Calculate Mean and Median from the following frequency distribution :
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|---|
| No. of students | 5 | 12 | 15 | 25 | 8 | 3 | 2 |
OR
Calculate Mode from the following frequency distribution :
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|
| No. of students | 12 | 18 | 27 | 20 | 17 | 6 |
You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
Start your 14-day free trial to unlock the full solution →Mean ≈ 30.14 and Median = 31.2 for the given frequency distribution. [OR: Mode = 25.625 for the alternative distribution.]
Part A — Mean and Median:
| Marks | No. of students (f) | Mid-point (x) | fx | Cumulative frequency (cf) |
|---|---|---|---|---|
| 0–10 | 5 | 5 | 25 | 5 |
| 10–20 | 12 | 15 | 180 | 17 |
| 20–30 | 15 | 25 | 375 | 32 |
| 30–40 | 25 | 35 | 875 | 57 |
| 40–50 | 8 | 45 | 360 | 65 |
| 50–60 | 3 | 55 | 165 | 68 |
| 60–70 | 2 | 65 | 130 | 70 |
| Total | N = 70 | Σfx = 2110 |
Mean (Direct method): Mean = Σfx / N = 2110 / 70 = 30.14 (approx.)
Median: N/2 = 70/2 = 35. The cumulative frequency first reaching/exceeding 35 is 57, in the class 30–40 — so 30–40 is the Median class.
L (lower limit of median class) = 30; cf (cumulative frequency of the class before median class) = 32; f (frequency of median class) = 25; h (class width) = 10
Median = L + [(N/2 − cf)/f] × h = 30 + [(35 − 32)/25] × 10 = 30 + (3/25) × 10 = 30 + 1.2 = 31.2
OR — Part B: Mode (for the alternative frequency distribution):
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | …
Unlock everything free for 14 days
- Full step-by-step solutions
- Concept-first explanations
- Methods, shortcuts & mistakes
- PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.