 Hello and welcome to the session. In this session we discussed the following question with phase. The mode of the following series is 35, find the missing frequency F in it. So we are given this series in which we have the class intervals and the corresponding frequencies. We have to find this missing frequency F. First let's recall the formula to find out the mode. Mode is equal to L class F1 minus F0 this whole upon 2 F1 minus F0 minus F2 and this whole multiplied by H. Now here this L is the lower limit of the modal class. H is the class size and we assume the class size to be equal. F1 is the frequency of the modal class. F0 is the frequency of the class preceding the modal class. F2 is the frequency of the class succeeding the modal class. This is the key idea that we use for this question. Now let's see the solution of this question. In the question we have that the mode is 35. First let's find out the modal class. Now consider the series in this the maximum frequency is 6 and it lies in the class interval 30 to 40. That is we have maximum frequency 6 lies in the class interval 30 to 40. Therefore the class interval 30 to 40 is the modal class. So we have got the modal class. Now what would be L in this case? We know that L is the lower limit of the modal class and modal class is 30 to 40. It's lower limit is 30 so L is equal to 30. Then we have H that is the class size which is 10. Then F1 which is the frequency of the modal class and from the table we have that the frequency of the modal class is 6. So F1 is equal to 6. Then F0 it is the frequency of the class proceeding the modal class. Now from the table modal class is 30 to 40. The class proceeding the modal class is 20 to 30 and its frequency is given by F which we have to find out. So we have F0 is equal to F. Then we have F2 which is the frequency of the class succeeding the modal class. Now the class succeeding the modal class is 40 to 50 and its frequency is 5. So F2 is equal to 5. Now let's write the formula for the mode. Mode is equal to L plus F1 minus F0 this whole upon 2 F1 minus F0 minus F2 and this whole multiplied by H. Now let's substitute the values for mode L F1 F0 F2 and H. So we get 35 is equal to 30 plus 6 minus F this upon 2 into 6 minus F minus 5 and this whole multiplied by 10. Further we get 35 is equal to 30 plus 6 minus F this whole upon 12 minus F minus 5 and this multiplied by 10. Next we have 35 minus 30 is equal to 6 minus F upon 7 minus F and this whole multiplied by 10. So we get 5 upon 10 is equal to 6 minus F upon 7 minus F. Now 5 2 times is 10 so we have 7 minus F is equal to 2 into 6 minus F. Further we have 7 minus F is equal to 12 minus 2F so we get 12 minus F is equal to 12 minus 7 or you can say we have F is equal to 5. And we were supposed to find this missing frequency F this comes out to be equal to 5 so this is our final answer. So this completes the session hope you have understood the solution of this question.