 Hello, I am welcome to the session, I am Deepika here. Let's discuss a question which says 100 surnames were randomly picked up from local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows. Now if the number of letters is 1 to 4 then the number of surnames is 6. If number of letters is 4 to 7 then number of surnames is 30 and for 7 to 10 it is 40, for 10 to 13 it is 16 and for 13 to 16 it is 4 and for 16 to 19 it is also 4. If letters in the surnames find the mean number of letters in the surname also find the modal size of the surnames in median. Now we know that the class mark is equal to of the modal class and f0 in modal class is greater than number of observations. As we will apply the formula median is equal to the lower limit of median class observations and see accumulative frequency of class the idea to solve the above questions. So let's start the solution. Let us is 1 to 4 to 7 then the number of surnames is 30 the number of surnames 4 and again to 19 it is 4. Now to find column as 1 to 4 is 1 to 19 the class mark is 16 plus 19 upon 2 which is 17.5 with the corresponding frequency f100 and sigma f832 is equal to equal to 8.32. Now we will find the modal sign the modal class maximum frequency lower limit of the modal class 16 and h is equal to 3. So minus f is values 3 minus 30 upon 2 into 40 minus 30 minus 16 equal to 7 plus now 30 upon the modal size of the surnames 7.88 column for cumulative frequency as 1 to 4 the frequency of the class 4 to 7 is 6 plus 30 that is 36 as it is 36 plus 40 which is our cumulative frequency for the class 16 is 92 and see of the class 13 to 16 is 92 plus 4 which is 6 plus 4 that is 100 equal to 36 as 40 and h is equal to 3 is equal to 21 upon 20 is 1.0 so median is equal to 105. Now question is the limit is equal to