 Hello and welcome to the session. The given question says the mean of the following distribution as 62.8 and the sum of all the frequencies is 50. Compute the missing frequencies F1 and F2. So this is the following given frequency table. Let's start with the state solution and first let us make a table in which we shall write down the class in the first column frequency in the second column and in the third column we shall write down the midpoints of these class intervals and then we shall write Fy into the midpoints of each of these class intervals. So let us make a table first. So this is the required table in the first column we have written the given class intervals. In the second column we have written down the values of the frequencies. Now in the third column we shall write the midpoint of each of these class intervals. Midpoint of 0 to 20 we get on adding these two and dividing it by 2. So 0 plus 20 is 20 and dividing by 2 we get 10. Midpoint of 20 and 40 similarly adding 20 with 40 and dividing by 2 we get 30 and so on. Midpoint of 40 and 60 is 50. Midpoint of 60 and 80 is 70. Midpoint of 80 and 100 is 90 and midpoint of 120 is 110. Now we shall write here the value of Fy into xi. So in the first row we shall write F1 into X1 that is 5 into 10 gives 50. 30 into F1 is the value F2 X2. So we have 30 F1 then we have 500 then we have 78 too similarly and here we have 630 and here we have 880. Now we are given that the sum of all the frequencies is 60. So summation Fy is already given to us and this is equal to 50 and now let us find summation Fy xi that is we shall add all these 6 values to get summation Fy xi and on adding we get 2060 plus 30 F1 plus 70 F2. So we have to find the missing frequencies F1 and F2. Now we are given that summation of Fy that is sum of all these frequencies is equal to 50. We are given that summation Fy is equal to 50 so this implies 5 plus F1 plus 10 plus F2 plus 7 plus 8 is equal to 50 which further implies that F1 plus F2 plus on adding 5, 10, 7 and 8 we get 30 and this is equal to 50. So this implies F1 plus F2 is equal to 50 minus 30 which is equal to 20. Therefore we have F1 plus F2 equals to 20. Let this be equation number 1. Also we have summation Fy xi is equal to 2060 plus 30 F1 plus 70 F2. Now we have 2060 plus 2060 plus 2060 plus 2060 plus 2060 from equation number 1 we have F2 is equal to 20 minus F1. So let us substitute the value of F2 in summation of Fy xi we get 2060 plus 30 F1 plus 70 into 20 minus F1. So this is equal to 2060 plus 30 F1 plus 1400 minus 70 F1 plus 2060 plus 2060 plus 2060 plus 70 F1 which further implies that on adding 1400 with 2060 we get 3460 and on simplifying 30 F1 and minus 70 F1 we get minus 40 F1. Therefore we have summation Fy xi is equal to 3460 minus 40 F1. Let this be equation number 2. Now we know that mean is equal to summation Fy xi divided by summation Fy. Now we are given that the mean of the given frequency distribution is 62.8 and summation Fy xi is 3460 minus 40 F1 and summation Fy that is the sum of all the frequencies we are given is equal to 50 or we have 62.8 into 50 is equal to 3460 minus 40 F1 which further implies that 40 F1 is equal to 3460 minus on multiplying 62.8 with 50 we get 3140 which further implies that 40 F1 is equal to 3460 minus on multiplying F1 is equal to 320 or F1 is equal to 320 divided by 40 which gives 8. So this implies that F1 is equal to 8. Now we know that F1 plus F2 is equal to 20. So from equation 1 we have F2 is equal to 20 minus F1 that is 20 minus 8 which is equal to 12. Therefore F2 is equal to 12. Hence our answer is the missing frequencies F1 and F2 are 8 and 12. So this completes the session. Bye and take care.