The following table shows the number of patients of different age group who were discharged from the hospital in a particular month :
Find the 'mean' and the 'mode' of the above data.
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Mean:
| Age | $f_i$ | $x_i$ (mid-value) | $f_i x_i$ |
|---|---|---|---|
| 5–15 | 6 | 10 | 60 |
| 15–25 | 11 | 20 | 220 |
| 25–35 | 21 | 30 | 630 |
| 35–45 | 23 | 40 | 920 |
| 45–55 | 14 | 50 | 700 |
| 55–65 | 5 | 60 | 300 |
| Total | 80 | | 2830 |
$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{2830}{80} = \textbf{35.375 years}$$
Mode:
The class with the highest frequency is 35–45 (frequency = 23).
So, Modal class = 35–45, $l = 35$, $f_1 = 23$, $f_0 = 21$, $f_2 = 14$, $h = 10$
$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$
$$= 35 + \frac{23 - 21}{2(23) - 21 - 14} \times 10 = 35 + \frac{2}{11} \times 10 = 35 + 1.82 = \textbf{36.82 years}$$
Source: Exercise 13.2, Q.1; Chapter 13
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Explanation
- Mean: Use Direct Method — find mid-value ($x_i$) of each class, multiply by frequency, sum up, then divide by total frequency (80).
- Mode: Identify the modal class (highest frequency = 23, class 35–45). Apply the formula with $l$ = lower limit, $f_1$ = modal class frequency, $f_0$ = preceding class frequency, $f_2$ = succeeding class frequency, $h$ = class width.
- Examiners expect the full table for mean (awarded ~2 marks) and correct substitution in mode formula (~2 marks) + final answers (~1 mark). Never skip the formula step.