 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that construct a histogram for the given distribution, find the mode graphically, verify the result with the help of interpolation. We know that mode is given by the formula l1 plus fm minus f1 upon twice of fm minus f1 minus f2 into i, where l1 is the lower limit of the modal class frequency of the modal class f1 is the frequency of the class preceding the modal class f2 is the frequency of the class succeeding the modal class width of the modal class. Key idea, let us proceed with the solution. By inspection, find that the modal class 12 to 16 it has the maximum frequency. Now we take marks on the x-axis and frequency on the y-axis. For x-axis we have taken 1 square is equal to 4 units and for y-axis 1 square is equal to 5 units. Now we will draw 3 rectangles of the histogram for the given distribution. First of modal class 12 to 16 having frequency 28, second rectangle of the class preceding the modal class that is 8 to 12 having frequency 15 and the third rectangle of the class succeeding the modal class that is 16 to 20 having frequency 20. Now draw lines ac and bd and name their point of intersection as p. From p draw a perpendicular on the x-axis and the point where this perpendicular meets the x-axis gives the mode that is 14.5. Therefore mode is equal to 14.5. Now we will do the verification. We know that mode is given by the formula l1 plus fm minus f1 upon twice of fm minus f1 minus f2 into i. Here l1 is the lower limit of the modal class that is 12. fm is the frequency of the modal class that is 28. f1 is the frequency of the class preceding the modal class that is 15. f2 is the frequency of the class succeeding the modal class that is 20 and i is the width of the modal class that is 16 minus 12 which is equal to 4 substituting all these values in the formula we get mode m0 is equal to l1 that is 12 plus fm minus f1 that is 28 minus 15 upon twice of fm that is 2 into 28 minus f1 minus f2 minus 15 minus 20 into i that is 4 which is equal to 12 plus 13 upon 56 minus 35 into 4 that is 12 plus 13 upon 21 into 4 which is equal to 12 plus 52 upon 21 on taking the LCM we get 252 plus 52 upon 21 which is equal to 304 by 21 that is 14.47 or can be approximately written as 14.5 which is same as what we have in the graphical method hence the mode of the given distribution is 14.5 approximately which is the required answer this completes our session hope you enjoyed this session