This formula is the result of a linear interpolation, which identifies the median under the assumption that data are uniformly distributed within the median class.

To derive the formula, we can note that since $N/2$ is the number of observations below the median, then $N/2 - F_{m-1}$ is the number of observations that are within the median class and that are below the median ($F_{m-1}$ is the cumulative frequency of the interval below the median class, i.e. of all classes below the median class).

As a result, the fraction $\displaystyle\frac {N/2 - F_{m-1}}{f_m}$ (where $f_m$ is the frequency of the median class) represents the proportion of data values in the median class that are below the median.

Now if we assume that data are uniformly distributed (i.e., equally spaced) within the median class, multiplying the last fraction by $c$ (total width of the median class) we obtain the fraction of median class width corresponding to the position of the median. Adding the result to $L_m$ (lower limit of the median class), we get the final formula $\displaystyle L_m + \left [ \frac { \frac{N}{2} - F_{m-1} }{f_m} \right ] \cdot c$, which identifies the median.

Because this is essentially a duplicate, I address a few issues that are do not explicitly overlap the related question or answer:

If a class has cumulative frequency .5, then the median is at the boundary of that class and the next larger one.

If $N$ is large (really the only case where this method is generally successful), there is little difference between $N/2$ and $(N+1)/2$ in the formula. All references I checked use $N/2$.

Before computers were widely available, large datasets were customarily reduced to categories (classes) and plotted as histograms. Then the histograms were used to approximate the mean, variance, median, and other descriptive measures. Nowadays, it is best just to use a statistical computer package to find exact values of all measures.

One remaining application is to try to re-claim the descriptive measures from grouped data or from a histogram published in a journal. These are cases in which the original data are no longer available.

This procedure to approximate the sample median from grouped data $assumes$ that data are distributed in roughly a uniform fashion throughout the median interval. Then it uses interpolation to approximate the median. (By contrast, methods to approximate the sample mean and sample variance from grouped data one assumes that all obseervations are concentrated at their class midpoints.)

Hello NAFISA, I see that you are trying to calculate the median of grouped data using MATLAB's median function. However, the 'median' function in MATLAB is designed for raw data inputs and does not directly compute the median for grouped data. you can expand your grouped data as follows: class_intervals = [0 5; 5 10; 10 15; 15 20; 20 25; 25 30; 30 35; 35 40; 40 45; 45 50]; frequencies = [14, 8, 20, 7, 11, 10, 5, 16, 21, 9]; midpoints = (class_intervals(:, 1) + class_intervals(:, 2)) / 2; expanded_data = []; for i = 1:length(frequencies) expanded_data = [expanded_data, repmat(midpoints(i), 1, frequencies(i))]; end median_value = median(expanded_data); disp(['The median is: ', num2str(median_value)]); Using this method, you will find that the output is 27.5 instead of the expected 25.25. This discrepancy occurs because: When you expand grouped data into individual data points, you assume that all data points within a class interval are located at the midpoint of that interval. For example, if a class interval is [20, 25] with a frequency of 11, you assume there are 11 data points all exactly at 22.5. This assumption can lead to inaccuracies because the actual data points could be spread across the entire interval [20, 25]. So, as Muskan mentioned , you can use the grouped data median formula (L + ((N/2 - cf) / f) * h). If you frequently work with frequency distribution tables and find it cumbersome to use this formula manually, you can use MATLAB functions to simplify the process. Here is an example: class_intervals1 = [0 5; 5 10; 10 15; 15 20; 20 25; 25 30; 30 35; 35 40; 40 45; 45 50]; frequencies1 = [14, 8, 20, 7, 11, 10, 5, 16, 21, 9]; class_intervals2 = [420 430; 430 440; 440 450; 450 460; 460 470; 470 480; 480 490; 490 500]; frequencies2 = [336, 2112, 2336, 1074, 1553, 1336, 736, 85]; % Calculate the median using the custom function median_value1 = groupedMedian(class_intervals1, frequencies1); median_value2 = groupedMedian(class_intervals2, frequencies2); disp(['The median of the grouped data1 is: ', num2str(median_value1)]); disp(['The median of the grouped data2 is: ', num2str(median_value2)]); function median_value = groupedMedian(class_intervals, frequencies) % Calculate cumulative frequency cum_frequencies = cumsum(frequencies); % Total number of observations N = sum(frequencies); % Find the median class (first class where cumulative frequency >= N/2) median_class_index = find(cum_frequencies >= N/2, 1); % Extract the median class boundaries and frequency L = class_intervals(median_class_index, 1); f = frequencies(median_class_index); CF = cum_frequencies(median_class_index - 1); if isempty(CF) CF = 0; end h = class_intervals(median_class_index, 2) - L; % Calculate the median median_value = L + ((N/2 - CF) / f) * h; end You can also try referring to these file exchange functions which might help you https://www.mathworks.com/matlabcentral/fileexchange/38238-gmedian https://www.mathworks.com/matlabcentral/fileexchange/38228-gprctile I hope this helps you moving forward