Mathematics — Standard (041) — CBSE Class 10 board question

Frequency Distribution Table with Cumulative Frequency:

| Class | Frequency (f) | cf |
|-------|---------------|----|
| 0–15 | 15 | 15 |
| 15–30 | 10 | 25 |
| 30–45 | 12 | 37 |
| 45–60 | 9 | 46 |
| 60–75 | 8 | 54 |
| 75–90 | 10 | 64 |
| 90–105 | 6 | 70 |

Total n = 70

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Median:

$\dfrac{n}{2} = 35$

Cumulative frequency just greater than 35 is 37 (class 30–45).

∴ Median class = 30–45, $l = 30$, $cf = 25$, $f = 12$, $h = 15$

$$\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h = 30 + \left(\frac{35 - 25}{12}\right) \times 15$$

$$= 30 + \frac{150}{12} = 30 + 12.5 = \mathbf{42.5}$$

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Mode:

Maximum frequency = 15 → Modal class = 0–15

$l = 0$, $f_1 = 15$, $f_0 = 0$ (no preceding class), $f_2 = 10$, $h = 15$

$$\text{Mode} = 0 + \left(\frac{15 - 0}{2(15) - 0 - 10}\right) \times 15 = \frac{15}{20} \times 15 = \mathbf{11.25}$$

Source: Chapter 13, Sections 13.3 and 13.4

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• Median: Find n/2, locate the class whose cumulative frequency first exceeds n/2 — that is the median class. Then apply the formula. Here cf = 25 (class before 30–45) and f = 12.
• Mode: The modal class has the highest frequency (15 for 0–15). Since it is the first class, $f_0 = 0$. Substitute carefully in the mode formula.
• Common mistakes: using wrong cf (must be of the class preceding median class), and forgetting $f_0 = 0$ when modal class is the first interval.