is equal to the sum of the frequencies denoted by summation of f and is given by 62. Now we shall find the value of the lower quartile. Lower quartile q1 is given by the size of n by fourth item. The value of m is equal to 62. Therefore q1 is given by the size of 62 by fourth item which is equal to the size of 15.5th item. Mark this point along the y axis. We name it as b. From b draw a line parallel to the x axis corresponding to y is equal to 15.5. Let it meet the curve in the point q. Draw a perpendicular from q on the x axis. The point where this perpendicular meets the x axis gives the value of the lower quartile and is equal to 14.59. Therefore lower quartile q1 is equal to the size of 16.5th item is equal to 14.59. Now we shall find the value of the upper quartile. Upper quartile q3 is given by the size of 3n by fourth item. The value of n is equal to 62. So we have the size of 3 into 62 by fourth item which is equal to the size of 186 by fourth item. We can write it as the size of 36.5th item. We mark this point along the y axis as point c. From point c draw a line parallel to x axis corresponding to y is equal to 36.5. Let it meet the lift and curve at point r. From point r draw a perpendicular on the x axis and the point where this perpendicular meets the x axis gives the value of the upper quartile which is equal to 51.4. Therefore upper quartile which is equal to the size of 36.5th item is equal to 51.4. Therefore we get median envy is equal to 30. Lower quartile q1 is equal to 14.59. And upper quartile q3 is equal to 51.4 which is the required answer. This completes our session. Hope you enjoyed this session.