To address this problem effectively, let's break it down step-by-step and create a histogram based on the provided frequency table.
### Step 1: Create the Frequency Table
The frequency table given is:
| 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 |
### Step 2: Visualize the Data in a Histogram
We need to imagine placing this data into a histogram.
The x-axis represents the range of bedroom areas (in sq. ft):
- 60–80
- 80–100
- 100–120
- 120–140
- 140–160
The y-axis represents the number of bedrooms.
### Step 3: Draw the Histogram
Here’s a conceptual outline of the histogram:
```
Number of |
Bedrooms |
|
|
|
|
|
+------+--+-------------+--+-------+--+
60 80 100 120 140 160
≤ A < 80 ≤ A < 100 ≤ A <120 ≤ A <140 ≤ A < 160
```
The number of asterisks () in each column represents the number of bedrooms in each area range.
### Step 4: Analyze the Shape of the Histogram
Looking at the histogram, most of the data is clustered on the left side (60–120 sq. ft), and it tapers off towards the right (120-160 sq. ft). This pattern is characteristic of a right-skewed distribution because there is a longer tail on the right side.
### Step 5: Determine the Relationship Between Mean and Median
In a right-skewed distribution:
- The mean is affected by the higher values (the tail part) and will be greater than the median because the few larger values pull the mean towards the right.
- The mean is not likely to have the central majority values (like the median does), making it higher.
### Conclusion
Based on the shape of the histogram and the properties of skewed distributions:
- The histogram is right-skewed, so the mean is greater than the median.
This matches with the second option provided in the question.
### Final Answer:
The histogram is right-skewed, so the mean is greater than the median.
