Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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."
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.