 Hello and welcome to the session. In this session we will discuss about median. Median can be defined as the mid value of series when arranged either in ascending order or in descending order. It can also be defined as that value of the series which divides the given arranged series into two equal parts. It is denoted by m. Frequency distribution can be of two types. One, ungrouped frequency distribution in the regional series and the other is grouped frequency distribution. Now we will learn how median can be calculated in each of the frequency distribution. First we shall discuss ungrouped frequency distribution. Ungrouped frequency distribution or in individual series the median can be calculated by the formula m is equal to n plus 1 by towards item where n is the total number of items. For this we have to first arrange the series in ascending or descending order then locate the median item and the value of this item is the median for all series. Mid value is the median for even series that is having even number of items. Median is the mean of the two median values for example if the series is 1, 2, 3, 4, 5 then median is given by n plus 1 by towards item. The number of items is 5 so we have 5 plus 1 by towards item that is equal to 6 by towards item which is equal to third item. And the value of the third item is 3 therefore median is equal to 3 but if the series is 5, 6, 7, 8, 9, 10 that is the even number of items then median is given by the mean of the two median values that is the size of n by towards item and n is 6 therefore the size of third item plus the size of n by 2 plus 1th item that is the size of fourth item by 2 which is equal to the size of third item. The third item is 7 therefore 7 plus the size of fourth item and the fourth item is 8 that is 8 by 2 which is equal to 15 by 2 that is 7.5. Now we shall discuss good frequency distribution. Good frequency distribution can be further classified into two types one is discrete series and the other is continuous series. A discrete series first arranged a series in ascending or descending order then we have to find the cumulative frequency then we apply the formula m is equal to n plus 1 by towards item here. n is given by summation of f that is the frequency after this we locate the plus 1 by towards item in the cumulative frequency column. Now median is the value of the variable for example let us find the median for the given distribution where the x series is given by 1, 2, 3, 4 and the corresponding frequency is given by 2, 3, 1, 5. Since it is already arranged in the ascending order now we shall find the cumulative frequency. Now in the cumulative frequency column the first entry will be same as that of frequency that is 2. The next entry would be 2 plus 3 that is 5, 5 plus 1 that is 6 and then 6 plus 5 that is 11. Here n is given by the summation of the frequency that is 2 plus 3 plus 1 plus 5 which is given by 11. Now we apply the formula m is equal to n plus 1 by towards item which is equal to n is 11 so we have 11 plus 1 by towards item that is given by 12 by towards item which is equal to 6th item, 6th item lies against 6 whose corresponding x value is 3. Therefore median is equal to 3. Now we shall discuss the case of continuous series. In this first we find the cumulative frequency then we will find the median size by the formula n by 2 where n is the sum of the frequencies. Given by summation f with this median number we will be able to predict the median class and then we shall apply the formula median m is equal to n plus i upon f into n by 2 minus c where n by 2 is the median number If the cumulative frequency of the class just lower than the median class is the class interval of the median class f is the frequency of the median class l is the lower limit of the median class. Let's take the following example to calculate median for the continuous series. First of all we shall calculate the cumulative frequency. The first entry in the cumulative frequency column will be same as that of frequency that is 2. Next we have 2 plus 4 that is 6 then 6 plus 6 that is 12 then 12 plus 8 that is 20. Here n is equal to sum of the frequencies which is given by 2 plus 4 plus 6 plus 8 that is 20 and we know that median number is given by n by 2 that is 20 by 2 which is equal to 10. Now cumulative frequency just greater than 10 is 12 so median number lies in class 10 to 15 therefore median class is 10 to 15. Now we know that median is given by l plus i upon f into n by 2 minus c which is equal to l that is the lower limit of the median class and is given by 10 plus i that is the class interval which is given by 15 minus 10 that is 5 by f that is the frequency of the median class given by 6 that is 6 into n by 2 that is given by 10 minus c. c is the cumulative frequency of the class just lower than the median class and is given by 6 which is equal to 10 plus 5 upon 6 into 4 which is equal to 10 plus 10 upon 3 On taking the LCM we get 30 plus 10 by 3 which is equal to 40 by 3 that is 13.33 which is the required median. This completes our session hope you enjoyed this session.