Mathematics — CBSE Class 10 board question

Step 1: Calculate Mean (Direct Method)

| Weight (kg) | f | x (mid-value) | fx |
|---|---|---|---|
| 38–40 | 3 | 39 | 117 |
| 40–42 | 2 | 41 | 82 |
| 42–44 | 4 | 43 | 172 |
| 44–46 | 5 | 45 | 225 |
| 46–48 | 14 | 47 | 658 |
| 48–50 | 4 | 49 | 196 |
| 50–52 | 3 | 51 | 153 |
| Total | 35 | | 1603 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{1603}{35} = 45.8 \text{ kg}$$

Step 2: Calculate Median

Cumulative frequencies: 3, 5, 9, 14, 28, 32, 35.

$n = 35$, so $\dfrac{n}{2} = 17.5$

The cumulative frequency just greater than 17.5 is 28 → Median class = 46–48

Here: $l = 46$, $cf = 14$, $f = 14$, $h = 2$

$$\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h = 46 + \left(\frac{17.5 - 14}{14}\right) \times 2$$

$$= 46 + \frac{3.5}{14} \times 2 = 46 + 0.5 = \textbf{46.5 kg}$$

Step 3: Difference

$$\text{Difference} = \text{Median} - \text{Mean} = 46.5 - 45.8 = \boxed{0.7 \text{ kg}}$$

Source: Statistics, Chapter 13 (Mean of Grouped Data §13.2; Median of Grouped Data §13.4)

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• Mean uses the direct formula $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$ with midpoints as $x_i$.
• Median requires building cumulative frequency, finding $n/2 = 17.5$, identifying the median class (where cf first exceeds 17.5), then applying the formula.
• The answer is Median − Mean = 0.7 kg (positive, since the distribution is skewed left by the dominant 46–48 class).
• Examiners award marks at each step: the table (1–2 marks), mean calculation (1 mark), median calculation with correct cf and formula (2 marks), and final difference (stated clearly).