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Sagot :
Let's first understand the data provided in the table and create a histogram to represent it.
### Frequency Table
| Area (sq. ft) | Number of Bedrooms |
|---------------------|-------------------|
| 60 ≤ A < 80 | 4 |
| 80 ≤ A < 100 | 6 |
| 100 ≤ A < 120 | 5 |
| 120 ≤ A < 140 | 3 |
| 140 ≤ A < 160 | 1 |
### Histogram Representation
To plot a histogram, we can use the midpoints of each interval as the x-values, and the frequencies as the y-values.
- Midpoint of 60 ≤ A < 80: [tex]\((60 + 80)/2 = 70\)[/tex]
- Midpoint of 80 ≤ A < 100: [tex]\((80 + 100)/2 = 90\)[/tex]
- Midpoint of 100 ≤ A < 120: [tex]\((100 + 120)/2 = 110\)[/tex]
- Midpoint of 120 ≤ A < 140: [tex]\((120 + 140)/2 = 130\)[/tex]
- Midpoint of 140 ≤ A < 160: [tex]\((140 + 160)/2 = 150\)[/tex]
So, the midpoints and their corresponding frequencies are:
| Midpoints ([tex]\(A\)[/tex]) | Frequencies |
|-------------------|-------------|
| 70 | 4 |
| 90 | 6 |
| 110 | 5 |
| 130 | 3 |
| 150 | 1 |
### Creating the Histogram
By plotting the midpoints on the x-axis and the frequencies on the y-axis, we observe the following:
- The frequency is higher for the smallest midpoints and decreases as the area increases.
Visually, this would result in a histogram where the bars are highest on the left side and gradually decrease towards the right side, indicating that the histogram is right-skewed.
### Analysis of Mean and Median
For a right-skewed distribution:
- The mean is pulled towards the higher end (right side) due to the presence of higher values.
- The median is less affected by the extreme values.
Thus, the mean tends to be greater than the median in a right-skewed distribution.
### Conclusion
Based on the shape of the histogram and the nature of right-skewed distributions, the most likely true statement is:
The histogram is right-skewed, so the mean is greater than the median.
### Frequency Table
| Area (sq. ft) | Number of Bedrooms |
|---------------------|-------------------|
| 60 ≤ A < 80 | 4 |
| 80 ≤ A < 100 | 6 |
| 100 ≤ A < 120 | 5 |
| 120 ≤ A < 140 | 3 |
| 140 ≤ A < 160 | 1 |
### Histogram Representation
To plot a histogram, we can use the midpoints of each interval as the x-values, and the frequencies as the y-values.
- Midpoint of 60 ≤ A < 80: [tex]\((60 + 80)/2 = 70\)[/tex]
- Midpoint of 80 ≤ A < 100: [tex]\((80 + 100)/2 = 90\)[/tex]
- Midpoint of 100 ≤ A < 120: [tex]\((100 + 120)/2 = 110\)[/tex]
- Midpoint of 120 ≤ A < 140: [tex]\((120 + 140)/2 = 130\)[/tex]
- Midpoint of 140 ≤ A < 160: [tex]\((140 + 160)/2 = 150\)[/tex]
So, the midpoints and their corresponding frequencies are:
| Midpoints ([tex]\(A\)[/tex]) | Frequencies |
|-------------------|-------------|
| 70 | 4 |
| 90 | 6 |
| 110 | 5 |
| 130 | 3 |
| 150 | 1 |
### Creating the Histogram
By plotting the midpoints on the x-axis and the frequencies on the y-axis, we observe the following:
- The frequency is higher for the smallest midpoints and decreases as the area increases.
Visually, this would result in a histogram where the bars are highest on the left side and gradually decrease towards the right side, indicating that the histogram is right-skewed.
### Analysis of Mean and Median
For a right-skewed distribution:
- The mean is pulled towards the higher end (right side) due to the presence of higher values.
- The median is less affected by the extreme values.
Thus, the mean tends to be greater than the median in a right-skewed distribution.
### Conclusion
Based on the shape of the histogram and the nature of right-skewed distributions, the most likely true statement is:
The histogram is right-skewed, so the mean is greater than the median.
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