Statistics · Lesson 3 of 4
Median of Grouped Data
“Use cumulative frequency to locate the middle of grouped data without unpacking every observation.”
• Understand cumulative frequency. • Construct cumulative frequency tables. • Identify the median class using n/2. • Use the grouped-data median formula. • Solve problems with cumulative data and missing frequencies.
Median
Imagine a school records the daily travel time of a large group of students. Instead of writing down every student's exact travel time, the data is grouped into intervals such as 0–10 minutes, 10–20 minutes, 20–30 minutes, 30–40 minutes, and so on.
Now suppose we want to know the travel time that divides the students into two equal groups — about half the students take less time than this value, while the other half take more time. The value that does this is called the median.
The median is the value that divides an ordered data set into two equal halves. In simple data, we can usually find it by arranging the values from smallest to largest and locating the middle observation.
In grouped data, however, the individual values are hidden inside class intervals. So we cannot directly see the exact middle value. Instead, we first use the cumulative frequencies to identify the class interval containing the middle observation. This interval is called the median class.
Once the median class is known, we use a formula to estimate the median within that interval.
So, the main idea is simple: first locate where the middle observation lies, then estimate its value inside that class interval.
The running total of frequencies up to and including a class.
| Class interval | Frequency f | Cumulative frequency cf |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 3 | 8 |
| 20–30 | 4 | 12 |
| 30–40 | 3 | 15 |
| 40–50 | 3 | 18 |
| 50–60 | 4 | 22 |
| 60–70 | 7 | 29 |
| 70–80 | 9 | 38 |
| 80–90 | 7 | 45 |
| 90–100 | 8 | 53 |
Finding the Median Class
Calculate n/2, then scan the cumulative-frequency column. The first cumulative frequency greater than n/2 identifies the median class.
| Symbol | Meaning |
|---|---|
| l | Lower limit of median class |
| n | Total frequency |
| cf | Cumulative frequency before the median class |
| f | Frequency of the median class |
| h | Class size |
Problem
Find the median of the distribution shown above.
- 1.n = 53, so n/2 = 26.5.
- 2.The first cumulative frequency greater than 26.5 is 29.
- 3.Therefore median class = 60–70.
- 4.l = 60, cf = 22, f = 7, h = 10.
- 5.Median = 60 + [(26.5 − 22)/7] × 10.
- 6.Median ≈ 66.43.
A median near 66.4 means roughly half the observations lie below 66.4 and roughly half lie above it.
Less-Than Cumulative Data
If a table gives values such as 'less than 145' and 'less than 150', those are cumulative totals. Recover ordinary class frequencies by subtracting consecutive cumulative frequencies.
| Statement | Cumulative frequency | Recovered interval | Frequency |
|---|---|---|---|
| Less than 140 | 4 | Below 140 | 4 |
| Less than 145 | 11 | 140–145 | 7 |
| Less than 150 | 29 | 145–150 | 18 |
| Less than 155 | 40 | 150–155 | 11 |
| Less than 160 | 46 | 155–160 | 6 |
| Less than 165 | 51 | 160–165 | 5 |
Problem
Find the median height from the cumulative table above.
- 1.n = 51, so n/2 = 25.5.
- 2.The first cumulative frequency above 25.5 is 29.
- 3.Median class = 145–150.
- 4.l = 145, cf = 11, f = 18, h = 5.
- 5.Median = 145 + [(25.5 − 11)/18] × 5.
- 6.Median ≈ 149.03 cm.
Missing Frequencies from a Known Median
| Class interval | Frequency | Cumulative frequency |
|---|---|---|
| 0–100 | 2 | 2 |
| 100–200 | 5 | 7 |
| 200–300 | x | 7 + x |
| 300–400 | 12 | 19 + x |
| 400–500 | 17 | 36 + x |
| 500–600 | 20 | 56 + x |
| 600–700 | y | 56 + x + y |
| 700–800 | 9 | 65 + x + y |
| 800–900 | 7 | 72 + x + y |
| 900–1000 | 4 | 76 + x + y |
Problem
Median = 525 and total frequency = 100. Find x and y.
- 1.Total frequency gives 76 + x + y = 100, so x + y = 24.
- 2.Since 525 lies in 500–600, this is the median class.
- 3.l = 500, f = 20, h = 100, cf = 36 + x, n/2 = 50.
- 4.525 = 500 + [(50 − (36 + x))/20] × 100.
- 5.25 = 5(14 − x).
- 6.x = 9.
- 7.Then 9 + y = 24, so y = 15.
• Using ordinary frequency instead of cumulative frequency to locate the median class, • Using the cumulative frequency of the median class as cf, • Forgetting that n is total frequency, • Treating a less-than table as an ordinary frequency table, • Using an incorrect class width h,
Quiz
Which expression gives the median of grouped data?
How is the median class identified in a grouped frequency distribution?
The frequencies of the classes 0–10, 10–20, 20–30, 30–40 and 40–50 are 5, 9, 12, 8 and 6 respectively. Which is the median class?
For a grouped distribution, l = 20, N = 40, cf = 14, f = 12 and h = 10. What is the median?
In the median formula, what does cf represent?
Practice Problems
- Electricity use 65–85, 85–105, 105–125, 125–145, 145–165, 165–185, 185–205 has frequencies 4, 5, 13, 20, 14, 8, 4. Find the median.
- Intervals 0–10, 10–20, 20–30, 30–40, 40–50, 50–60 have frequencies 5, x, 20, 15, y, 5. Total = 60 and median = 28.5. Find x and y.
- Lamp lifetimes 1500–2000, 2000–2500, 2500–3000, 3000–3500, 3500–4000, 4000–4500, 4500–5000 have frequencies 14, 56, 60, 86, 74, 62, 48. Find the median.
- Weights 40–45, 45–50, 50–55, 55–60, 60–65, 65–70, 70–75 have frequencies 2, 3, 8, 6, 6, 3, 2. Find the median.
Key Takeaways
• Median is based on position, so cumulative frequency is essential. • Calculate n/2 and locate the first cumulative frequency greater than it. • In the formula, cf belongs to the class before the median class. • Less-than cumulative frequencies can be converted by subtraction.