The following frequency table shows monthly rainfall (in cm) for a village over six months: | Class | Frequency | |-------|-----------| | 10-20 | 2 | | 20-30 | 5 | | 30-40 | 9 | | 40-50 | 7 | | 50-60 | 3 | | 60-70 | 4 | Find the class mark of the modal class and compute the approximate mean rainfall using the mid‑point method.
Answer
35 cm, 40.33 cm
The correct answer is 35 cm, 40.33 cm — Note: The class mark of the modal class is 35 cm and the approximate mean is 40.33 cm.
Step-by-step solution
- Identify the modal class: it is the class with the highest frequency, which is 30‑40 cm with a frequency of 9.
- Calculate the class mark (mid‑point) of the modal class: (30+40) ÷ 2 = 35 cm. This is the required class mark.
- Compute the product of class mark and frequency for each class: - 10‑20: 15 × 2 = 30 - 20‑30: 25 × 5 = 125 - 30‑40: 35 × 9 = 315 - 40‑50: 45 × 7 = 315 - 50‑60: 55 × 3 = 165 - 60‑70: 65 × 4 = 260 These products are needed for the mid‑point mean calculation.
- Sum all the products: 30+125+315+315+165+260 = 1210. This is the numerator of the mean.
- Sum all frequencies: 2+5+9+7+3+4 = 30. This is the denominator of the mean.
- Divide the numerator by the denominator: 1210 ÷ 30 ≈ 40.33 cm. This is the approximate mean rainfall.
- Note: The class mark of the modal class is 35 cm and the approximate mean is 40.33 cm.
Common mistake: Using the lower or upper bound as the class mark, or mis‑adding the products or frequencies
Concept tested: Mid‑point mean and modal class
Published · How this solution was prepared and checked
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