 Hello and welcome to the session. In this session we discussed the following question that says, find the missing frequencies f1 and f2 in the falling frequency distribution if it is known that the mean of the distribution is 1.46. So this is the frequency distribution given to us in which we are given the observations x and the corresponding frequencies by f and we are also given that the total of the frequencies is 200. Suppose we are given that x1, x2 and so on up to xn are the observations with respective frequencies f1, f2 and so on up to fn then we have the mean is given by x bar is equal to summation f i x i where i varies from 1 to n this upon summation f i where again i varies from 1 to n. So this is the key idea to be used in this question. Now let's move on to its solution. We prepare this table in which we have the observations x i and the corresponding frequencies f i. Now we need to find f1 and f2. First let's find out x i f i. Now 0 multiplied by 46 is 0 then 1 multiplied by f1 is f1, 2 multiplied by f2 is 2f2, 3 multiplied by 25 is 75, 4 multiplied by 10 is 45 multiplied by 5 is 25. Now summation f i is equal to the sum of all these frequencies. So this would be equal to 86 plus f1 plus f2 then summation x i f i is equal to sum of all these terms which is equal to 140 plus f1 plus 2f2. In the given frequency distribution we were given that the sum total of all the frequencies is 200 that is we have summation f i is equal to 200 as given to us. So this would mean 86 plus f1 plus f2 is equal to 200 which further gives us f1 plus f2 is equal to 200 minus 86 and this is equal to 114. Thus we have f1 plus f2 is equal to 114. We take this as equation 1. Now we have found out summation x i f i is equal to 140 plus f1 plus 2f2. In the question we were given that for the given frequency distribution the mean is 1.46 that is we have x bar is equal to 1.46 we know that x bar that is the mean is equal to summation x i f i upon summation f i. So this means we have x bar is equal to 140 plus 2f2 plus f1 upon 200 or you can say 1.46 is equal to 140 plus 2f2 plus f1 upon 200 this gives us 1.46 into 200 is equal to 140 plus 2f2 plus f1 or you can say we have 292 is equal to 140 plus 2f2 plus f1 this gives us is equal to 292 minus 140. So further we have 2f2 plus f1 is equal to 152 or you can also say that f1 plus 2f2 is equal to 152. So we have got another equation in f1 and f2 let this be equation 2. Now the two equations that we have got are f1 plus f2 is equal to 114 this was equation 1 then f1 plus 2f2 is equal to 152 this was equation 2. Now we will solve both these equations for f1 and f2. So subtracting equation 1 from equation 2 we get f1 plus 2f2 minus f1 plus f2 is equal to 152 minus 114 or you can say we get f2 is equal to 38. Now we have got the value for f2 now substituting f2 equal to 38 in equation 1 we get f1 plus 38 is equal to 114 or you can say we get f1 is equal to 114 minus 38 which gives us f1 equal to 76. So we have got the values for f1 also. So we have found out the missing frequencies f1 and f2 where we have f1 is equal to 76 and f2 is equal to 38. So this is our final answer this completes the session hope you have understood the solution of this question.