Median of Grouped Data
The median of grouped data is the value that divides the data into two equal halves -- 50% of observations lie below it and 50% above it. This topic is covered in CBSE Class 10 Mathematics, Chapter 14 (Statistics).
Unlike the mean, the median is not affected by extreme values (outliers), making it a more robust measure of central tendency for skewed distributions.
For grouped data, the median cannot be found by simple inspection. It is calculated using a formula that involves identifying the median class -- the class interval in which the median lies -- and then interpolating within that class.
What is Median of Grouped Data - Formula, Median Class & Solved Examples?
Definition: The median of a data set is the middle value when all observations are arranged in ascending order. For grouped data, it is the value below which 50% of the total observations fall.
Key terms:
- Cumulative frequency (cf) -- the running total of frequencies up to and including each class.
- n -- total number of observations = Sigma(f_i).
- n/2 -- half the total frequency (the position of the median).
- Median class -- the class interval whose cumulative frequency is the first to exceed n/2.
- l -- lower limit of the median class.
- f -- frequency of the median class.
- cf -- cumulative frequency of the class just BEFORE the median class.
- h -- class size (width) of the median class.
Median of Grouped Data Formula
Median of Grouped Data:
Median = l + [(n/2 - cf) / f] x h
Where:
- l = lower limit of the median class
- n = total number of observations (Sigma f_i)
- cf = cumulative frequency of the class BEFORE the median class
- f = frequency of the median class
- h = class size (upper limit - lower limit of the median class)
Steps to find the median class:
- Calculate n/2.
- Build the cumulative frequency column.
- The median class is the first class whose cumulative frequency is greater than or equal to n/2.
Derivation and Proof
Derivation of the Median Formula:
- The median is the (n/2)-th observation.
- From the cumulative frequency table, the median lies in the class whose cumulative frequency first reaches or exceeds n/2.
- Let this be the class [l, l + h) with frequency f.
- The cumulative frequency up to the class before the median class is cf.
- So (n/2 - cf) observations within the median class need to be counted to reach the median.
- Assuming observations are uniformly distributed within each class:
- The fraction of the class to be covered = (n/2 - cf)/f.
- The median is located at a distance of [(n/2 - cf)/f] x h from the lower limit l.
- Median = l + [(n/2 - cf)/f] x h. Hence derived.
Assumption: The formula assumes that observations within each class interval are uniformly (evenly) distributed. This is a reasonable approximation for most practical data sets.
Types and Properties
Problems on median of grouped data include:
Type 1: Direct Calculation
- Given a frequency distribution, find the median using the formula.
Type 2: Finding Missing Frequency
- Given the median and all but one frequency, find the missing frequency.
Type 3: Median from Cumulative Frequency Table
- Data is given in "less than" or "more than" cumulative form. Convert to class intervals and find the median.
Type 4: Ogive-Based Median
- Draw the ogive (cumulative frequency curve), mark n/2 on the y-axis, draw a horizontal line to the curve, and drop a perpendicular to the x-axis to read the median.
Type 5: Comparing Mean, Median, and Mode
- Calculate all three measures for the same data set and compare.
Methods
Step-by-step method to find median of grouped data:
- Construct the cumulative frequency table.
- Find n/2 where n = Sigma(f_i).
- Identify the median class -- the class whose cumulative frequency first exceeds n/2.
- Note the values:
- l = lower limit of median class
- f = frequency of median class
- cf = cumulative frequency of the class just before the median class
- h = class size
- Apply the formula: Median = l + [(n/2 - cf)/f] x h.
- Simplify to get the answer.
Finding median from an Ogive:
- Plot the ogive (cumulative frequency vs. upper class boundary).
- Mark n/2 on the y-axis.
- Draw a horizontal line from n/2 to the ogive curve.
- From the point of intersection, drop a perpendicular to the x-axis.
- The x-coordinate of this point is the median.
Common Mistakes:
- Using the wrong cf -- it must be the cumulative frequency of the class BEFORE the median class, not the median class itself.
- Confusing n/2 with (n+1)/2. For grouped data, always use n/2.
- Taking the upper limit instead of the lower limit for l.
Solved Examples
Example 1: Finding Median of Grouped Data
Problem: Find the median of the following distribution:
| Class | Frequency |
|---|---|
| 0-10 | 5 |
| 10-20 | 8 |
| 20-30 | 12 |
| 30-40 | 15 |
| 40-50 | 10 |
Solution:
| Class | f_i | cf |
|---|---|---|
| 0-10 | 5 | 5 |
| 10-20 | 8 | 13 |
| 20-30 | 12 | 25 |
| 30-40 | 15 | 40 |
| 40-50 | 10 | 50 |
n = 50, n/2 = 25.
Median class: cf first reaching 25 is 25 (class 20-30).
- l = 20, f = 12, cf = 13 (class before), h = 10
Median = 20 + [(25 - 13)/12] x 10 = 20 + (12/12) x 10 = 20 + 10 = 30.
