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Part 1

Listed below are prices in dollars for one night at different hotels in a certain region. Find the​ range, variance, and standard deviation for the given sample data. Include appropriate units in the results. How useful are the measures of variation for someone searching for a​ room?

206206

127127

100100

156156

281281

171171

8989

196196
 
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Part 1

The range of the sample data is 

enter your response here 

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dollars.dollars.

nights squared .nights2.

nights.nights.

dollars squared .dollars2.

​(Round to one decimal place as​ needed.)

Let's start by calculating the **range**, **variance**, and **standard deviation** of the given hotel prices.

### Step 1: Data Overview
The given hotel prices in dollars for one night are:

\[
206, 127, 100, 156, 281, 171, 89, 196
\]

### Step 2: Range
The range is calculated as:

\[
\text{Range} = \text{Maximum value} - \text{Minimum value}
\]

\[
\text{Range} = 281 - 89 = 192 \text{ dollars}
\]

### Step 3: Variance
To calculate the sample variance, follow these steps:

1. Find the mean (average) of the dataLet's start by calculating the **range**, **variance**, and **standard deviation** of the given hotel prices.

### Step 1: Data Overview
The given hotel prices in dollars for one night are:

\[
206, 127, 100, 156, 281, 171, 89, 196
\]

### Step 2: Range
The range is calculated as:

\[
\text{Range} = \text{Maximum value} - \text{Minimum value}
\]

\[
\text{Range} = 281 - 89 = 192 \text{ dollars}
\]

### Step 3: Variance
To calculate the sample variance, follow these steps:

1. Find the mean (average) of the 
\[
\text{Mean} = \frac{206 + 127 + 100 + 156 + 281 + 171 + 89 + 196}{8} = 165.75
\]

2. Subtract the mean from each data point and square the result:
\[
(206 - 165.75)^2 = 1613.06
\]
\[
(127 - 165.75)^2 = 1514.06
\]
\[
(100 - 165.75)^2 = 4332.06
\]
\[
(156 - 165.75)^2 = 95.06
\]
\[
(281 - 165.75)^2 = 13235.06
\]
\[
(171 - 165.75)^2 = 27.56
\]
\[
(89 - 165.75)^2 = 5942.06
\]
\[
(196 - 165.75)^2 = 915.06
\]

3. Find the sum of these squared differences:
\[
1613.06 + 1514.06 + 4332.06 + 95.06 + 13235.06 + 27.56 + 5942.06 + 915.06 = 27673
\]

4. Divide by \( n - 1 \) (since this is a sample variance, where \( n = 8 \)):
\[
\text{Sample variance} = \frac{27673}{8 - 1} = \frac{27673}{7} \approx 3953.3 \text{ dollars}^2
\]

### Step 4: Standard Deviation
The standard deviation is the square root of the variance:
\[
\text{Standard deviation} = \sqrt{3953.3} \approx 62.9 \text{ dollars}
\]

### Conclusion:
- **Range**: 192 dollars
- **Variance**: 3953.3 dollars²
- **Standard Deviation**: 62.9 dollars

### Usefulness of Measures:
The range provides a quick sense of the spread between the cheapest and most expensive hotel. However, the standard deviation is more useful for understanding how most prices vary from the average, giving a better sense of price consistency.

Do you need more details or have any questions?

---

Here are 5 related questions for further practice:
1. How would the range change if an additional hotel at 300 dollars were included?
2. What is the difference between population and sample variance?
3. Can the variance ever be negative?
4. How would removing the highest price affect the standard deviation?
5. Why is the standard deviation more informative than the range?

**Tip**: Variance gives you an idea of the data's spread, but standard deviation is easier to interpret as it has the same units as the data itself.

Listed below are prices in dollars for one night at different hotels in a certain region. Find the​ range, variance, and standard deviation for the given sample data. Include appropriate units in the results. How useful are the measures of variation for someone searching for a​ room?

206, 127, 100, 156, 281, 171, 89, 196

Learn how to calculate the range, variance, and standard deviation of hotel prices in a given sample. This problem involves key statistical concepts like measures of variation, including detailed steps to solve. Suitable for high school students in Grades 10-12.

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