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Sure! Let's carefully go through the steps to solve the problem.
### Step 1: Organize the Data
The data is already given in a table format:
| Temperature [tex]\( \left({ }^{\circ} C \right) \)[/tex] | Frequency |
|----------------------------------------------|-----------|
| [tex]\(14 < t \leq 18\)[/tex] | 4 |
| [tex]\(18 < t \leq 20\)[/tex] | 10 |
| [tex]\(20 < t \leq 22\)[/tex] | 8 |
| [tex]\(22 < t \leq 24\)[/tex] | 5 |
| [tex]\(24 < t \leq 28\)[/tex] | 3 |
### Step 2: Calculate the Total Number of Observations
To find the total number of observations, we sum up all the frequencies:
[tex]\[ 4 + 10 + 8 + 5 + 3 = 30 \][/tex]
So, the total number of observations is 30.
### Step 3: Calculate the Cumulative Frequency
Next, we need to calculate the cumulative frequency for each class interval:
| Class Interval | Frequency | Cumulative Frequency |
|---------------------|-----------|----------------------|
| [tex]\(14 < t \leq 18\)[/tex] | 4 | 4 |
| [tex]\(18 < t \leq 20\)[/tex] | 10 | 4 + 10 = 14 |
| [tex]\(20 < t \leq 22\)[/tex] | 8 | 14 + 8 = 22 |
| [tex]\(22 < t \leq 24\)[/tex] | 5 | 22 + 5 = 27 |
| [tex]\(24 < t \leq 28\)[/tex] | 3 | 27 + 3 = 30 |
### Step 4: Find the Median Class Interval
The median position is calculated as:
[tex]\[ \frac{\text{Total Number of Observations}}{2} = \frac{30}{2} = 15 \][/tex]
Now, we look for the class interval in which the cumulative frequency just surpasses the median position of 15:
- Cumulative frequency of the first class interval (4) is less than 15.
- Cumulative frequency of the second class interval (14) is still less than 15.
- Cumulative frequency of the third class interval (22) is greater than 15.
So, the median class interval is [tex]\(20 < t \leq 22\)[/tex].
### Step 5: Calculate an Estimate for the Mean Maximum Temperature
To estimate the mean, we first need the midpoints of each class interval. These midpoints are calculated as follows:
- For [tex]\(14 < t \leq 18\)[/tex]: midpoint is [tex]\( \frac{14+18}{2} = 16.0\)[/tex]
- For [tex]\(18 < t \leq 20\)[/tex]: midpoint is [tex]\( \frac{18+20}{2} = 19.0\)[/tex]
- For [tex]\(20 < t \leq 22\)[/tex]: midpoint is [tex]\( \frac{20+22}{2} = 21.0\)[/tex]
- For [tex]\(22 < t \leq 24\)[/tex]: midpoint is [tex]\( \frac{22+24}{2} = 23.0\)[/tex]
- For [tex]\(24 < t \leq 28\)[/tex]: midpoint is [tex]\( \frac{24+28}{2} = 26.0\)[/tex]
Next, we calculate the total sum of the products of frequencies and their corresponding midpoints:
[tex]\[ (4 \times 16.0) + (10 \times 19.0) + (8 \times 21.0) + (5 \times 23.0) + (3 \times 26.0) \][/tex]
[tex]\[ = 64.0 + 190.0 + 168.0 + 115.0 + 78.0 = 615.0 \][/tex]
Finally, the estimated mean maximum temperature is:
[tex]\[ \frac{\text{Total Sum of the Products}}{\text{Total Number of Observations}} = \frac{615.0}{30} = 20.5^\circ C \][/tex]
### Summary of Results
- The class interval that contains the median is [tex]\(20 < t \leq 22\)[/tex].
- The estimated mean maximum temperature is [tex]\(20.5^\circ C\)[/tex].
### Step 1: Organize the Data
The data is already given in a table format:
| Temperature [tex]\( \left({ }^{\circ} C \right) \)[/tex] | Frequency |
|----------------------------------------------|-----------|
| [tex]\(14 < t \leq 18\)[/tex] | 4 |
| [tex]\(18 < t \leq 20\)[/tex] | 10 |
| [tex]\(20 < t \leq 22\)[/tex] | 8 |
| [tex]\(22 < t \leq 24\)[/tex] | 5 |
| [tex]\(24 < t \leq 28\)[/tex] | 3 |
### Step 2: Calculate the Total Number of Observations
To find the total number of observations, we sum up all the frequencies:
[tex]\[ 4 + 10 + 8 + 5 + 3 = 30 \][/tex]
So, the total number of observations is 30.
### Step 3: Calculate the Cumulative Frequency
Next, we need to calculate the cumulative frequency for each class interval:
| Class Interval | Frequency | Cumulative Frequency |
|---------------------|-----------|----------------------|
| [tex]\(14 < t \leq 18\)[/tex] | 4 | 4 |
| [tex]\(18 < t \leq 20\)[/tex] | 10 | 4 + 10 = 14 |
| [tex]\(20 < t \leq 22\)[/tex] | 8 | 14 + 8 = 22 |
| [tex]\(22 < t \leq 24\)[/tex] | 5 | 22 + 5 = 27 |
| [tex]\(24 < t \leq 28\)[/tex] | 3 | 27 + 3 = 30 |
### Step 4: Find the Median Class Interval
The median position is calculated as:
[tex]\[ \frac{\text{Total Number of Observations}}{2} = \frac{30}{2} = 15 \][/tex]
Now, we look for the class interval in which the cumulative frequency just surpasses the median position of 15:
- Cumulative frequency of the first class interval (4) is less than 15.
- Cumulative frequency of the second class interval (14) is still less than 15.
- Cumulative frequency of the third class interval (22) is greater than 15.
So, the median class interval is [tex]\(20 < t \leq 22\)[/tex].
### Step 5: Calculate an Estimate for the Mean Maximum Temperature
To estimate the mean, we first need the midpoints of each class interval. These midpoints are calculated as follows:
- For [tex]\(14 < t \leq 18\)[/tex]: midpoint is [tex]\( \frac{14+18}{2} = 16.0\)[/tex]
- For [tex]\(18 < t \leq 20\)[/tex]: midpoint is [tex]\( \frac{18+20}{2} = 19.0\)[/tex]
- For [tex]\(20 < t \leq 22\)[/tex]: midpoint is [tex]\( \frac{20+22}{2} = 21.0\)[/tex]
- For [tex]\(22 < t \leq 24\)[/tex]: midpoint is [tex]\( \frac{22+24}{2} = 23.0\)[/tex]
- For [tex]\(24 < t \leq 28\)[/tex]: midpoint is [tex]\( \frac{24+28}{2} = 26.0\)[/tex]
Next, we calculate the total sum of the products of frequencies and their corresponding midpoints:
[tex]\[ (4 \times 16.0) + (10 \times 19.0) + (8 \times 21.0) + (5 \times 23.0) + (3 \times 26.0) \][/tex]
[tex]\[ = 64.0 + 190.0 + 168.0 + 115.0 + 78.0 = 615.0 \][/tex]
Finally, the estimated mean maximum temperature is:
[tex]\[ \frac{\text{Total Sum of the Products}}{\text{Total Number of Observations}} = \frac{615.0}{30} = 20.5^\circ C \][/tex]
### Summary of Results
- The class interval that contains the median is [tex]\(20 < t \leq 22\)[/tex].
- The estimated mean maximum temperature is [tex]\(20.5^\circ C\)[/tex].
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