Answer: The median is 30.
Example 2: Median of Student Marks
Problem: Find the median marks from the following data:
| Marks | Students |
|---|---|
| 0-20 | 6 |
| 20-40 | 8 |
| 40-60 | 14 |
| 60-80 | 10 |
| 80-100 | 2 |
Solution:
| Marks | f_i | cf |
|---|---|---|
| 0-20 | 6 | 6 |
| 20-40 | 8 | 14 |
| 40-60 | 14 | 28 |
| 60-80 | 10 | 38 |
| 80-100 | 2 | 40 |
n = 40, n/2 = 20.
Median class: cf first exceeding 20 is 28 (class 40-60).
- l = 40, f = 14, cf = 14, h = 20
Median = 40 + [(20 - 14)/14] x 20 = 40 + (6/14) x 20 = 40 + 8.57 = 48.57.
Answer: The median marks are 48.57.
Example 3: Median of Daily Earnings
Problem: Find the median from the data:
| Earnings (Rs.) | Workers |
|---|---|
| 100-150 | 12 |
| 150-200 | 20 |
| 200-250 | 25 |
| 250-300 | 18 |
| 300-350 | 5 |
Solution:
| Earnings | f_i | cf |
|---|---|---|
| 100-150 | 12 | 12 |
| 150-200 | 20 | 32 |
| 200-250 | 25 | 57 |
| 250-300 | 18 | 75 |
| 300-350 | 5 | 80 |
n = 80, n/2 = 40.
Median class: cf first exceeding 40 is 57 (class 200-250).
- l = 200, f = 25, cf = 32, h = 50
Median = 200 + [(40 - 32)/25] x 50 = 200 + (8/25) x 50 = 200 + 16 = 216.
Answer: The median earning is Rs. 216.
Example 4: Median from Less-Than Cumulative Frequency
Problem: Find the median from the following "less than" distribution:
| Less than | Cumulative Frequency |
|---|---|
| 30 | 10 |
| 40 | 22 |
| 50 | 40 |
| 60 | 55 |
| 70 | 60 |
Solution:
Convert to class intervals:
| Class | f_i | cf |
|---|---|---|
| 20-30 | 10 | 10 |
| 30-40 | 12 | 22 |
| 40-50 | 18 | 40 |
| 50-60 | 15 | 55 |
| 60-70 | 5 | 60 |
n = 60, n/2 = 30.
Median class: cf first exceeding 30 is 40 (class 40-50).
- l = 40, f = 18, cf = 22, h = 10
Median = 40 + [(30 - 22)/18] x 10 = 40 + (8/18) x 10 = 40 + 4.44 = 44.44.
Answer: The median is 44.44.
Example 5: Finding Missing Frequency Using Median
Problem: The median of the following data is 46. Find the value of x.
| Class | Frequency |
|---|---|
| 10-20 | 12 |
| 20-30 | 30 |
| 30-40 | x |
| 40-50 | 65 |
| 50-60 | 45 |
| 60-70 | 25 |
| 70-80 | 18 |
Solution:
n = 195 + x, n/2 = (195 + x)/2.
Since median = 46 lies in 40-50, median class is 40-50.
- l = 40, f = 65, h = 10
- cf = 12 + 30 + x = 42 + x
46 = 40 + [((195 + x)/2 - (42 + x))/65] x 10
- 6 = [(195 + x - 84 - 2x)/130] x 10
- 6 = [(111 - x)/130] x 10
- 78 = 111 - x
- x = 33
Answer: The missing frequency x = 33.
Example 6: Median of Heights Data
Problem: Find the median height of 51 students:
| Height (cm) | Students |
|---|---|
| 140-145 | 4 |
| 145-150 | 7 |
| 150-155 | 18 |
| 155-160 | 11 |
| 160-165 | 6 |
| 165-170 | 5 |
Solution:
| Height | f_i | cf |
|---|---|---|
| 140-145 | 4 | 4 |
| 145-150 | 7 | 11 |
| 150-155 | 18 | 29 |
| 155-160 | 11 | 40 |
| 160-165 | 6 | 46 |
| 165-170 | 5 | 51 |
n = 51, n/2 = 25.5.
Median class: cf first exceeding 25.5 is 29 (class 150-155).
- l = 150, f = 18, cf = 11, h = 5
Median = 150 + [(25.5 - 11)/18] x 5 = 150 + (14.5/18) x 5 = 150 + 4.03 = 154.03.
Answer: The median height is 154.03 cm.
Example 7: Median of Boundary Case
Problem: Find the median from: 0-10 (8), 10-20 (12), 20-30 (10), 30-40 (6), 40-50 (4).
Solution:
| Class | f_i | cf |
|---|---|---|
| 0-10 | 8 | 8 |
| 10-20 | 12 | 20 |
| 20-30 | 10 | 30 |
| 30-40 | 6 | 36 |
| 40-50 | 4 | 40 |
n = 40, n/2 = 20.
Median class: cf first reaching 20 is 20 (class 10-20).
