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प्रश्न
Find the mode for the following frequency distribution.
| Class | 1 – 10 | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 |
| Frequency | 2 | 3 | 5 | 7 | 1 |
Use formula: Mode = `L = (f_0 - f_1)/(2f_0 - f_1 - f_2) xx h`
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उत्तर
Step 1: Convert discontinuous classes to continuous classes
Since the class intervals are discontinuous (1 – 10, 11 – 20, etc.), we must make them continuous by subtracting 0.5 from the lower limits and adding 0.5 to the upper limits.
| Original Class | Continuous Class | Frequency |
| 1 – 10 | 0.5 – 10.5 | 2 |
| 11 – 20 | 10.5 – 20.5 | 3 |
| 21 – 30 | 20.5 – 30.5 | 5 |
| 31 – 40 | 30.5 – 40.5 | 7 |
| 41 – 50 | 40.5 – 50.5 | 1 |
Step 2: Identify the variables for the formula
From the table, the highest frequency is 7, which makes 30.5 – 40.5 our modal class.
Using the notation provided in your formula:
L (Lower boundary of the modal class) = 30.5
f0 (Frequency of the modal class) = 7
f1 (Frequency of the preceding class) = 5
f2 (Frequency of the succeeding class) = 1
h (Class width) = 40.5 – 30.5 = 10
Substitute these values into the standard mode formula:
Mode = `L + ((f_0 - f_1)/(2f_0 - f_1 - f_2)) xx h`
Mode = `30.5 + ((7 - 5)/(2(7) - 5 - 1)) xx 10`
Mode = `30.5 + (2/(14 - 6)) xx 10`
Mode = `30.5 + (2/8) xx 10`
Mode = 30.5 + 0.25 × 10
Mode = 30.5 + 2.5
Mode = 33
