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

Mean (using assumed mean method, a = 52.5, h = 15):

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i - 52.5}{15}$ | $f_i u_i$ |
|-------|--------|--------|-------------------------------|-----------|
| 0–15 | 4 | 7.5 | –3 | –12 |
| 15–30 | 8 | 22.5 | –2 | –16 |
| 30–45 | 11 | 37.5 | –1 | –11 |
| 45–60 | 14 | 52.5 | 0 | 0 |
| 60–75 | 10 | 67.5 | 1 | 10 |
| 75–90 | 7 | 82.5 | 2 | 14 |
| 90–105| 6 | 97.5 | 3 | 18 |
| Total | 60 | | | 3 |

$$\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 52.5 + \frac{3}{60} \times 15 = 52.5 + 0.75 = \textbf{53.25}$$

Mode:

The highest frequency is 14, so the modal class = 45–60.

$l = 45,\ f_1 = 14,\ f_0 = 11,\ f_2 = 10,\ h = 15$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 45 + \frac{14-11}{28-11-10} \times 15 = 45 + \frac{3}{7} \times 15 = 45 + 6.43 = \textbf{51.43}$$

Mean = 53.25, Mode = 51.43

Source: Chapter 13, Section 13.2 (Mean) and Section 13.3 (Mode)

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• Mean: The step-deviation (assumed mean) method is most efficient here since all class widths are equal (h = 15). Take a = 52.5 (midpoint of middle class). Calculate $u_i$ and $f_i u_i$, sum them, and apply the formula.
• Mode: Identify the modal class as the one with the highest frequency (14 → class 45–60). Then substitute $l, f_1, f_0, f_2, h$ into the mode formula. Examiners award marks for correctly identifying the modal class, writing the formula with correct substitution, and the final answer.
• Show the table neatly — 1–2 marks are often for working, not just the final answer.