1. **Problem Statement:** Given the rainfall data for different sub-divisions in 2023, find: (i) The modal class. (ii) (a) The median rainfall or (b) the mean rainfall. (iii) The number of sub-divisions with rainfall at least 800 mm. 2. **Data Table:** | Rainfall (mm) | Frequency (f) | |---------------|--------------| | 200 – 400 | 3 | | 400 – 600 | 4 | | 600 – 800 | 7 | | 800 – 1000 | 4 | | 1000 – 1200 | 3 | | 1200 – 1400 | 3 | 3. **(i) Modal Class:** The modal class is the class interval with the highest frequency. Here, maximum frequency is 7 corresponding to 600 – 800 mm. **Answer:** Modal class = 600 – 800 mm. 4. **(ii) (a) Median Calculation:** - Total frequency $N = 3+4+7+4+3+3 = 24$. - Median class is the class where cumulative frequency just exceeds $\frac{N}{2} = 12$. Cumulative frequencies: - 200–400: 3 - 400–600: 3+4=7 - 600–800: 7+7=14 (exceeds 12, so median class) - Median class = 600 – 800 - Lower boundary $l = 600$ - Frequency of median class $f_m = 7$ - Cumulative frequency before median class $F = 7$ - Class width $h = 200$ Median formula: $$\text{Median} = l + \left(\frac{\frac{N}{2} - F}{f_m}\right) \times h$$ Substitute values: $$= 600 + \left(\frac{12 - 7}{7}\right) \times 200 = 600 + \frac{5}{7} \times 200 = 600 + 142.86 = 742.86$$ **Answer:** Median rainfall = 742.86 mm. 5. **(ii) (b) Mean Calculation:** - Find midpoints $x_i$ of each class: - 200–400: 300 - 400–600: 500 - 600–800: 700 - 800–1000: 900 - 1000–1200: 1100 - 1200–1400: 1300 - Calculate $f_i x_i$: - $3 \times 300 = 900$ - $4 \times 500 = 2000$ - $7 \times 700 = 4900$ - $4 \times 900 = 3600$ - $3 \times 1100 = 3300$ - $3 \times 1300 = 3900$ - Sum of $f_i x_i = 900 + 2000 + 4900 + 3600 + 3300 + 3900 = 18600$ - Total frequency $N = 24$ Mean formula: $$\text{Mean} = \frac{\sum f_i x_i}{N} = \frac{18600}{24} = 775$$ **Answer:** Mean rainfall = 775 mm. 6. **(iii) Number of good rainfall sub-divisions:** - Good rainfall means rainfall $\geq 800$ mm. - Classes with rainfall $\geq 800$ mm: 800–1000, 1000–1200, 1200–1400. - Frequencies: 4 + 3 + 3 = 10 **Answer:** 10 sub-divisions had good rainfall.