Note: Please make sure to properly format your answers. All dollar figures in the answers need to include the dollar sign and any amount over 1,000 should include the comma ($2,354.67). All percentage values in the answers need to include a percentage sign (%). For all items without specific rounding instructions, round your answers to two decimal places, show both decimal places (5.06).


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 $400. 
 
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 the problem step by step.

### Part (a): Finding the Z-Score for a Per-Day Expense of $400

The **z-score** is calculated using the formula:

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

Where:
- $ X $ is the value of interest (the per-day expense),
- $ \mu $ is the mean (average) of the data,
- $ \sigma $ is the standard deviation.

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

Substitute these values into the formula:

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

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

---

### Part (b): Percent of Hotels with Per-Day Expenses Greater Than $268

To find the percentage of hotels with per-day expenses greater than $268, we need to calculate the **z-score** for $268 and then refer to a standard normal distribution (Z-table) to find the cumulative probability.

1. **Z-Score Calculation** for $268:

\[
z = \frac{X - \mu}{\sigma} = \frac{268 - 348}{80} = \frac{-80}{80} = -1.00
\]

2. **Find the Cumulative Probability**:

The z-score of **-1.00** corresponds to a cumulative probability of approximately **0.1587** (or **15.87%**). This means that **15.87%** of the hotels have per-day expenses less than $268.

3. **Find the Percentage Above $268**:

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

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

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

---

### Summary of Answers:
- (a) The z-score for a per-day expense of $400 is **0.65**.
- (b) **84.13%** of the hotels have per-day expenses greater than $268.

---

Would you like more details or have any questions? 

Here are 5 related questions to think about:
1. How would the z-score change if the per-day expense was $500 instead of $400?
2. What percentage of hotels have per-day expenses less than $400?
3. How would increasing the standard deviation to $100 affect the z-score for $400?
4. What is the z-score for a per-day expense of $200?
5. How do you interpret a z-score of 0 in this context?

**Tip:** When interpreting z-scores, remember that a positive z-score indicates a value above the mean, while a negative z-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.
a. If the standard deviation is $80, find the z-score for a per-day expense of $400. 
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?

Learn how to calculate the z-score for a per-day expense of $400 at Disney World hotels, given a mean of $348 and standard deviation of $80. Understand how to use the normal distribution to find the percentage of hotels with expenses greater than $268. Suitable for students in Grades 10-12.

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