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प्रश्न
Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included).
319, 320, 300, 290, 311, 242, 272, 220, 268, 258, 242, 210, 268, 215, 306, 316, 280, 240, 210, 278, 254, 304, 302, 318, 306, 292, 236, 256.
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उत्तर
Given:
Data (monthly wages, in rupees) for 28 labourers: 319, 320, 300, 290, 311, 242, 272, 220, 268, 258, 242, 210, 268, 215, 306, 316, 280, 240, 210, 278, 254, 304, 302, 318, 306, 292, 236, 256.
Step-wise calculation:
1. Arrange/find extremes:
Minimum = 210
Maximum = 320
Range = 320 – 210 = 110
2. Class width:
One class is given as 210 – 230 (230 not included) class width h = 20.
3. Number of classes:
Required classes
= `"Range"/h`
= `110/20`
= 5.5
Take 6 classes.
4. Choose classes starting at 210 (lower limit inclusive) with width 20 (upper limit exclusive):
210 – 230, 230 – 250, 250 – 270, 270 – 290, 290 – 310, 310 – 330
5. Tally frequencies (include value x if lower ≤ x < upper):
210 – 230: 210, 210, 215, 220 → Frequency = 4
230 – 250: 236, 240, 242, 242 → Frequency = 4
250 – 270: 254, 256, 258, 268, 268 → Frequency = 5
270 – 290: 272, 278, 280 → Frequency = 3 (290 is excluded here)
290 – 310: 290, 292, 300, 302, 304, 306, 306 → Frequency = 7
310 – 330: 311, 316, 318, 319, 320 → Frequency = 5
Check: 4 + 4 + 5 + 3 + 7 + 5 = 28 (total observations)
Frequency table (equal class intervals; upper limits not included):
| Class | Frequency | Cumulative frequency |
| 210 – 230 | 4 | 4 |
| 230 – 250 | 4 | 8 |
| 250 – 270 | 5 | 13 |
| 270 – 290 | 3 | 16 |
| 290 – 310 | 7 | 23 |
| 310 – 330 | 5 | 28 |
