Question
The marks of 50 students in a test are given by the frequency distribution below. Find the mean, median, and mode.
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 5 | 12 | 18 | 10 | 5 |
CBSE 2024 boards 5-mark question.
Solution — Step by Step
Take assumed mean (midpoint of the modal class). Class size . Compute deviations where is the class midpoint.
| 5 | 15 | 25 | 35 | 45 | |
|---|---|---|---|---|---|
| -2 | -1 | 0 | 1 | 2 | |
| 5 | 12 | 18 | 10 | 5 | |
| -10 | -12 | 0 | 10 | 10 |
. .
Cumulative frequencies: 5, 17, 35, 45, 50. . The class containing the 25th item is 20–30 (cf reaches 35 here). So median class is 20–30, with , (cf before median class), , .
Modal class is the one with highest frequency: 20–30 (frequency 18). Here , , (previous class), (next class), .
Final answers: Mean , Median , Mode .
Why This Works
For grouped data, we cannot find exact mean/median/mode — we estimate using class midpoints (for mean) and linear interpolation within the relevant class (for median and mode). The formulas come from assuming uniform distribution within each class.
For roughly symmetric distributions, mean ≈ median ≈ mode, which is what we observe here. For skewed distributions, the relationship is mode = 3·median − 2·mean (the empirical Pearson formula).
Alternative Method
Direct mean calculation: . Same answer, more arithmetic.
In the median formula, refers to the cumulative frequency of the class just before the median class — not including the median class itself. Confusing this with cumulative frequency including the median class is the most common error.
For CBSE Class 10 boards, this exact pattern (frequency table with 5 classes, find mean/median/mode) appears almost every year. Practice the assumed mean shortcut — it cuts arithmetic by 70%.