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

Cumulative Frequency Table:

| Class | Frequency | Cumulative Frequency (cf) |
|-------|-----------|--------------------------|
| 1400–1550 | 6 | 6 |
| 1550–1700 | 13 | 19 |
| 1700–1850 | 25 | 44 |
| 1850–2000 | 10 | 54 |

Here, $n = 54$, so $\dfrac{n}{2} = 27$.

The cumulative frequency just greater than 27 is 44, which belongs to class 1700–1850.

∴ Median class = 1700–1850

Here, $l = 1700$, $cf = 19$, $f = 25$, $h = 150$

$$\text{Median} = l + \left(\frac{\dfrac{n}{2} - cf}{f}\right) \times h = 1700 + \left(\frac{27 - 19}{25}\right) \times 150$$

$$= 1700 + \frac{8 \times 150}{25} = 1700 + 48 = \mathbf{1748}$$

Source: Chapter 13, Section 13.4 Median of Grouped Data

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• Always build the cf column first, then find $n/2$.
• The median class is the one whose cf first exceeds $n/2$ — a common error is picking the class where cf equals $n/2$ exactly.
• Correctly identify $l$ (lower limit of median class), $cf$ (cf of the preceding class), $f$ (frequency of median class), and $h$ (class width = 150 here).
• Examiners award marks for: cf table (1 mark), identifying median class with correct values (1 mark), and correct formula application giving 1748 (1 mark).