 Hello and welcome to the session. I am Deepika and I am going to help you to solve the following question. The question says the mean of the following frequency distribution is 62.8 find the missing frequency x for the class 0 to 20 Frequency is 5 for the class 20 to 40 Frequency is 8 and for the class 40 to 60 It is x for 60 to 80. It is 10 for 80 to 100 It is 7 and for 100 to 120. It is 8 Now we know that by assumed mean method the mean of the group data is given by that is x bar is equal to a plus sigma fi ti over sigma fi Where a is the assumed mean and di is the difference between a and each of the x i's that is it is a deviation of a from each of the x i's and x i is the midpoint or class mark of the i th class and class mark is given by upper class limit plus lower class limit over 2 So this is a key idea behind our question We will take the help of this key idea to solve the above question So let's start the solution Now the mean of the following frequency distribution is given to us which is 62.8 and We have to find the missing frequency x Now here we will find the mean by the assumed mean method So for this let us first find out the class mark of each class Now we know that The class mark is given by upper class limit plus lower class limit upon 2 So the class mark of the class 0 to 20 is 0 plus 20 over 2 which is 10 similarly the class mark of the class 20 to 40 is 20 plus 40 over 2 which is 30 Similarly for the class 40 to 60 it is 50 For the class 60 to 80 it is 70 and for 80 to 100 It is 90 and for the class 100 to 120 it is 110 Now our next step is to choose one among the x i's as assumed mean So let us choose a is equal to 50 That is let us choose the assumed mean is equal to 50 Now we will find the difference Di between a and each of the x i's that is the deviation of a from each of the x i's So di is equal to x i Minus a now we have chosen a is equal to 50 So di is equal to x i minus 50 Now for the class 0 to 20 D1 is equal to x1 minus 50 that is 10 minus 50 which is equal to minus 40 Now for the class 20 to 40 D2 is equal to 30 minus 50 Which is minus 20 now for the class 40 to 60 D3 is equal to 0 and for the class 60 to 80 D4 is equal to 70 minus 50 which is 20 and For the fifth class D5 is equal to 90 minus 50 which is 40 and For the last class D6 is equal to 110 minus 50 which is equal to 60 Now we will find the product of di with the corresponding fi that is we will find out fi ti So for the class 0 to 20 F1 T1 is equal to minus 200 Now for the second class F2 D2 is equal to minus 116 for the third class it is 0 for the fourth class It is 240 for the fifth class. It is 280 and for the sixth class It is 480 Now we will find The sum of all the fi di's that is we will find out sigma fi di and this is equal to 640 now sigma fi is Equal to x plus 40 so we have Assumed being a is equal to 50 Sigma fi is Equal to x plus 40 and Sigma fi di is Equal to 640 So according to our key idea x bar is equal to a plus Sigma fi di over sigma fi Now in the question mean is given to us which is 62.8 So 62.8 is equal to a Which is 50 plus sigma fi di which is 640 Over sigma fi which is x plus 40 Now we will solve this equation so 62.8 is Equal to 50 Into x plus 40 Plus 640 over x plus 40 Or this can be written as 62.8 into x plus 40 is equal to 50 x plus 2000 plus 640 or This can be written as 62.8 x Plus 2500 12 is equal to 50 x plus 2640 or this can be written as 62.8 x minus 50 x is equal to 2640 minus 2500 12 or This can be written as 12.8 x is equal to 128 or x is equal to 128 over 12.8 Which is equal to 10 So the missing frequency x is equal to 10 Hence the answer for the above question is 10 So this completes our session. I hope the solution is clear to you. Bye and have a nice day