At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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].
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
