Question 1184005: Use the given frequency distribution to find the
Class width
Class midpoints
Class boundaries
Temperature 35-39,40-44,45-49,50-54,55-59,60-64,65-69
Frequency 1,3,5,11,7,7,7,1
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's the breakdown of how to find the class width, midpoints, and boundaries for the given frequency distribution:
**1. Class Width:**
The class width is the difference between the upper and lower limits of consecutive class intervals. It should be consistent across all classes. In your example:
* 40 - 35 = 5
* 45 - 40 = 5
* And so on...
Therefore, the class width is **5**.
**2. Class Midpoints:**
The class midpoint is the average of the upper and lower limits of each class interval.
* 35-39: (35 + 39) / 2 = **37**
* 40-44: (40 + 44) / 2 = **42**
* 45-49: (45 + 49) / 2 = **47**
* 50-54: (50 + 54) / 2 = **52**
* 55-59: (55 + 59) / 2 = **57**
* 60-64: (60 + 64) / 2 = **62**
* 65-69: (65 + 69) / 2 = **67**
**3. Class Boundaries:**
Class boundaries are the values that separate the classes. They are halfway between the upper limit of one class and the lower limit of the next class.
* The lower boundary of the first class is typically found by subtracting half the class width from the lowest limit of the first class: 35 - (5/2) = 32.5. However, since your data starts at 35, we can use 34.5 to ensure all data points are included.
* The upper boundary of the last class is found by adding half the class width to the highest limit of the last class: 69 + (5/2) = 71.5. However, since your data ends at 69, we can use 69.5 to ensure all data points are included.
Here's a complete table summarizing the results:
| Temperature | Frequency | Class Midpoint | Class Boundaries |
|---|---|---|---|
| 35-39 | 1 | 37 | 34.5 - 39.5 |
| 40-44 | 3 | 42 | 39.5 - 44.5 |
| 45-49 | 5 | 47 | 44.5 - 49.5 |
| 50-54 | 11 | 52 | 49.5 - 54.5 |
| 55-59 | 7 | 57 | 54.5 - 59.5 |
| 60-64 | 7 | 62 | 59.5 - 64.5 |
| 65-69 | 7 | 67 | 64.5 - 69.5 |
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