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प्रश्न
Construct a frequency distribution table from the following cumulative frequency distribution:
Class Interval | Cumulative Frequency |
10 - 19 | 8 |
20 - 29 | 19 |
30- 39 | 23 |
40- 49 | 30 |
उत्तर
The frequency distribution table is
C. I | c.f |
10 - 19 | 8 |
20 - 29 | 11 |
30 - 39 | 4 |
40 - 49 | 7 |
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संबंधित प्रश्न
For class interval 20-25 write the lower class limit and the upper class limit.
Complete the Following Table.
Classes (age) | Tally marks | Frequency (No. of students) |
12-13 | `cancel(bb|bb|bb|bb|)` | `square` |
13-14 | `cancel(bb|bb|bb|bb|)` `cancel(bb|bb|bb|bb|)` `bb|bb|bb|bb|` | `square` |
14-15 | `square` | `square` |
15-16 | `bb|bb|bb|bb|` | `square` |
N = ∑f = 35 |
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3.14159265358979323846264338327950288419716939937510
From this information prepare an ungrouped frequency distribution table of digits appearing after the decimal point.
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Class width | Frequency |
5 | 3 |
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35 | 13 |
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Class width | Frequency |
22 | 6 |
24 | 7 |
26 | 13 |
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The monthly maximum temperature of a city is given in degree celcius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5, 29.0, 29.5, 29.9, 33.2, 30.2
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3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution table of digits from 0 to 9 after the decimal place.
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Construct a cumulative frequency distribution table from the frequency table given below:
( i )
Class Interval | Frequency |
0 - 8 | 9 |
8 - 16 | 13 |
16 - 24 | 12 |
24 - 32 | 7 |
32 - 40 | 15 |
( ii )
Class Interval | Frequency |
1 - 10 | 12 |
11 - 20 | 18 |
21 - 30 | 23 |
31 - 40 | 15 |
41 - 50 | 10 |
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1, 3, 0, 2, 5, 2, 3, 4, 1, 0, 5, 4, 3, 1, 3, 2, 5, 2, 1, 1, 2, 6, 2, 1, 4
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Frequency | 5 | 10 | 8 | 6 | 12 | 9 |
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Using the following frequency table.
Marks (obtained out of 10) | 4 | 5 | 7 | 8 | 9 | 10 |
Frequency | 5 | 10 | 8 | 6 | 12 | 9 |
10 marks the highest frequency.