- A10
- B9
- C9.5
- D11
Step-by-Step Solution
In continuous form 20–29 becomes 19.5–29.5, so the class size = 29.5 − 19.5 = 10. Subtracting the inclusive limits (29 − 20 = 9) misses the half-unit gap on each side.
Answer: A) 10
A table uses the inclusive classes 20–29, 30–39 and 40–49. After converting them to continuous form, what is the class size?
Class 9StatisticsStatistics and Probability
Key idea: Class intervals and class marks
When data has many distinct values (like heights or weights), group it into **class intervals**.
Example table:
| Class interval | Frequency |
|---|---|
| 0–10 | 5 |
| 10–20 | 8 |
| 20–30 | 4 |
**Class mark** of an interval = (lower limit + upper limit) / 2.
For 0–10: mark = 5. For 10–20: mark = 15.
**Inclusive vs exclusive intervals**:
- Exclusive (continuous) form like 0–10, 10–20: 10 belongs to the SECOND interval, not the first.
- Inclusive form 0–9, 10–19: rare in modern textbooks — convert to continuous form (subtract 0.5 from lower, add 0.5 to upper) before drawing histograms.
Example table:
| Class interval | Frequency |
|---|---|
| 0–10 | 5 |
| 10–20 | 8 |
| 20–30 | 4 |
**Class mark** of an interval = (lower limit + upper limit) / 2.
For 0–10: mark = 5. For 10–20: mark = 15.
**Inclusive vs exclusive intervals**:
- Exclusive (continuous) form like 0–10, 10–20: 10 belongs to the SECOND interval, not the first.
- Inclusive form 0–9, 10–19: rare in modern textbooks — convert to continuous form (subtract 0.5 from lower, add 0.5 to upper) before drawing histograms.
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