Consider grouped data with frequencies and more specifically data that are discrete.
If I have for example different xi with the same frequencies, what is the right answer: there is no mode or all the modalities are modes?
Example value/ frequency , 0 / 42, 1 / 42, 2 / 42, 3 /42, 4 / 42, 5 / 42, 6 / 42, so what is the right answer no mode or multi modes??
For the median, some literature computes the median as the average of both xi between $F<0.5$ and $F>0.5$ OR gives the interval $[a,b]$. Others say that the median is the first xi having $F \ge 0.5$ because in the discrete case, the xi is an integer, so maybe the average did not give an integer. Also the median is the middle value so why do we have to create a fictitious value (average)? to develop this point a)To determine the median of a discrete statistical series: It stores values of the variable in ascending order. When the total number N is odd, the median is the value of the series of row (N+1/2) When the total number N is even, the median is the half sum of the ranks (N/2) and (N/2 +1)
it is clear point
b)To determine the statistical median of a discrete series where each value x n is assigned frequencies, We can calculate the cumulative increasing numbers to exceed or equal to the half of the total N. (i want to verify this last point)
From the previous example
The average in the case even number and serie of data, it is clear that we compute the average of the both middle values, in the case of grouped data value/ frequency , 0 / 42, 1 / 42, 2 / 42, 3 /42, 4 / 42, 5 / 42, 6 / 42, the median is the average of the both modalities (2 and3) or the first modality having (Ni>=N/2(=126)) in the example it will be 'x= me=2 or 3 '(42+42+42=126) also let's suppose that We get an no integer value (1.5 for example, it is logical to say we have mode 1.5 children??) that is why i prefer this definition "To determine the statistical median of a discrete series where each value x n is assigned frequencies, We can calculate the cumulative increasing numbers to exceed half of the total N. "