1. Chapter at a Glance
When data is large — marks of 200 students, daily wages of 500 workers — we group it into class intervals (0–10, 10–20, …). How do we find the mean, median and mode when individual values are no longer listed? This chapter gives formulas for the three measures of central tendency of grouped data: the mean (by direct, assumed-mean and step-deviation methods), the mode and the median. Statistics regularly appears as a 5-mark board question.
- Class intervals, class marks and frequency tables.
- Mean of grouped data: direct, assumed mean, step-deviation methods.
- Mode of grouped data: modal class and formula.
- Cumulative frequency and median of grouped data.
- Empirical relation: 3 Median = Mode + 2 Mean.
- Choosing the right measure.
Note: all data tables in this chapter are practice data for learning.
2. Grouped Data Basics
- Class interval: e.g. 10–20 (lower limit 10, upper limit 20). In continuous classes, a value equal to the upper limit goes to the next class.
- Class size (h): upper − lower limit.
- Class mark (xᵢ): (lower + upper)/2 — the representative value of the class.
3. Mean of Grouped Data
- Direct: x̄ = Σfᵢxᵢ / Σfᵢ
- Assumed mean: dᵢ = xᵢ − a; x̄ = a + Σfᵢdᵢ / Σfᵢ
- Step deviation: uᵢ = (xᵢ − a)/h; x̄ = a + h × Σfᵢuᵢ / Σfᵢ
Practice data — marks of 40 students:
| Marks | fᵢ | xᵢ | uᵢ = (xᵢ − 35)/10 | fᵢuᵢ |
|---|---|---|---|---|
| 10–20 | 4 | 15 | −2 | −8 |
| 20–30 | 8 | 25 | −1 | −8 |
| 30–40 | 12 | 35 | 0 | 0 |
| 40–50 | 10 | 45 | 1 | 10 |
| 50–60 | 6 | 55 | 2 | 12 |
| Total | 40 | 6 |
x̄ = 35 + 10 × 6/40 = 35 + 1.5 = 36.5.
Check with the direct method: Σfᵢxᵢ = 60 + 200 + 420 + 450 + 330 = 1460; 1460/40 = 36.5 ✓.
Take the assumed mean a as a class mark near the middle of the table — it makes the numbers small.
4. Mode of Grouped Data
The modal class is the class with the highest frequency. Then:
l = lower limit of modal class, h = class size, f₁ = frequency of modal class, f₀ = frequency of the class before, f₂ = frequency of the class after.
For the marks data: modal class 30–40, l = 30, f₁ = 12, f₀ = 8, f₂ = 10, h = 10.
Mode = 30 + [(12 − 8)/(24 − 8 − 10)] × 10 = 30 + (4/6) × 10 = 36.67.
5. Median of Grouped Data
Make a cumulative frequency (cf) column (running total). Let n = Σf. The median class is the first class whose cf is greater than or equal to n/2.
l = lower limit of median class, cf = cumulative frequency of the class before it, f = frequency of the median class, h = class size.
| Marks | f | cf |
|---|---|---|
| 10–20 | 4 | 4 |
| 20–30 | 8 | 12 |
| 30–40 | 12 | 24 |
| 40–50 | 10 | 34 |
| 50–60 | 6 | 40 |
n/2 = 20 → median class 30–40 (cf 24 ≥ 20). l = 30, cf = 12, f = 12, h = 10.
Median = 30 + [(20 − 12)/12] × 10 = 30 + 6.67 = 36.67.
For moderately skewed data: 3 Median = Mode + 2 Mean.
Check: 3 × 36.67 ≈ 110; 36.67 + 73 ≈ 109.7 ✓ (close).
6. Which Measure to Use?
| Measure | Best for |
|---|---|
| Mean | Comparing groups; uses every value (e.g. average marks) |
| Median | Data with extreme values (e.g. incomes, house prices) |
| Mode | Most common value (e.g. shoe size, popular product) |
7. Solved Problems (Board Pattern)
The mean of the data below is 18. Find p.
| Class | 11–13 | 13–15 | 15–17 | 17–19 | 19–21 | 21–23 | 23–25 |
|---|---|---|---|---|---|---|---|
| f | 7 | 6 | 9 | 13 | p | 5 | 4 |
Solution: xᵢ: 12, 14, 16, 18, 20, 22, 24. Take a = 18, dᵢ: −6, −4, −2, 0, 2, 4, 6. Σfd = −42 − 24 − 18 + 0 + 2p + 20 + 24 = 2p − 40. Mean = 18 → Σfd = 0 → p = 20.
The median of 60 observations is 28.5. The table has f: 5, x, 20, 15, y, 5 for classes 0–10 … 50–60. Find x and y.
Solution: 45 + x + y = 60 → x + y = 15. Median class 20–30 (l = 20, f = 20, cf = 5 + x). 28.5 = 20 + [(30 − 5 − x)/20] × 10 → 8.5 = (25 − x)/2 → x = 8, y = 7.
If mode = 24.5 and mean = 29.75, estimate the median.
Solution: 3 Median = 24.5 + 59.5 = 84 → Median = 28.
