 Hello and welcome to the session. I am Shashin and I am going to help you with the following question. Question says, A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. This is the given table. First of all let us understand that mean is equal to assume mean plus summation f i d i upon summation f i. Here in this formula x bar is the mean a is assume mean f i is frequency and d i is equal to x i minus a that is deviation of assume mean from class mark of the interval x i is the class mark of any interval and a is the assume mean here. So, d i is deviation of assume mean that is a from class mark x i. This is the key idea to solve the given question. Let us now start with the solution. First of all we will rewrite the data given in the question. We are given number of days and number of students. We know number of students represent frequency and we denote it by f i. Now let us find out x i that is class mark for every interval. Now we know class mark is equal to upper class limit plus lower class limit upon 2. For this interval class mark is 6 plus 0 upon 2 that is 3. Similarly for this interval class mark is equal to 8. For this interval class mark is 12 and it is 17, 24, 33 and 39 for next four intervals. Now we will choose one among x i's as assume mean. Now let it be 3. Now we will find out d i that is deviation of assume mean from class mark. So, it is equal to x i minus a. Now here x i is equal to 3 and a is also equal to 3. So, 3 minus 3 is equal to 0. Here x i is equal to 8 and a is equal to 3. So, 8 minus 3 is equal to 5. Similarly, 12 minus 3 is equal to 9. 17 minus 3 is equal to 14, 24 minus 3 is equal to 21. 33 minus 3 is equal to 30 and 39 minus 3 is equal to 36. Now we will find out the product f i d i. 11 multiplied by 0 is equal to 0. 10 multiplied by 5 is equal to 50. 7 multiplied by 9 is equal to 63. 4 multiplied by 14 is equal to 56. 4 multiplied by 21 is equal to 84. 3 multiplied by 30 is equal to 90. 1 multiplied by 36 is equal to 36. Now we will find out summation f i. Summation f i is equal to sum of all these frequencies. That is it is equal to 40. Similarly, we will find out summation f i d i. Summation f i d i is equal to sum of all these products. This is equal to 379. We know assumed mean that is a is equal to 3 and from key idea we know mean is equal to a plus summation f i d i upon summation f i. Now substituting corresponding values of a summation f i and summation f i d i in this formula we get mean is equal to 3 plus 379 upon 40. Now this implies mean is equal to 3 plus 9.475. We know 379 divided by 40 is equal to 9.475. Now adding these two terms we get mean is equal to 12.475. Now it can be further written as 12.48 rounding of this value up to two places of decimal we get mean is equal to 12.48. So the mean number of days our student was absent is equal to 12.48. Remember that we have used in this solution assumed mean method for finding the mean. Assumed mean method is used when the numerical values of x i and f i are large. So 12.48 days is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.