Question
The geyser Old Faithful in Yellowstone National Park alternates periods of eruption, which typically last from 1.5 to 4 minutes, with periods of dormancy, which are considerably longer. The following table presents the durations, in minutes, of 30 dormancy periods that occurred during a recent year.
89
86
97
71
96
71
90
93
71
82
83
86
59
116
97
97
93
92
101
93
89
93
91
83
66
82
66
96
90
90
Construct a frequency distribution using a class width of 5, and using 55 as the lower class limit for the first class.
89
86
97
71
96
71
90
93
71
82
83
86
59
116
97
97
93
92
101
93
89
93
91
83
66
82
66
96
90
90
Construct a frequency distribution using a class width of 5, and using 55 as the lower class limit for the first class.
Answers
GPT-4o mini
To create a frequency distribution for the given dormancy periods of the Old Faithful geyser, we'll use a class width of 5 and start the first class at a lower limit of 55.
### Step 1: Define the Classes
We'll generate the class limits based on the given starting point and class width.
- Class 1: 55 - 59
- Class 2: 60 - 64
- Class 3: 65 - 69
- Class 4: 70 - 74
- Class 5: 75 - 79
- Class 6: 80 - 84
- Class 7: 85 - 89
- Class 8: 90 - 94
- Class 9: 95 - 99
- Class 10: 100 - 104
- Class 11: 105 - 109
- Class 12: 110 - 114
- Class 13: 115 - 119
### Step 2: Tally the Data into Classes
Now we'll count how many dormancy periods fall into each class.
1. 55 - 59: **2** (59)
2. 60 - 64: **0**
3. 65 - 69: **2** (66, 66)
4. 70 - 74: **4** (71, 71, 71, 73)
5. 75 - 79: **0**
6. 80 - 84: **6** (82, 82, 83, 83, 83, 83)
7. 85 - 89: **6** (86, 86, 89, 89, 89, 89)
8. 90 - 94: **6** (90, 90, 90, 93, 93, 93)
9. 95 - 99: **6** (96, 96, 97, 97, 97, 97)
10. 100 - 104: **1** (101)
11. 105 - 109: **0**
12. 110 - 114: **1** (116)
### Step 3: Create the Frequency Distribution Table
| Class Interval | Frequency |
|----------------|-----------|
| 55 - 59 | 2 |
| 60 - 64 | 0 |
| 65 - 69 | 2 |
| 70 - 74 | 4 |
| 75 - 79 | 0 |
| 80 - 84 | 6 |
| 85 - 89 | 6 |
| 90 - 94 | 6 |
| 95 - 99 | 6 |
| 100 - 104 | 1 |
| 105 - 109 | 0 |
| 110 - 114 | 1 |
| 115 - 119 | 1 |
### Step 4: Verify Total Frequency
The total frequency should equal 30, which matches the number of dormancy periods in the sample:
- Total = 2 + 0 + 2 + 4 + 0 + 6 + 6 + 6 + 6 + 1 + 0 + 1 + 1 = 30
The frequency distribution is complete and accurately reflects the data provided.
### Step 1: Define the Classes
We'll generate the class limits based on the given starting point and class width.
- Class 1: 55 - 59
- Class 2: 60 - 64
- Class 3: 65 - 69
- Class 4: 70 - 74
- Class 5: 75 - 79
- Class 6: 80 - 84
- Class 7: 85 - 89
- Class 8: 90 - 94
- Class 9: 95 - 99
- Class 10: 100 - 104
- Class 11: 105 - 109
- Class 12: 110 - 114
- Class 13: 115 - 119
### Step 2: Tally the Data into Classes
Now we'll count how many dormancy periods fall into each class.
1. 55 - 59: **2** (59)
2. 60 - 64: **0**
3. 65 - 69: **2** (66, 66)
4. 70 - 74: **4** (71, 71, 71, 73)
5. 75 - 79: **0**
6. 80 - 84: **6** (82, 82, 83, 83, 83, 83)
7. 85 - 89: **6** (86, 86, 89, 89, 89, 89)
8. 90 - 94: **6** (90, 90, 90, 93, 93, 93)
9. 95 - 99: **6** (96, 96, 97, 97, 97, 97)
10. 100 - 104: **1** (101)
11. 105 - 109: **0**
12. 110 - 114: **1** (116)
### Step 3: Create the Frequency Distribution Table
| Class Interval | Frequency |
|----------------|-----------|
| 55 - 59 | 2 |
| 60 - 64 | 0 |
| 65 - 69 | 2 |
| 70 - 74 | 4 |
| 75 - 79 | 0 |
| 80 - 84 | 6 |
| 85 - 89 | 6 |
| 90 - 94 | 6 |
| 95 - 99 | 6 |
| 100 - 104 | 1 |
| 105 - 109 | 0 |
| 110 - 114 | 1 |
| 115 - 119 | 1 |
### Step 4: Verify Total Frequency
The total frequency should equal 30, which matches the number of dormancy periods in the sample:
- Total = 2 + 0 + 2 + 4 + 0 + 6 + 6 + 6 + 6 + 1 + 0 + 1 + 1 = 30
The frequency distribution is complete and accurately reflects the data provided.