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

Step 1: Find x

Total families = 200
∴ 24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 200
172 + x = 200 → x = 28

Step 2: Median (using step-deviation / cumulative frequency table, $h = 500$)

| Class | $f$ | $cf$ |
|-------|-----|------|
| 1000–1500 | 24 | 24 |
| 1500–2000 | 40 | 64 |
| 2000–2500 | 33 | 97 |
| 2500–3000 | 28 | 125 |
| 3000–3500 | 30 | 155 |
| 3500–4000 | 22 | 177 |
| 4000–4500 | 16 | 193 |
| 4500–5000 | 7 | 200 |

$\dfrac{n}{2} = 100$. The cf just exceeding 100 is 125 (class 2500–3000).

$l = 2500,\ cf = 97,\ f = 28,\ h = 500$

$$\text{Median} = 2500 + \frac{100 - 97}{28} \times 500 = 2500 + \frac{1500}{28} = 2500 + 53.57 \approx \textbf{₹2553.57}$$

Step 3: Mean (step-deviation method, $a = 2750,\ h = 500$)

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i-2750}{500}$ | $f_i u_i$ |
|-------|--------|--------|-------------------------------|-----------|
| 1000–1500 | 24 | 1250 | –3 | –72 |
| 1500–2000 | 40 | 1750 | –2 | –80 |
| 2000–2500 | 33 | 2250 | –1 | –33 |
| 2500–3000 | 28 | 2750 | 0 | 0 |
| 3000–3500 | 30 | 3250 | 1 | 30 |
| 3500–4000 | 22 | 3750 | 2 | 44 |
| 4000–4500 | 16 | 4250 | 3 | 48 |
| 4500–5000 | 7 | 4750 | 4 | 28 |
| Total | 200 | | | –35 |

$$\bar{x} = 2750 + \frac{-35}{200} \times 500 = 2750 - 87.5 = \textbf{₹2662.50}$$

Source: Chapter 13, Sections 13.2 & 13.4

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• Finding x: always start by using $\Sigma f = n$ (total). This fetches 1 mark.
• Median: identify median class where cumulative frequency first exceeds $n/2 = 100$; cf up to 2000–2500 is 97, so the next class 2500–3000 is the median class. Apply the standard formula carefully.
• Mean: step-deviation method is easiest with large $x_i$; choose $a$ as the middle class mark. Keep signs of $u_i$ correct — here $\Sigma f_i u_i$ is negative, so mean < assumed mean.
• Show the full table; examiners award marks for the working, not just the final answer.