Advertisements
Advertisements
प्रश्न
Construct a frequency table for the following ages (in years) of 30 students using equal class-intervals, one of them being 9 – 12, where 12 is not included.
18, 12, 7, 6, 11, 15, 21, 9, 8, 13, 15, 17, 22, 19, 14, 21, 23, 8, 12, 16, 11, 21, 9, 16, 22, 23, 18, 6, 15, 17
Advertisements
उत्तर
Given: Data (30 ages): 18, 12, 7, 6, 11, 15, 21, 9, 8, 13, 15, 17, 22, 19, 14, 21, 23, 8, 12, 16, 11, 21, 9, 16, 22, 23, 18, 6, 15, 17.
Step-wise calculation:
1. Sort/inspect range:
Minimum = 6, Maximum = 23, so all values lie between 6 and 23.
2. Use the given class 9 – 12 (12 not included). Interpret classes as left-inclusive, right-exclusive [a, b):
Choose class width = 3 so classes are: 6 – 9, 9 – 12, 12 – 15, 15 – 18, 18 – 21, 21 – 24 (each includes the lower bound, excludes the upper).
3. Assign observations to classes and tally:
6 – 9 (includes 6, 7, 8):
6, 6, 7, 8, 8
Frequency = 5
9 – 12 (includes 9, 10, 11):
9, 9, 11, 11
Frequency = 4 (12 is excluded here)
12 – 15 (includes 12, 13, 14):
12, 12, 13, 14
Frequency = 4
15 – 18 (includes 15, 16, 17):
15, 15, 15, 16, 16, 17, 17
Frequency = 7
18 – 21 (includes 18, 19, 20):
18, 18, 19
Frequency = 3
21 – 24 (includes 21, 22, 23):
21, 21, 21, 22, 22, 23, 23
Frequency = 7
4. Check total: 5 + 4 + 4 + 7 + 3 + 7 = 30 (matches number of students).
Frequency table (class intervals are [lower, upper) with upper not included):
| Class | Number of students |
| 6 – 9 | 5 |
| 9 – 12 | 4 |
| 12 – 15 | 4 |
| 15 – 18 | 7 |
| 18 – 21 | 3 |
| 21 – 24 | 7 |
| Total | 30 |
