Median of Grouped Data
Learning Objective: Students will be able to find the median of grouped data
Hook Activity: The Class Performance Puzzle
Imagine you're the class representative and want to know the typical test performance of your class. You have the following data:
| Marks Range | Number of Students |
|---|---|
| 0 - 20 | 4 |
| 20 - 40 | 10 |
| 40 - 60 | 14 |
| 60 - 80 | 8 |
| 80 - 100 | 4 |
Challenge: How would you find the "middle" score? What problems do you encounter when trying to find the median with this grouped data?
Introduction: Median of Ungrouped Data
The median is a measure of central tendency that represents the middle value of a dataset when arranged in order.
When number of observations (n) is odd
The median is the middle value, which is at position:
Example: Find the median of: 78, 85, 92, 64, 80, 72, 90
Step 1: Arrange in ascending order: 64, 72, 78, 80, 85, 90, 92
Step 2: n = 7 (odd)
Step 3: Position = (7+1)/2 = 4
Step 4: The 4th value is 80
Median = 80
When number of observations (n) is even
The median is the average of the two middle values:
Example: Find the median of: 78, 85, 92, 64, 80, 72
Step 1: Arrange in ascending order: 64, 72, 78, 80, 85, 92
Step 2: n = 6 (even)
Step 3: Positions = 3 and 4
Step 4: Values are 78 and 80
Median = (78 + 80)/2 = 79
Key Vocabulary:
- Median: The middle value in a dataset when arranged in order
- Cumulative Frequency: The running total of frequencies
- Median Class: The class interval containing the median
- Class Interval: A range of values grouped together
- Frequency: The number of times a value or class appears
Median of Grouped Data
When data is grouped, we can't identify the exact median, but we can estimate it using this formula:
Where:
- I = Lower limit of the median class
- n = Total number of observations
- cf = Cumulative frequency of the class before the median class
- f = Frequency of the median class
- h = Class size (interval width)
Example: Test Scores
Let's find the median for our hook activity data:
| Marks Range | Number of Students (f) | Cumulative Frequency (cf) |
|---|---|---|
| 0 - 20 | 4 | 4 |
| 20 - 40 | 10 | 14 |
| 40 - 60 | 14 | 28 |
| 60 - 80 | 8 | 36 |
| 80 - 100 | 4 | 40 |
\( I = 40 \), \( \frac{n}{2} = 20 \), \( cf = 14 \), \( f = 14 \), \( h = 20 \)
\[ \text{Median} = 40 + \frac{(20 - 14)}{14} \times 20 \] \[ = 40 + \frac{6}{14} \times 20 \] \[ = 40 + \frac{120}{14} \] \[ = 40 + 8.57 = 48.57 \]
Interpretation: The median test score is approximately 48.57, meaning half the students scored below this value and half scored above.
Group Work (We Do)
Let's work through this problem together:
| Height (cm) | Number of Students |
|---|---|
| 150-155 | 5 |
| 155-160 | 8 |
| 160-165 | 15 |
| 165-170 | 10 |
| 170-175 | 2 |
Steps:
- Calculate cumulative frequencies
- Find N/2
- Identify the median class
- Apply the median formula
Solution:
Total students (N) = 5+8+15+10+2 = 40
\( \frac{N}{2} = \frac{40}{2} = 20 \)
| Height (cm) | Frequency | Cumulative Frequency |
|---|---|---|
| 150-155 | 5 | 5 |
| 155-160 | 8 | 13 |
| 160-165 | 15 | 28 |
| 165-170 | 10 | 38 |
| 170-175 | 2 | 40 |
Median class: 160-165 (where CF first exceeds 20)
\( I = 160 \), \( cf = 13 \), \( f = 15 \), \( h = 5 \)
\[ \text{Median} = 160 + \frac{(20 - 13)}{15} \times 5 \]
\[ = 160 + \frac{7}{15} \times 5 \]
\[ = 160 + \frac{35}{15} \]
\[ = 160 + 2.33 = 162.33 \text{ cm} \]
Independent Work (You Do)
Now try this problem on your own:
| Daily Wage (₹) | Number of Workers |
|---|---|
| 200-300 | 5 |
| 300-400 | 8 |
| 400-500 | 12 |
| 500-600 | 10 |
| 600-700 | 5 |
Find the median daily wage.
Solution:
Total workers (N) = 5+8+12+10+5 = 40
\( \frac{N}{2} = \frac{40}{2} = 20 \)
| Daily Wage (₹) | Frequency | Cumulative Frequency |
|---|---|---|
| 200-300 | 5 | 5 |
| 300-400 | 8 | 13 |
| 400-500 | 12 | 25 |
| 500-600 | 10 | 35 |
| 600-700 | 5 | 40 |
Median class: 400-500 (where CF first exceeds 20)
\( I = 400 \), \( cf = 13 \), \( f = 12 \), \( h = 100 \)
\[ \text{Median} = 400 + \frac{(20 - 13)}{12} \times 100 \]
\[ = 400 + \frac{7}{12} \times 100 \]
\[ = 400 + \frac{700}{12} \]
\[ = 400 + 58.33 = ₹458.33 \]
Assessment
Question 1: The median of a grouped data is calculated using the formula:
a) \( I + \frac{\frac{N}{2} - cf}{f} \times h \)
b) \( I + \frac{N - cf}{f} \times h \)
c) \( I + \frac{\frac{N}{2}}{f} \times h \)
Question 2: For the following data, find the median class:
| Class Interval | Frequency | Cumulative Frequency |
|---|---|---|
| 10-20 | 5 | 5 |
| 20-30 | 7 | 12 |
| 30-40 | 10 | 22 |
| 40-50 | 8 | 30 |
Question 3: Calculate the median for the data in Question 2.
Answers:
1) a) \( I + \frac{\frac{N}{2} - cf}{f} \times h \)
2) The median class is 30-40 (where CF first exceeds \( \frac{N}{2} = 15 \))
3) \( I = 30 \), \( \frac{N}{2} = 15 \), \( cf = 12 \), \( f = 10 \), \( h = 10 \)
\[ \text{Median} = 30 + \frac{(15 - 12)}{10} \times 10 \]
\[ = 30 + \frac{3}{10} \times 10 \]
\[ = 30 + 3 = 33 \]
Homework
Problem 1: Find the median for the following data:
| Class Interval | Frequency |
|---|---|
| 0-10 | 5 |
| 10-20 | 10 |
| 20-30 | 15 |
| 30-40 | 8 |
| 40-50 | 2 |
Problem 2: The median of the following distribution is 28.5. Find the missing frequencies if the total number of observations is 60:
| Class Interval | Frequency |
|---|---|
| 0-10 | 5 |
| 10-20 | ? |
| 20-30 | 20 |
| 30-40 | 15 |
| 40-50 | ? |
Conclusion
Today we learned how to find the median for both ungrouped and grouped data:
- For ungrouped data: Arrange values and find the middle
- For grouped data: Calculate cumulative frequencies, find N/2, identify median class, apply formula
The median is a valuable measure of central tendency that is not affected by extreme values, making it useful for understanding the typical value in a dataset.