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

Mean (Direct Method):

| Class | $x_i$ | $f_i$ | $f_i x_i$ |
|-------|--------|--------|------------|
| 0–10 | 5 | 7 | 35 |
| 10–20 | 15 | 14 | 210 |
| 20–30 | 25 | 13 | 325 |
| 30–40 | 35 | 12 | 420 |
| 40–50 | 45 | 20 | 900 |
| 50–60 | 55 | 11 | 605 |
| 60–70 | 65 | 15 | 975 |
| 70–80 | 75 | 8 | 600 |
| Total | | 100 | 4070 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{4070}{100} = \textbf{40.7}$$

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Median:

| Class | $f_i$ | $cf$ |
|-------|--------|------|
| 0–10 | 7 | 7 |
| 10–20 | 14 | 21 |
| 20–30 | 13 | 34 |
| 30–40 | 12 | 46 |
| 40–50 | 20 | 66 |
| 50–60 | 11 | 77 |
| 60–70 | 15 | 92 |
| 70–80 | 8 | 100 |

$n = 100$, so $\dfrac{n}{2} = 50$.

The cumulative frequency just greater than 50 is 66 → Median class = 40–50.

$l = 40,\ cf = 46,\ f = 20,\ h = 10$

$$\text{Median} = l + \left(\frac{\dfrac{n}{2} - cf}{f}\right) \times h = 40 + \left(\frac{50 - 46}{20}\right) \times 10 = 40 + 2 = \textbf{42}$$

Source: Chapter 13, Sections 13.2 and 13.4

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• Mean: Use the Direct Method — multiply each class mark $x_i$ by its frequency $f_i$, sum the products, divide by $\Sigma f_i = 100$.
• Median: Build the cumulative frequency column, find $n/2 = 50$, identify the class whose cf first exceeds 50 (that's 40–50 with cf = 66), then apply the formula. Examiners expect the full table, identification of median class, and substitution into the formula with correct values of $l$, $cf$, $f$, $h$.