- l = 10, f = 12, cf = 8, h = 10
Median = 10 + [(20 - 8)/12] x 10 = 10 + (12/12) x 10 = 10 + 10 = 20.
Answer: The median is 20.
Example 8: Median of Annual Rainfall
Problem: Annual rainfall (mm) for 60 years. Find the median.
| Rainfall (mm) | Years |
|---|---|
| 200-400 | 8 |
| 400-600 | 12 |
| 600-800 | 20 |
| 800-1000 | 15 |
| 1000-1200 | 5 |
Solution:
| Rainfall | f_i | cf |
|---|---|---|
| 200-400 | 8 | 8 |
| 400-600 | 12 | 20 |
| 600-800 | 20 | 40 |
| 800-1000 | 15 | 55 |
| 1000-1200 | 5 | 60 |
n = 60, n/2 = 30.
Median class: cf first exceeding 30 is 40 (class 600-800).
- l = 600, f = 20, cf = 20, h = 200
Median = 600 + [(30 - 20)/20] x 200 = 600 + (10/20) x 200 = 600 + 100 = 700.
Answer: The median rainfall is 700 mm.
Real-World Applications
Economics:
- Median income is more meaningful than mean income because it is not distorted by extremely high earners.
Real Estate:
- Median house prices are reported instead of mean prices to avoid distortion from very expensive properties.
Education:
- Median marks give a better sense of typical performance than the mean, especially with extreme scores.
Government:
- Poverty line calculations, median household size, and median age are used for policy planning.
Healthcare:
- Median survival time in clinical trials is preferred over mean because it is unaffected by outliers.
Key Points to Remember
- Median divides the data into two equal halves (50% below, 50% above).
- Formula: Median = l + [(n/2 - cf)/f] x h.
- Median class is the first class whose cumulative frequency exceeds n/2.
- cf in the formula is the cumulative frequency of the class BEFORE the median class.
- For grouped data, use n/2 (not (n+1)/2 which is for ungrouped data).
- The median is NOT affected by extreme values (unlike the mean).
- The median can also be found graphically from the ogive (cumulative frequency curve).
- If two ogives (less-than and more-than) are drawn, the median is the x-coordinate of their intersection.
- The median always lies within the median class interval.
- Empirical relationship: Mode = 3 x Median - 2 x Mean (approximate).
Practice Problems
- Find the median of: 0-10 (5), 10-20 (15), 20-30 (20), 30-40 (8), 40-50 (2).
- Find the median of: 100-200 (10), 200-300 (18), 300-400 (25), 400-500 (12), 500-600 (5).
- If the median of 0-10 (5), 10-20 (p), 20-30 (20), 30-40 (10) is 22.5, find p.
- Draw a less-than ogive for 0-10 (4), 10-20 (8), 20-30 (12), 30-40 (10), 40-50 (6) and find the median graphically.
- The median age of 100 people is 36.5. Which class is the median class if classes are 0-10, 10-20, ..., 60-70?
- Find the median of: 1000-1500 (3), 1500-2000 (5), 2000-2500 (10), 2500-3000 (8), 3000-3500 (4).
Frequently Asked Questions
Q1. What is the median of grouped data?
The median is the value that divides the distribution into two equal halves. For grouped data: Median = l + [(n/2 - cf)/f] x h.
Q2. How do you find the median class?
Calculate n/2. Build the cumulative frequency column. The median class is the first class whose cumulative frequency is greater than or equal to n/2.
Q3. What is cf in the median formula?
cf is the cumulative frequency of the class BEFORE the median class (not the median class itself). This is a common source of errors.
Q4. Why use n/2 and not (n+1)/2?
For grouped (continuous) data, use n/2. The formula (n+1)/2 applies to ungrouped (discrete) data arranged in order.
Q5. Can the median be found graphically?
Yes. Plot the ogive (cumulative frequency curve), mark n/2 on the y-axis, draw a horizontal line to the curve, drop a perpendicular to the x-axis. The x-value is the median.
Q6. What is the advantage of median over mean?
The median is not affected by extreme values (outliers). It gives a better measure of central tendency for skewed distributions.
Q7. What is the empirical relationship between mean, median, and mode?
Mode = 3 x Median - 2 x Mean (approximately). This holds for moderately skewed distributions.
Q8. What if two classes have the same cumulative frequency at n/2?
Use the lower class among them as the median class and apply the formula. The formula will interpolate correctly within that class.
Related Topics
- Mean of Grouped Data
- Mode of Grouped Data
- Cumulative Frequency Distribution
- Ogive (Cumulative Frequency Curve)
- Statistics - Collection and Presentation
- Frequency Distribution Table
- Histogram of Grouped Data
- Frequency Polygon
- Mean of Ungrouped Data
- Median of Ungrouped Data
- Empirical Relationship Between Mean, Median, Mode
- Mean by Assumed Mean Method
- Mean by Step-Deviation Method
- Statistics in Real Life










