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Sagot :
To answer the question step-by-step, we need to analyze the data provided and create a histogram. From the histogram, we can determine the skewness of the data and estimate the relationship between the mean and the median.
### Data Summary
| Area (sq. ft) | Number of Bedrooms |
|-----------------------|-------------------|
| [tex]\(60 \leq A < 80\)[/tex] | 4 |
| [tex]\(80 \leq A < 100\)[/tex] | 6 |
| [tex]\(100 \leq A < 120\)[/tex] | 5 |
| [tex]\(120 \leq A < 140\)[/tex] | 3 |
| [tex]\(140 \leq A < 160\)[/tex] | 1 |
### Step 1: Create a Histogram
A histogram groups data into bins (ranges) and plots the frequency of each bin. Here is the histogram for the given data ranges:
| Area Range (sq. ft) | Frequency |
|-----------------------|-----------|
| [tex]\(60 \leq A < 80\)[/tex] | 4 |
| [tex]\(80 \leq A < 100\)[/tex] | 6 |
| [tex]\(100 \leq A < 120\)[/tex] | 5 |
| [tex]\(120 \leq A < 140\)[/tex] | 3 |
| [tex]\(140 \leq A < 160\)[/tex] | 1 |
```
Frequency
6 | x
5 | x x x
4 | x x x x
3 | x x x
2 | x
1 | x
------------------------------------------------------
60-80 80-100 100-120 120-140 140-160 (Area in sq. ft.)
```
### Step 2: Determine the Mean
The mean can be estimated by calculating the weighted average of the midpoint of each range:
- Midpoint of [tex]\(60 \leq A < 80\)[/tex] is [tex]\(70\)[/tex]
- Midpoint of [tex]\(80 \leq A < 100\)[/tex] is [tex]\(90\)[/tex]
- Midpoint of [tex]\(100 \leq A < 120\)[/tex] is [tex]\(110\)[/tex]
- Midpoint of [tex]\(120 \leq A < 140\)[/tex] is [tex]\(130\)[/tex]
- Midpoint of [tex]\(140 \leq A < 160\)[/tex] is [tex]\(150\)[/tex]
Calculate the weighted mean:
[tex]\[ \text{Weighted Mean} = \frac{(70 \times 4) + (90 \times 6) + (110 \times 5) + (130 \times 3) + (150 \times 1)}{4 + 6 + 5 + 3 + 1} \][/tex]
[tex]\[ = \frac{280 + 540 + 550 + 390 + 150}{19} \][/tex]
[tex]\[ = \frac{1910}{19} \][/tex]
[tex]\[ \approx 100.53 \][/tex]
### Step 3: Determine the Median
To find the median, locate the middle value when all frequency values are sorted. The cumulative frequency will help:
- [tex]\(60 \leq A < 80\)[/tex]: 4
- [tex]\(80 \leq A < 100\)[/tex]: 10 (4+6)
- [tex]\(100 \leq A < 120\)[/tex]: 15 (10+5)
- [tex]\(120 \leq A < 140\)[/tex]: 18 (15+3)
- [tex]\(140 \leq A < 160\)[/tex]: 19 (18+1)
Total number of bedrooms: 19. The median position is at the [tex]\( \left(\frac{19 + 1}{2}\right) \)[/tex]-th value, which is the 10th value.
The 10th value lies within the range [tex]\(80 \leq A < 100\)[/tex], with the midpoint of 90.
Thus, the median is 90.
### Step 4: Determine the Skewness
Compare the mean and the median:
- Mean ≈ 100.53
- Median = 90
Since the mean is greater than the median, the histogram is right-skewed.
### Conclusion
The histogram is right-skewed, so the mean is greater than the median.
Thus, the correct statement is:
"The histogram is right-skewed, so the mean is greater than the median."
### Data Summary
| Area (sq. ft) | Number of Bedrooms |
|-----------------------|-------------------|
| [tex]\(60 \leq A < 80\)[/tex] | 4 |
| [tex]\(80 \leq A < 100\)[/tex] | 6 |
| [tex]\(100 \leq A < 120\)[/tex] | 5 |
| [tex]\(120 \leq A < 140\)[/tex] | 3 |
| [tex]\(140 \leq A < 160\)[/tex] | 1 |
### Step 1: Create a Histogram
A histogram groups data into bins (ranges) and plots the frequency of each bin. Here is the histogram for the given data ranges:
| Area Range (sq. ft) | Frequency |
|-----------------------|-----------|
| [tex]\(60 \leq A < 80\)[/tex] | 4 |
| [tex]\(80 \leq A < 100\)[/tex] | 6 |
| [tex]\(100 \leq A < 120\)[/tex] | 5 |
| [tex]\(120 \leq A < 140\)[/tex] | 3 |
| [tex]\(140 \leq A < 160\)[/tex] | 1 |
```
Frequency
6 | x
5 | x x x
4 | x x x x
3 | x x x
2 | x
1 | x
------------------------------------------------------
60-80 80-100 100-120 120-140 140-160 (Area in sq. ft.)
```
### Step 2: Determine the Mean
The mean can be estimated by calculating the weighted average of the midpoint of each range:
- Midpoint of [tex]\(60 \leq A < 80\)[/tex] is [tex]\(70\)[/tex]
- Midpoint of [tex]\(80 \leq A < 100\)[/tex] is [tex]\(90\)[/tex]
- Midpoint of [tex]\(100 \leq A < 120\)[/tex] is [tex]\(110\)[/tex]
- Midpoint of [tex]\(120 \leq A < 140\)[/tex] is [tex]\(130\)[/tex]
- Midpoint of [tex]\(140 \leq A < 160\)[/tex] is [tex]\(150\)[/tex]
Calculate the weighted mean:
[tex]\[ \text{Weighted Mean} = \frac{(70 \times 4) + (90 \times 6) + (110 \times 5) + (130 \times 3) + (150 \times 1)}{4 + 6 + 5 + 3 + 1} \][/tex]
[tex]\[ = \frac{280 + 540 + 550 + 390 + 150}{19} \][/tex]
[tex]\[ = \frac{1910}{19} \][/tex]
[tex]\[ \approx 100.53 \][/tex]
### Step 3: Determine the Median
To find the median, locate the middle value when all frequency values are sorted. The cumulative frequency will help:
- [tex]\(60 \leq A < 80\)[/tex]: 4
- [tex]\(80 \leq A < 100\)[/tex]: 10 (4+6)
- [tex]\(100 \leq A < 120\)[/tex]: 15 (10+5)
- [tex]\(120 \leq A < 140\)[/tex]: 18 (15+3)
- [tex]\(140 \leq A < 160\)[/tex]: 19 (18+1)
Total number of bedrooms: 19. The median position is at the [tex]\( \left(\frac{19 + 1}{2}\right) \)[/tex]-th value, which is the 10th value.
The 10th value lies within the range [tex]\(80 \leq A < 100\)[/tex], with the midpoint of 90.
Thus, the median is 90.
### Step 4: Determine the Skewness
Compare the mean and the median:
- Mean ≈ 100.53
- Median = 90
Since the mean is greater than the median, the histogram is right-skewed.
### Conclusion
The histogram is right-skewed, so the mean is greater than the median.
Thus, the correct statement is:
"The histogram is right-skewed, so the mean is greater than the median."
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