 Hello and welcome to the session. In this session, we are going to discuss about computation of mode for continuous series. In case of continuous series, first we locate the modal class by inspection or by dripping method. The value of mode is then determined by applying the following formula. The symbol m naught or z is used to denote mode and is equal to l 1 plus f m minus f 1 upon twice of f n minus f 1 minus f 2 into i where l 1 is the lower limit of the modal class f n is the frequency of the modal class f 1 is the frequency of the class preceding the modal class f 2 is the frequency of the class succeeding the modal class and i is the width of the modal class. The formula is applied in the following forms depending on whether the series is given in ascending form or descending form. If the series is given in ascending order mode m naught is given by l 1 plus f n minus f 1 upon twice of f n minus f 1 minus f 2 into i also. If the value of twice of f n minus f 1 minus f 2 is equal to 0 f 1 is equal to 0 upon twice of f n minus f 1 minus f 2 into i also. If the value of twice of f n minus f 1 minus f 2 is equal to 0 f 1 is greater than f n f 2 is greater than f n or both f 1 and f 2 are greater than f n then mode m naught is given by l 1 plus f 2 upon f 1 plus f 2 into i. Here we should note that if the modal class happens to be the first or the last class interval then we may respectively take f 1 f m equal to 0 also it is necessary that the class intervals are of uniform width throughout if they are unequal they should be made equal. Let us take an example determining the mode also construct its histogram and the distribution is given as the class intervals are given as 0 to 5 5 to 10 10 to 15 15 to 20 20 to 25 25 to 30 and I represented by x with the corresponding frequency represented by f is given by 2 3 7 12 5 and 1. Since the given series is regular therefore we can determine modal class by inspection which is equal to 15 to 20. Therefore the modal class is 15 to 20 since it has the maximum frequency that is 12 and we know that mode m naught is given by l 1 plus f n minus f 1 upon twice of f n minus f 1 minus f 2 into i. So we have mode m naught is equal to l 1 that is the lower limit of the modal class which is equal to 15 plus f m that is the frequency of the modal class which is equal to 12 minus f 1 that is the frequency of the class preceding the modal class which is equal to 7 upon twice of f n that is 2 into 12 minus f 1 that is 7 minus f 2 that is the frequency of the class succeeding the modal class and is given by 5 into i that is the width of the class interval which is given by 20 minus 15 that is 5 which is equal to 15 plus 12 minus 7 that is 5 upon 2 into 12 that is 24 minus 7 minus 5 is minus 12 into 5 that is 15 plus 5 upon 24 minus 12 that is 12 into 5 which is equal to 15 plus 5 into 5 that is 25 upon 12 on taking the ACM we get 180 plus 25 upon 12 which is equal to 205 upon 12 that is 17.08 so mode m is given by 17.08 now we shall study the procedure for the graphical location of mode and the first step is to draw a histogram of the data for convenience only three rectangles may be drawn that is rectangles for modal pre-modal and first modal class second step is to draw two lines diagonally inside the modal class rectangle starting from each upper corner of the modal rectangle to the upper corner of the adjacent rectangle step 3 is from the point of intersection of both these lines draw a perpendicular to the x axis the point where this perpendicular meets the x axis is the value of the mode we should note that mode cannot be determined graphically if two or more classes have the same highest frequency now we shall construct the histogram for the same distribution to draw the histogram we take class along the x axis and frequency along the y axis and for x axis 1 square is equal to 5 units and for y axis 1 square is equal to 2 units now in the first step we shall construct three rectangles for modal pre-modal and first modal class to draw the rectangle for the modal class we have the class interval that is 15 to 20 with the frequency as 12 this is the rectangle of the modal class now for pre-modal class we have the class interval 10 to 15 with the corresponding frequency as 7 this is the rectangle for the pre-modal class for post-modal class we have the class interval 20 to 25 with frequency 5 this is the rectangle for the post-modal class now we draw two lines diagonally inside the modal class rectangle starting from each upper corner of the modal rectangle to the upper corner of the adjacent rectangle let the two diagonals intersect at point P from P we draw a perpendicular on the x axis and the point where this perpendicular meets the x axis is the value of the required mode which is equal to 17.08 therefore we can say the mode is equal to 17.08 this completes our session hope you enjoyed this session