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Median of Grouped Data

Class 10Statistics

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:

  1. Calculate n/2.
  2. Build the cumulative frequency column.
  3. 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:

  1. The median is the (n/2)-th observation.
  2. From the cumulative frequency table, the median lies in the class whose cumulative frequency first reaches or exceeds n/2.
  3. Let this be the class [l, l + h) with frequency f.
  4. The cumulative frequency up to the class before the median class is cf.
  5. So (n/2 - cf) observations within the median class need to be counted to reach the median.
  6. Assuming observations are uniformly distributed within each class:
  7. The fraction of the class to be covered = (n/2 - cf)/f.
  8. The median is located at a distance of [(n/2 - cf)/f] x h from the lower limit l.
  9. 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

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:

  1. Construct the cumulative frequency table.
  2. Find n/2 where n = Sigma(f_i).
  3. Identify the median class -- the class whose cumulative frequency first exceeds n/2.
  4. 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
  5. Apply the formula: Median = l + [(n/2 - cf)/f] x h.
  6. Simplify to get the answer.

Finding median from an Ogive:

  1. Plot the ogive (cumulative frequency vs. upper class boundary).
  2. Mark n/2 on the y-axis.
  3. Draw a horizontal line from n/2 to the ogive curve.
  4. From the point of intersection, drop a perpendicular to the x-axis.
  5. 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:

ClassFrequency
0-105
10-208
20-3012
30-4015
40-5010

Solution:

Classf_icf
0-1055
10-20813
20-301225
30-401540
40-501050

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:

MarksStudents
0-206
20-408
40-6014
60-8010
80-1002

Solution:

Marksf_icf
0-2066
20-40814
40-601428
60-801038
80-100240

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
100-15012
150-20020
200-25025
250-30018
300-3505

Solution:

Earningsf_icf
100-1501212
150-2002032
200-2502557
250-3001875
300-350580

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 thanCumulative Frequency
3010
4022
5040
6055
7060

Solution:

Convert to class intervals:

Classf_icf
20-301010
30-401222
40-501840
50-601555
60-70560

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.

ClassFrequency
10-2012
20-3030
30-40x
40-5065
50-6045
60-7025
70-8018

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
140-1454
145-1507
150-15518
155-16011
160-1656
165-1705

Solution:

Heightf_icf
140-14544
145-150711
150-1551829
155-1601140
160-165646
165-170551

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:

Classf_icf
0-1088
10-201220
20-301030
30-40636
40-50440

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
200-4008
400-60012
600-80020
800-100015
1000-12005

Solution:

Rainfallf_icf
200-40088
400-6001220
600-8002040
800-10001555
1000-1200560

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

  1. Find the median of: 0-10 (5), 10-20 (15), 20-30 (20), 30-40 (8), 40-50 (2).
  2. Find the median of: 100-200 (10), 200-300 (18), 300-400 (25), 400-500 (12), 500-600 (5).
  3. If the median of 0-10 (5), 10-20 (p), 20-30 (20), 30-40 (10) is 22.5, find p.
  4. 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.
  5. The median age of 100 people is 36.5. Which class is the median class if classes are 0-10, 10-20, ..., 60-70?
  6. 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.

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