Answer :
Sure! Let's go through the steps to complete the grouped frequency distribution and draw the frequency polygon for the hottest recorded temperatures in the sixteen cities.
### (a) Completing the Grouped Frequency Distribution
We have the hottest recorded temperatures:
- 93, 105, 110, 107, 103, 106, 94, 102,
- 99, 101, 110, 108, 109, 94, 104, 112
The task is to categorize these temperatures into groups (class intervals) with a class width of 4. The intervals provided are:
1. 92.5 to 96.5
2. 96.5 to 100.5
3. 100.5 to 104.5
4. 104.5 to 108.5
5. 108.5 to 112.5
Each of these class intervals has a specific range, and we need to count how many temperatures fall into each range.
- 92.5 to 96.5: This interval includes 93 and 94 (total of 3 temperatures).
- 96.5 to 100.5: This interval includes only 99 (total of 1 temperature).
- 100.5 to 104.5: This interval includes 103, 102, and 104 (total of 4 temperatures).
- 104.5 to 108.5: This interval includes 105, 107, 106, and 108 (total of 4 temperatures).
- 108.5 to 112.5: This interval includes 110, 110, 109, and 112 (total of 4 temperatures).
Here is the completed frequency distribution:
| Temperatures (in °F) | Frequency |
|----------------------|-----------|
| 92.5 to 96.5 | 3 |
| 96.5 to 100.5 | 1 |
| 100.5 to 104.5 | 4 |
| 104.5 to 108.5 | 4 |
| 108.5 to 112.5 | 4 |
### (b) Drawing the Frequency Polygon
To draw the frequency polygon, we need the midpoints of each class interval:
- Midpoint of 92.5 to 96.5: [tex]\( \frac{92.5 + 96.5}{2} = 94.5 \)[/tex]
- Midpoint of 96.5 to 100.5: [tex]\( \frac{96.5 + 100.5}{2} = 98.5 \)[/tex]
- Midpoint of 100.5 to 104.5: [tex]\( \frac{100.5 + 104.5}{2} = 102.5 \)[/tex]
- Midpoint of 104.5 to 108.5: [tex]\( \frac{104.5 + 108.5}{2} = 106.5 \)[/tex]
- Midpoint of 108.5 to 112.5: [tex]\( \frac{108.5 + 112.5}{2} = 110.5 \)[/tex]
Now plot these midpoints against their corresponding frequencies on a graph and connect the points with straight lines to form the frequency polygon. Make sure to label each class with its midpoint on the graph.
This completes the solution for the problem!
### (a) Completing the Grouped Frequency Distribution
We have the hottest recorded temperatures:
- 93, 105, 110, 107, 103, 106, 94, 102,
- 99, 101, 110, 108, 109, 94, 104, 112
The task is to categorize these temperatures into groups (class intervals) with a class width of 4. The intervals provided are:
1. 92.5 to 96.5
2. 96.5 to 100.5
3. 100.5 to 104.5
4. 104.5 to 108.5
5. 108.5 to 112.5
Each of these class intervals has a specific range, and we need to count how many temperatures fall into each range.
- 92.5 to 96.5: This interval includes 93 and 94 (total of 3 temperatures).
- 96.5 to 100.5: This interval includes only 99 (total of 1 temperature).
- 100.5 to 104.5: This interval includes 103, 102, and 104 (total of 4 temperatures).
- 104.5 to 108.5: This interval includes 105, 107, 106, and 108 (total of 4 temperatures).
- 108.5 to 112.5: This interval includes 110, 110, 109, and 112 (total of 4 temperatures).
Here is the completed frequency distribution:
| Temperatures (in °F) | Frequency |
|----------------------|-----------|
| 92.5 to 96.5 | 3 |
| 96.5 to 100.5 | 1 |
| 100.5 to 104.5 | 4 |
| 104.5 to 108.5 | 4 |
| 108.5 to 112.5 | 4 |
### (b) Drawing the Frequency Polygon
To draw the frequency polygon, we need the midpoints of each class interval:
- Midpoint of 92.5 to 96.5: [tex]\( \frac{92.5 + 96.5}{2} = 94.5 \)[/tex]
- Midpoint of 96.5 to 100.5: [tex]\( \frac{96.5 + 100.5}{2} = 98.5 \)[/tex]
- Midpoint of 100.5 to 104.5: [tex]\( \frac{100.5 + 104.5}{2} = 102.5 \)[/tex]
- Midpoint of 104.5 to 108.5: [tex]\( \frac{104.5 + 108.5}{2} = 106.5 \)[/tex]
- Midpoint of 108.5 to 112.5: [tex]\( \frac{108.5 + 112.5}{2} = 110.5 \)[/tex]
Now plot these midpoints against their corresponding frequencies on a graph and connect the points with straight lines to form the frequency polygon. Make sure to label each class with its midpoint on the graph.
This completes the solution for the problem!
