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A recent Internet survey showed per-day lodging expenses at Disney World in Orlando range from less than $100 for a campsite to more than $2,000 for a top luxury room. The cost-per-day average of the top 10 most popular hotels
at Disney World is $348.
a. If the standard deviation is $80, find the z-score for a per-day expense of b. If the data are approximately normally distributed with mean $348 and standard deviation $80, what percent of the hotels have per-day expenses
greater than $268?

Let's work through question 6 on lodging expenses at Disney World:

**Part a: Finding the z-score for a per-day expense of $400**

The formula for calculating the z-score is:

\[
z = \frac{x - \mu}{\sigma}
\]

Where:
- $ x $ is the value (per-day expense),
- $ \mu $ is the mean (average per-day cost),
- $ \sigma $ is the standard deviation.

Given:
- $ x = 400 $,
- $ \mu = 348 $,
- $ \sigma = 80 $,

Substituting into the formula:

\[
z = \frac{400 - 348}{80} = \frac{52}{80} = 0.65
\]

So, the z-score for a per-day expense of $400 is **0.65**.

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**Part b: Finding the percent of hotels with per-day expenses greater than $268**

To find the percent of hotels with per-day expenses greater than $268, we first calculate the z-score for $268 using the same z-score formula:

\[
z = \frac{268 - 348}{80} = \frac{-80}{80} = -1
\]

Now, using a standard normal distribution table, we find that the area to the left of $ z = -1 $ is approximately 0.1587, meaning that about 15.87% of hotels have expenses **less than** $268.

To find the percentage of hotels with expenses **greater than** $268, subtract this from 100%:

\[
100\% - 15.87\% = 84.13\%
\]

So, approximately **84.13%** of hotels have per-day expenses greater than $268.

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Would you like further details or clarifications on any part of this solution?

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Here are five related questions to explore further:

1. How do z-scores help compare values in different distributions?
2. What does a z-score of 1.5 indicate about a value's position relative to the mean?
3. How is the standard deviation useful in analyzing vacation costs?
4. What other applications could z-scores have beyond vacation expenses?
5. What are the limitations of using the normal distribution in real-world data?

**Tip**: When interpreting z-scores, remember that a positive score indicates a value above the mean, while a negative score indicates a value below the mean.

A recent Internet survey showed per-day lodging expenses at Disney World in Orlando range from less than $100 for a campsite to more than $2,000 for a top luxury room. The cost-per-day average of the top 10 most popular hotels at Disney World is $348. If the standard deviation is $80, find the z-score for a per-day expense of $400. What percent of the hotels have per-day expenses greater than $268?

Learn how to calculate the z-score for Disney World lodging expenses with a standard deviation of $80 and an average cost of $348. This problem covers statistical concepts, such as z-scores and normal distribution, making it suitable for high school students.

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