What is grouped data?
When there are many values, we put them into classes (groups) such as 0–10, 10–20, 20–30. The number of values in a class is its frequency (f). The total of all frequencies is n = Σf.
The class mark x is the middle of a class: x = (lower limit + upper limit) ÷ 2. We pretend every value in the class sits at this middle. This is a small, fair guess.
The class size h is the width of a class (for 20–30, h = 10).
Mean of grouped data: three methods
All three methods give the same answer. Pick the one that makes the numbers small.
1. Direct method
Mean x̄ = Σfx ÷ Σf. Multiply each frequency by its class mark, add, then divide by n.
2. Assumed mean method
Pick a class mark in the middle as a guess a. Find d = x − a. Then x̄ = a + Σfd ÷ Σf. The numbers d are small, so the sums are easier.
3. Step-deviation method
If all d share a factor h (the class size), use u = (x − a) ÷ h. Then x̄ = a + (Σfu ÷ Σf) × h. This gives the smallest numbers of all.
Why is the mean a balance point? If the bars were weights on a see-saw, the see-saw would balance exactly at the mean.
Median of grouped data
The median is the middle value when all values are in order. For grouped data:
- Make a cumulative frequency (cf) column: a running total of f.
- Find n/2.
- The first class whose cf is equal to or more than n/2 is the median class.
- Use: Median = l + ((n/2 − cf) ÷ f) × h
Here l = lower limit of the median class, cf = cumulative frequency of the class before it, f = frequency of the median class, h = class size.
Mode of grouped data
The modal class is the class with the highest frequency. The mode lies inside it, pulled toward the bigger neighbour.
Mode = l + ((f₁ − f₀) ÷ (2f₁ − f₀ − f₂)) × h
l = lower limit of the modal class, f₁ = its frequency, f₀ = frequency of the class before, f₂ = frequency of the class after, h = class size. If the modal class is the first class, take f₀ = 0.
Empirical relation and which average to use
For data that is not too lopsided: 3 Median = Mode + 2 Mean. You can use it to check an answer or find one average from the other two.
- Use the mean when values are fairly even (average marks).
- Use the median when a few values are very big or very small (salaries, house prices).
- Use the mode for the most popular choice (shoe size, favourite colour).
Board exam pattern
Statistics and Probability together carry 11 marks in CBSE Class 10. Expect a 3-mark mean or mode question, a median or missing-frequency question, and often a case-study table. Always draw the full table (x, f, fx or cf) — steps carry marks.
Try it: your class survey
Ask 20 friends how many minutes they walk each day. Group the answers into 0–10, 10–20, 20–30, 30–40. Make a table of f, find the mean, median and mode. Then open the last step of the 3D and set the same frequencies with the sliders. Predict first: will the mean be higher or lower than the mode?
Check your understanding
1. What is the class mark of 40–60? (50)
2. For the median, do you use cf of the median class or the class before it? (The class before it.)
Key formulas and definitions
- Class mark x = (lower limit + upper limit) ÷ 2
- Mean (direct) = Σfx ÷ Σf
- Mean (assumed mean) = a + Σfd ÷ Σf, d = x − a
- Mean (step deviation) = a + (Σfu ÷ Σf) × h, u = (x − a) ÷ h
- Median = l + ((n/2 − cf) ÷ f) × h
- Mode = l + ((f₁ − f₀) ÷ (2f₁ − f₀ − f₂)) × h
- Empirical relation: 3 Median = Mode + 2 Mean
Worked examples
1. Marks of 40 students: 0–10: 5, 10–20: 8, 20–30: 12, 30–40: 9, 40–50: 6. Find the mean by the direct method.
Class marks x = 5, 15, 25, 35, 45. fx = 25, 120, 300, 315, 270. Σfx = 1030, Σf = 40. Mean = 1030 ÷ 40 = 25.75.
2. Find the mean of the same data by the assumed mean method with a = 25.
d = x − 25 = −20, −10, 0, 10, 20. fd = −100, −80, 0, 90, 120. Σfd = 30. Mean = 25 + 30 ÷ 40 = 25 + 0.75 = 25.75.
3. Find the mean of the same data by the step-deviation method (a = 25, h = 10).
u = d ÷ 10 = −2, −1, 0, 1, 2. fu = −10, −8, 0, 9, 12. Σfu = 3. Mean = 25 + (3 ÷ 40) × 10 = 25 + 0.75 = 25.75. Same answer, smaller numbers.
4. Find the median of the same data.
cf = 5, 13, 25, 34, 40. n/2 = 20. The first cf ≥ 20 is 25, so the median class is 20–30. l = 20, cf (before) = 13, f = 12, h = 10. Median = 20 + ((20 − 13) ÷ 12) × 10 = 20 + 5.83 = 25.83.
5. Find the mode of the same data.
Modal class = 20–30 (f₁ = 12). f₀ = 8, f₂ = 9, l = 20, h = 10. Mode = 20 + ((12 − 8) ÷ (24 − 8 − 9)) × 10 = 20 + (4 ÷ 7) × 10 = 20 + 5.71 = 25.71. Check: 3 × 25.83 = 77.5 and 25.71 + 2 × 25.75 = 77.21, very close.
6. Daily wages (₹) of 40 workers: 100–120: 4, 120–140: 10, 140–160: 16, 160–180: 6, 180–200: 4. Find the mean, median and mode.
x = 110, 130, 150, 170, 190. Σfx = 440 + 1300 + 2400 + 1020 + 760 = 5920. Mean = 5920 ÷ 40 = ₹148. cf = 4, 14, 30, 36, 40; n/2 = 20 → median class 140–160: Median = 140 + ((20 − 14) ÷ 16) × 20 = ₹147.5. Modal class 140–160: Mode = 140 + ((16 − 10) ÷ (32 − 10 − 6)) × 20 = 140 + 7.5 = ₹147.5.
7. The mean of this data is 27. Find p. Classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 4, 6, p, 10, 5.
Σf = 25 + p. Σfx = 4×5 + 6×15 + 25p + 10×35 + 5×45 = 685 + 25p. (685 + 25p) ÷ (25 + p) = 27 → 685 + 25p = 675 + 27p → 10 = 2p → p = 5.
8. For some data, mean = 26 and median = 27. Estimate the mode.
3 Median = Mode + 2 Mean → Mode = 3 × 27 − 2 × 26 = 81 − 52 = 29.
Common mistakes
- Using the cf of the median class itself. The formula needs the cf of the class just before the median class.
- In the mode formula, taking the neighbours of the tallest class wrongly. f₀ is the class before, f₂ the class after the modal class.
- Dividing Σfx by the number of classes instead of Σf (the number of values).
- Forgetting h in the step-deviation method: mean = a + (Σfu ÷ Σf) × h, not a + Σfu ÷ Σf.