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The students in Marly's math class recorded the dimensions of their bedrooms in a frequency table.

Bedroom Areas
\begin{tabular}{|c|c|}
\hline
\begin{tabular}{c}
Area \\
(sq. ft)
\end{tabular} & \begin{tabular}{c}
Number of \\
Bedrooms
\end{tabular} \\
\hline [tex]$60 \leq A\ \textless \ 80$[/tex] & 4 \\
\hline [tex]$80 \leq A\ \textless \ 100$[/tex] & 6 \\
\hline [tex]$100 \leq A\ \textless \ 120$[/tex] & 5 \\
\hline [tex]$120 \leq A\ \textless \ 140$[/tex] & 3 \\
\hline [tex]$140 \leq A\ \textless \ 160$[/tex] & 1 \\
\hline
\end{tabular}

Create a histogram to represent the data. Which statement is most likely true about the mean and the median of the data?

A. The histogram is right-skewed, so the mean is less than the median.
B. The histogram is right-skewed, so the mean is greater than the median.
C. The histogram is left-skewed, so the mean is less than the median.
D. The histogram is left-skewed, so the mean is greater than the median.

Sagot :

GinnyAnswer
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."
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