Since you already know the formula, it should be easy enough to create a function to do the calculation for you.

Here, I've created a basic function to get you started. The function takes four arguments:

• frequencies: A vector of frequencies ("number" in your first example)
• intervals: A 2-row matrix with the same number of columns as the length of frequencies, with the first row being the lower class boundary, and the second row being the upper class boundary. Alternatively, "intervals" may be a column in your data.frame, and you may specify sep (and possibly, trim) to have the function automatically create the required matrix for you.
• sep: The separator character in your "intervals" column in your data.frame.
• trim: A regular expression of characters that need to be removed before trying to coerce to a numeric matrix. One pattern is built into the function: trim = "cut". This sets the regular expression pattern to remove (, ), [, and ] from the input.

Here's the function (with comments showing how I used your instructions to put it together):

GroupedMedian <- function(frequencies, intervals, sep = NULL, trim = NULL) {
  # If "sep" is specified, the function will try to create the 
  #   required "intervals" matrix. "trim" removes any unwanted 
  #   characters before attempting to convert the ranges to numeric.
  if (!is.null(sep)) {
    if (is.null(trim)) pattern <- ""
    else if (trim == "cut") pattern <- "\\[|\\]|\\(|\\)"
    else pattern <- trim
    intervals <- sapply(strsplit(gsub(pattern, "", intervals), sep), as.numeric)
  }

  Midpoints <- rowMeans(intervals)
  cf <- cumsum(frequencies)
  Midrow <- findInterval(max(cf)/2, cf) + 1
  L <- intervals[1, Midrow]      # lower class boundary of median class
  h <- diff(intervals[, Midrow]) # size of median class
  f <- frequencies[Midrow]       # frequency of median class
  cf2 <- cf[Midrow - 1]          # cumulative frequency class before median class
  n_2 <- max(cf)/2               # total observations divided by 2

  unname(L + (n_2 - cf2)/f * h)
}

---

Here's a sample data.frame to work with:

mydf <- structure(list(salary = c("1500-1600", "1600-1700", "1700-1800", 
    "1800-1900", "1900-2000", "2000-2100", "2100-2200", "2200-2300", 
    "2300-2400", "2400-2500"), number = c(110L, 180L, 320L, 460L, 
    850L, 250L, 130L, 70L, 20L, 10L)), .Names = c("salary", "number"), 
    class = "data.frame", row.names = c(NA, -10L))
mydf
#       salary number
# 1  1500-1600    110
# 2  1600-1700    180
# 3  1700-1800    320
# 4  1800-1900    460
# 5  1900-2000    850
# 6  2000-2100    250
# 7  2100-2200    130
# 8  2200-2300     70
# 9  2300-2400     20
# 10 2400-2500     10

---

Now, we can simply do:

GroupedMedian(mydf$number, mydf$salary, sep = "-")
# [1] 1915.294

---

Here's an example of the function in action on some made up data:

set.seed(1)
x <- sample(100, 100, replace = TRUE)
y <- data.frame(table(cut(x, 10)))
y
#           Var1 Freq
# 1   (1.9,11.7]    8
# 2  (11.7,21.5]    8
# 3  (21.5,31.4]    8
# 4  (31.4,41.2]   15
# 5    (41.2,51]   13
# 6    (51,60.8]    5
# 7  (60.8,70.6]   11
# 8  (70.6,80.5]   15
# 9  (80.5,90.3]   11
# 10  (90.3,100]    6

### Here's GroupedMedian's output on the grouped data.frame...
GroupedMedian(y$Freq, y$Var1, sep = ",", trim = "cut")
# [1] 49.49231

### ... and the output of median on the original vector
median(x)
# [1] 49.5

---

By the way, with the sample data that you provided, where I think there was a mistake in one of your ranges (all were separated by dashes except one, which was separated by a comma), since strsplit uses a regular expression by default to split on, you can use the function like this:

x<-c(110,180,320,460,850,250,130,70,20,10)
colnames<-c("numbers")
rownames<-c("[1500-1600]","(1600-1700]","(1700-1800]","(1800-1900]",
            "(1900-2000]"," (2000,2100]","(2100-2200]","(2200-2300]",
            "(2300-2400]","(2400-2500]")
y<-matrix(x,nrow=length(x),dimnames=list(rownames,colnames))
GroupedMedian(y[, "numbers"], rownames(y), sep="-|,", trim="cut")
# [1] 1915.294

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The median class is the class that actually contains the median. Cumulative frequency of a class tells you the total frequency of everything inside that class and all classes below it. Because of this, the cumulative frequency tells you the upper bound on the range of percentile of the class. You might find the answer to your question much clearer if you figure out the percentile range of each class, because the percentile range has 2 numbers so it tells you a more completely story. The cumulative frequency tell you only the upper bound of this range, so it can skew your intuition. If the cumulative frequency is =N/2, then that means if you start counting from the bottom to the (N/2)-th item, you will reach N/2 before you run out of items in that class and all classes below it. So how do you find the class that contains the median? Obviously, the cumulative frequency must be at least >=N/2 (so that the median is in this class or below), but the cumulative frequency of any classes below it must be

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

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 is large (really the only case where this method is generally successful), there is little difference between and in the formula. All references I checked use .

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.)