Tenth Grade Mathematics: Median of Grouped Data and Median Class
When analyzing skewed economic data—such as national household incomes or house prices—the arithmetic mean can be heavily distorted by a handful of extreme multi-millionaire outliers. In such contexts, statisticians turn to the Median. The median represents the exact physical center of an ordered distribution, partitioning the population into two equal halves such that 50 percent of the observations lie below it and 50 percent lie above it. In grouped frequency distributions, calculating the median requires tracking cumulative totals and interpolating within the median class.
The Concept of Cumulative Frequency
To locate the middle observation of a distribution, we must arrange frequencies sequentially into cumulative frequencies (cf):
Formal Definition:
The Cumulative Frequency of a class interval is the running sum of all
frequencies up to and including that class interval.
Cumulative Frequency Table Construction:
Class Interval Frequency (fi) Cumulative Frequency (cf)
10 - 20 5 5
20 - 30 8 5 + 8 = 13
30 - 40 12 13 + 12 = 25
40 - 50 15 25 + 15 = 40
50 - 60 6 40 + 6 = 46
Total Frequency N = Sum(fi) = 46. The last cumulative frequency MUST equal N!
Locating the Median Class
Let the total number of observations be N = Sum(fi). The middle position of the dataset is given by N / 2.
Protocol to Identify the Median Class:
1. Calculate N / 2.
2. Scan down the cumulative frequency (cf) column until you encounter the FIRST
class interval whose cumulative frequency is GREATER THAN OR EQUAL TO N / 2.
3. That interval is designated as the Median Class!
Derivation of the Grouped Median Formula
The median class contains the (N/2)-th observation. Within this interval, we assume that the frequency fi is distributed uniformly across the class interval of width h:
Interpolation Principle:
- Observations accumulated BEFORE the median class = cf
- Observations needed inside the median class to reach the center = (N / 2) - cf
- Total observations inside the median class = f
- Fraction of the interval needed = [ (N / 2) - cf ] / f
- Distance along the class interval = [ ( (N / 2) - cf ) / f ] * h
Adding this distance to the lower limit l of the median class yields the universal Median Formula:
Theorem (Median Formula for Grouped Data):
(N / 2) - cf
Median = l + [ ---------------- ] * h
f
Component Summary:
- l = Lower limit of the median class
- N = Total number of observations = Sum(fi)
- cf = Cumulative frequency of the class PRECEDING the median class
- f = Frequency of the median class itself
- h = Class size (width) of the median class
CRITICAL PITFALL: Notice that cf is the cumulative frequency of the PRECEDING class, while f is the simple frequency of the MEDIAN class itself!
Detailed Worked Example
The distribution below gives the weights of 30 students of a class. Find the median weight of the students:
Weight (in kg): 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75
Number of Students: 2 | 3 | 8 | 6 | 6 | 3 | 2
Step 1: Construct cumulative frequency table:
Class fi cf
40-45 2 2
45-50 3 5
50-55 8 13
55-60 6 19 <--- Median Class!
60-65 6 25
65-70 3 28
70-75 2 30
Step 2: Total N = 30. Calculate N / 2 = 30 / 2 = 15.
Step 3: The first class with cf >= 15 is 55-60 (cf = 19).
Therefore, Median Class is 55-60.
Step 4: Identify formula components:
Lower limit l = 55
Class size h = 60 - 55 = 5
N / 2 = 15
cf (preceding class 50-55) = 13
f (frequency of median class 55-60) = 6
Step 5: Apply formula:
Median = l + [ ((N/2) - cf) / f ] * h
Median = 55 + [ (15 - 13) / 6 ] * 5
Median = 55 + [ 2 / 6 ] * 5 = 55 + (1 / 3) * 5
Median = 55 + 5 / 3 = 55 + 1.667 = 56.67 kg.
The median weight of the students is 56.67 kg.
Chapter Exercises
Exercise 1: A life insurance agent found the following data for distribution of ages of 100 policy holders: Ages 20-25 (6 holders), 25-30 (18), 30-35 (21), 35-40 (27), 40-45 (15), 45-50 (8), 50-55 (5). Calculate the median age.
Exercise 2: The median of the following distribution of 60 observations is 28.5. Find the values of the two missing frequencies x and y: Class 0-10 (frequency 5), 10-20 (frequency x), 20-30 (frequency 20), 30-40 (frequency 15), 40-50 (frequency y), 50-60 (frequency 5).
Exercise 3: Explain how the median acts as a robust measure of central tendency compared to the mean when a dataset contains severe extreme values or open-ended class intervals.
Solutions and Explanations
Solution 1: We set up the cumulative frequency table: 20-25 (fi = 6, cf = 6); 25-30 (fi = 18, cf = 24); 30-35 (fi = 21, cf = 45); 35-40 (fi = 27, cf = 72); 40-45 (fi = 15, cf = 87); 45-50 (fi = 8, cf = 95); 50-55 (fi = 5, cf = 100). Total N = 100, so N/2 = 50. The first class with cf >= 50 is 35-40 (cf = 72). Thus, the median class is 35-40. Parameters: lower limit l = 35, class size h = 5, cf of preceding class = 45, frequency f = 27. Median = l + [((N/2) - cf) / f] * h = 35 + [(50 - 45) / 27] * 5 = 35 + [5 / 27] * 5 = 35 + 25 / 27 = 35 + 0.926 = 35.93 years. The median age is approximately 35.93 years.
Solution 2: Construct the cumulative frequency table: 0-10 (fi = 5, cf = 5); 10-20 (fi = x, cf = 5 + x); 20-30 (fi = 20, cf = 25 + x); 30-40 (fi = 15, cf = 40 + x); 40-50 (fi = y, cf = 40 + x + y); 50-60 (fi = 5, cf = 45 + x + y). Total frequency N = 60. Therefore: 45 + x + y = 60 ==> x + y = 15 (Equation 1). We are given Median = 28.5. Since 28.5 lies in the interval 20-30, the median class is 20-30. Parameters for class 20-30: lower limit l = 20, class size h = 10, frequency f = 20, and cf of preceding class = 5 + x. N/2 = 60/2 = 30. Apply median formula: Median = l + [((N/2) - cf) / f] * h ==> 28.5 = 20 + [(30 - (5 + x)) / 20] * 10 ==> 28.5 - 20 = [(25 - x) / 20] * 10 ==> 8.5 = (25 - x) / 2 ==> 17 = 25 - x ==> x = 25 - 17 = 8. Now find y from Equation 1: 8 + y = 15 ==> y = 15 - 8 = 7. Thus, the missing frequencies are x = 8 and y = 7.
Solution 3: The median is determined solely by the rank order of observations and the central observation at position N/2. If extreme outliers are added to either end of a dataset (such as an income of 50 million dollars), the physical values of those extreme observations do not enter the median computation; they merely count as one observation on the upper side, leaving the middle value almost completely unaffected. In contrast, the arithmetic mean incorporates the exact numeric value of every observation into its sum, causing the mean to be pulled significantly toward the outlier. Furthermore, in distributions with open-ended intervals (such as "Below 20" or "Above 80"), the mean cannot be computed because the class marks of the open ends are unknown. The median, however, can be calculated without difficulty provided the median class does not fall in an open-ended interval.