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प्रश्न
The water bills (in rupees) of 32 houses in a certain street for the period 1.1.98 to. 31.3.98 are given below:
56, 43, 32, 38, 56, 24, 68, 85, 52, 47, 35, 58, 63, 74, 27, 84, 69, 35, 44, 75, 55, 30, 54, 65, 45, 67, 95, 72, 43, 65, 35, 59.
Tabulate the data and resent the data as a cumulative frequency table using 70-79 as one of the class intervals.
उत्तर
The minimum bill is 24. Rs
Maximum bill is Rs 95.
Range = maximum bill – minimum bill
= 95-24
=71
Given class in terval is 70-79, So class size
= 79-70 = 9
∴ Number of classes =`"Range"/"Class size"=71/9`
=7.88
∴ Number of classes = 8
The cumulative frequency distribution table is as
| Bills | No. of houses (Frequency) | Cumulative frequency |
| 16-25 | 1 | 1 |
| 25-34 | 3 | 4 |
| 34-43 | 5 | 9 |
| 43-52 | 4 | 13 |
| 52-61 | 7 | 20 |
| 61-70 | 6 | 26 |
| 70-79 | 3 | 29 |
| 79-88 | 2 | 31 |
| 88-97 | 1 | 32 |
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