8. Master Formula Table
| Measure | Formula |
|---|---|
| Class mark | (lower + upper)/2 |
| Mean (direct) | Σfᵢxᵢ / Σfᵢ |
| Mean (assumed) | a + Σfᵢdᵢ / Σfᵢ |
| Mean (step deviation) | a + h × Σfᵢuᵢ / Σfᵢ |
| Mode | l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h |
| Median | l + [(n/2 − cf)/f] × h |
| Empirical | 3 Median = Mode + 2 Mean |
9. Master Summary Table
| Measure | Key class | Need |
|---|---|---|
| Mean | All classes | xᵢ, fᵢ (and dᵢ or uᵢ) |
| Mode | Highest frequency | f₀, f₁, f₂ |
| Median | First cf ≥ n/2 | cf of previous class, f |
Chart — the three measures for the practice marks data:
10. Mind Map
- Direct, assumed
- Step deviation
- Modal class
- f₀, f₁, f₂
- cf column
- n/2
- 3Md = Mo + 2x̄
- Choose measure
11. Tips and Tricks
- Step deviation is fastest when class sizes are equal and numbers are large.
- In the median formula, cf is of the class before the median class.
- Convert inclusive classes (10–19, 20–29) to continuous (9.5–19.5, …) before using formulas.
- Using the modal class's own cf in the median formula.
- Choosing the class with the highest cf as the median class.
- Forgetting to multiply by h in step deviation.
12. Important One-liners
- The class mark represents its whole class in mean calculations.
- All three mean methods give the same answer.
- The modal class has the highest frequency.
- The median class is the first with cf ≥ n/2.
- 3 Median = Mode + 2 Mean (empirical).
13. Exam-pattern Questions with Detailed Answers
The class mark of 25–35 is:
- 25
- 30
- 35
- 10
Show answer
(25 + 35)/2 = 30.
Answer: (B)
If mean = 26 and median = 27, find the mode (empirical relation).
Show answer
Mode = 3(27) − 2(26) = 81 − 52 = 29.
Answer: 29
Which measure is found graphically using cumulative frequency?
Show answer
The median (from an ogive).
Answer: Median
Practice data — daily pocket money (₹) of 50 children: 0–20: 6, 20–40: 10, 40–60: 14, 60–80: 12, 80–100: 8. Find the mean.
Show answer
a = 50, h = 20; u: −2, −1, 0, 1, 2. Σfu = −12 − 10 + 0 + 12 + 16 = 6. Mean = 50 + 20 × 6/50 = 52.4.
Answer: ₹52.40
Find the mode of the data in Q4.
Show answer
Modal class 40–60: l = 40, f₁ = 14, f₀ = 10, f₂ = 12. Mode = 40 + [4/(28 − 22)] × 20 = 40 + 13.33 = 53.33.
Answer: ₹53.33
Find the median of the data in Q4.
Show answer
cf: 6, 16, 30, 42, 50. n/2 = 25 → class 40–60, cf before = 16, f = 14. Median = 40 + (9/14) × 20 ≈ 52.86.
Answer: ≈ ₹52.86
Practice data — ages of 30 patients: 5–15: 6, 15–25: 11, 25–35: 7, 35–45: 4, 45–55: 2. Find the modal age.
Show answer
Modal class 15–25: l = 15, f₁ = 11, f₀ = 6, f₂ = 7. Mode = 15 + [5/(22 − 13)] × 10 = 15 + 5.56 = 20.56 years.
Answer: ≈ 20.56 years
Convert 10–19, 20–29, 30–39 into continuous classes and give the class size.
Show answer
Subtract and add 0.5: 9.5–19.5, 19.5–29.5, 29.5–39.5; h = 10.
Answer: 9.5–19.5, 19.5–29.5, 29.5–39.5; h = 10
The mean of the following practice data is 50. Find f₁ and f₂ if the total frequency is 120. Classes 0–20, 20–40, 40–60, 60–80, 80–100 with frequencies 17, f₁, 32, f₂, 19.
Show answer
f₁ + f₂ = 52. a = 50, u: −2, −1, 0, 1, 2. Σfu = −34 − f₁ + f₂ + 38 = f₂ − f₁ + 4. Mean 50 → Σfu = 0 → f₂ − f₁ = −4. So f₁ = 28, f₂ = 24.
Answer: f₁ = 28, f₂ = 24
Practice data — heights (cm) of 50 girls: 140–145: 5, 145–150: 10, 150–155: 20, 155–160: 9, 160–165: 6. Find the median height.
Show answer
cf: 5, 15, 35, 44, 50. n/2 = 25 → class 150–155, cf = 15, f = 20, h = 5. Median = 150 + (10/20) × 5 = 152.5 cm.
Answer: 152.5 cm
For the data in Q10, find the mean and verify the empirical relation approximately.
Show answer
a = 152.5, h = 5; u: −2, −1, 0, 1, 2. Σfu = −10 − 10 + 0 + 9 + 12 = 1. Mean = 152.5 + 5 × 1/50 = 152.6. Mode: l = 150, f₁ = 20, f₀ = 10, f₂ = 9 → 150 + [10/(40 − 19)] × 5 ≈ 152.38. 3 × 152.5 = 457.5; 152.38 + 305.2 = 457.58 ✓.
Answer: Mean 152.6; relation holds approximately
Explain why the median may be a better average than the mean for monthly incomes in a town.
Show answer
A few very high incomes raise the mean far above what most people earn. The median depends only on the middle position, so it is not pulled by extreme values and better represents a typical income.
Answer: Median resists extreme values
14. Quick Revision Summary
- Class mark xᵢ = (lower + upper)/2.
- Mean: Σfx/Σf, or a + Σfd/Σf, or a + h Σfu/Σf.
- Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] h.
- Median = l + [(n/2 − cf)/f] h, with cf of the previous class.
- 3 Median = Mode + 2 Mean.