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

Finding the missing frequency f:

| Class | $x_i$ | $f_i$ | $f_i x_i$ |
|-------|--------|--------|------------|
| 11–13 | 12 | 7 | 84 |
| 13–15 | 14 | 6 | 84 |
| 15–17 | 16 | 9 | 144 |
| 17–19 | 18 | 13 | 234 |
| 19–21 | 20 | f | 20f |
| 21–23 | 22 | 5 | 110 |
| 23–25 | 24 | 4 | 96 |
| Total | | $44 + f$ | $752 + 20f$ |

Given $\bar{x} = 18$:

$$18 = \frac{752 + 20f}{44 + f}$$

$$18(44 + f) = 752 + 20f$$

$$792 + 18f = 752 + 20f$$

$$40 = 2f \implies \boxed{f = 20}$$

Finding the Mode:

The highest frequency is 20, so the modal class is 19–21.

Here, $l = 19$, $h = 2$, $f_1 = 20$, $f_0 = 13$, $f_2 = 5$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 19 + \frac{20-13}{40-13-5} \times 2$$

$$= 19 + \frac{7}{22} \times 2 = 19 + \frac{7}{11} = 19 + 0.636 \approx \boxed{19.64}$$

Source: Chapter 13, Sections 13.2 and 13.3

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• Finding f: Use the direct method formula $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$. Set it equal to 18 and solve the linear equation for f.
• Modal class: The class with the highest frequency (here, 19–21 with f = 20) is the modal class.
• Mode formula: Examiners expect you to correctly identify $f_0$ (preceding class frequency = 13) and $f_2$ (succeeding class frequency = 5). Show substitution clearly for full marks.
• Write the table neatly — it earns method marks even if arithmetic slips occur.