 Hello and welcome to the session. In this session we discuss the following question which says if the median of the following frequency distribution is 46, find the missing frequencies f1 and f2. We are given this frequency distribution in which we are given the class intervals and the corresponding frequencies. We are also given that the total of all the frequencies is 229. We are supposed to find the frequencies f1 and f2. Let's first recall the formula to find out the median. This is equal to l plus n upon 2 minus cf this whole upon f and this whole multiplied by h where we have l is the lower limit of the median class n is the total frequency cf is the cumulative frequency of the class preceding the median class when f is the frequency of the median class and we have h is the class size where we assume the class size to be equal. This is the key idea that we use for this question. Now let's proceed with the solution. It's given that the sum total of all the frequencies is 229 so we will add up all these frequencies. So we have 12 plus 30 plus f1 plus 65 plus f2 plus 25 plus 18 is equal to 229. This gives us f1 plus f2 plus 150 is equal to 229. Further we get f1 plus f2 is equal to 229 minus 150 which is equal to 79. That is we get f1 plus f2 is equal to 79 and we take this as equation 1. Now we are already given that median is equal to 46. Now we make the cumulative frequency distribution in which we have the class intervals. The corresponding frequency is given by f and the corresponding cumulative frequency is given by cf which we have to find out. For the class interval 10 to 20 the frequency is 12. Its cumulative frequency is also 12. Now the cumulative frequency for the class interval 20 to 30 is given by adding its frequency and the cumulative frequency of the previous class. So it becomes 42. In the same way we get the cumulative frequency of the class interval 30 to 40 by adding this f1 to 42. So we have 42 plus f1. Now cumulative frequency for the class interval 40 to 50 is given by adding 65 to 42 plus f1. So that becomes 107 plus f1. Cumulative frequency for the class interval 50 to 60 is given by 107 plus f1 plus f2. Then the cumulative frequency for the class interval 60 to 70 is given by 132 plus f1 plus f2. Then the cumulative frequency for the class interval 70 to 80 is given by 150 plus f1 plus f2. Now we have median is equal to 46 and this lies in the class interval 42 50. Therefore we say that the class interval 40 to 50 is the median class. Now we have the formula for median is given by l plus l upon 2 minus cf this whole upon f and this whole multiplied by h. Now let's find out the values for l, n, cf, f and h. Now l is the lower limit of the median class. So l is equal to 40 since the class is 40 to 50 and slower limit is 40. So l is equal to 40. Then n is the total frequency and which is equal to 229 as already given to us. Then cf is the cumulative frequency of the class preceding the median class. Now the class preceding the median class is 30 to 40 and its cumulative frequency is 42 plus f1. So we have cf is equal to 42 plus f1. f is the frequency of the median class. Median classes 40 to 50 is frequency 65. So we have f is equal to 65. Next we have h which is the class size and this is equal to 10. Now we substitute the values for median l, n, cf, f and h. So we get 46 is equal to 40 plus 229 upon 2 minus 42 plus f1 this whole upon 65 and this whole multiplied by 10. Further we get 46 minus 40 is equal to 229 minus 84 minus 2f1 this whole upon 130 and this whole multiplied by 10. This 0 and this 0 cancel with each other. So next we get 6 is equal to 145 minus 2f1 and this whole upon 13. So further we have 145 minus 2f1 is equal to 13 into 6. So we get 145 minus 2f1 is equal to 78 this gives us 2f1 is equal to 145 minus 78. Further we have f1 is equal to 67 upon 2 that is equal to 33.5. Therefore we have f1 is equal to 33.5 or say 34. So we have got the value for f1 as 34. Now from equation one we get f1 plus f2 is equal to 79. Putting the value for f1 as 34 we get 34 plus f2 is equal to 79. So from here we get f2 is equal to 79 minus 34. So which is equal to 45. Therefore we get f2 equal to 45. So we have found out the missing frequencies f1 and f2 f1 is equal to 34 and f2 is equal to 45. So with this we complete the session hope you have understood the solution of this question.