 Hello and welcome to the session. In this session we are going to discuss the following question which says that calculate the mode from the following data graphically. The wages are represented by X and are given in dollars in the range of 0 to 5, 5 to 10, 10 to 15, 15 to 20, 20 to 25, 25 to 30 With the corresponding number of workers represented by F given by 3, 5, 10, 20, 12 and 6 Verify the results with the help of interpolation Formula for mode M0 is given by 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 Fm is the frequency of the modal class F1 is the frequency of the class The seeding the modal class F2 is the frequency of the class Succeeding the modal class is the size of the class interval With this key idea we shall proceed with the solution We are given the following distribution Here wages are given in dollars and are represented by X and F represents the number of workers Wages are given in the range of 0 to 5, 5 to 10, 10 to 15, 15 to 20, 20 to 25, 25 to 30 With the corresponding number of workers as 3, 5, 10, 20, 12 and 6 By inspection we find that the modal class is 15 to 20 As the class 15 to 20 has the maximum frequency that is 20 Now we draw 3 rectangles of the pictogram of the given distribution First of modal class 15 to 20 Now in the graph we have wages along the X axis and number of workers along the Y axis Now we shall draw first rectangle of the histogram corresponding to the modal class 15 to 20 For the class 15 to 20 there are 20 workers For the class 15 to 20 there are 20 workers In rectangle is of the class preceding the modal class that is 10 to 15 For the class 10 to 15 the number of workers is given by 10 That is for the class 10 to 15 there are 10 workers Next we have the class succeeding the modal class that is 20 to 25 For the class 20 to 25 the number of workers is given by 12 For the class interval 20 to 25 there are 12 workers Now draw the lines AC and BD diagonally Let P be their point of intersection From P draw a line perpendicular to the X axis Newton the X axis at the point Q It gives the value of the required mode that is 17.8 So mode is given by 17.8 Now we shall verify the above result with the help of the formula From the key idea we know that mode M not is given by L1 plus Fn minus F1 upon Christ of Fn minus F1 minus F2 into I Where L1 is the lower limit of the modal class Fn is the frequency of the modal class F1 is the frequency of the class preceding the modal class F2 is the frequency of the class preceding the modal class And I is the size of the class interval So we have mode M not is equal to L1 plus F1 minus F1 upon Christ of Fn minus F1 minus F2 into I Here the modal class is 15 to 20 With the corresponding frequency as 20 So we have M not is equal to L1 that is the lower limit of the modal class Which is equal to 15 plus Fn that is the frequency of the modal class That is 20 minus F1 That is the frequency of the class preceding the modal class that is 10 So we have 20 minus 10 upon Christ of Fn that is 2 into 20 minus F1 that is 10 minus F2 That is the frequency of the class preceding the modal class which is equal to 12 into I That is the size of the class interval which is given by 20 minus 15 that is 5 So we have 15 plus 20 minus 10 that is 10 upon 2 into 20 that is 40 minus 10 minus 12 Which is equal to minus 22 into 5 Which is equal to 15 plus 10 upon 30 minus 22 that is 18 into 5 Which is equal to 15 plus 5 into 5 that is 25 upon 9 On taking the LCM we get 135 plus 25 upon 9 which is equal to 160 upon 9 Which is equal to 17.77 or can be approximately written as 17.8 Which is the same value as we have got from the histogram Therefore the value of the modal is equal to 17.8 Which is the required answer This completes our session. Hope you enjoyed this session