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 $260

We are given a normally distributed data set with:

- Mean ($\mu$) = 348
- Standard deviation ($\sigma$) = 80

We are asked to find the percentage of hotels that have per day expenses greater than 260.

### Step-by-Step Solution:

1. **Standardize the value**: 
   To calculate the percentage, we first convert the expense of 260 to a z-score using the z-score formula:
   \[
   z = \frac{x - \mu}{\sigma}
   \]
   where $x$ is the value (260), $\mu$ is the mean (348), and $\sigma$ is the standard deviation (80).
   \[
   z = \frac{260 - 348}{80} = \frac{-88}{80} = -1.1
   \]

2. **Use the standard normal distribution**: 
   Now, we need to find the percentage of data greater than a z-score of -1.1. 

   - From z-tables or using a calculator, the area to the left of $z = -1.1$ is approximately 0.1357. 
   - This means that 13.57% of the hotels have per day expenses less than $260.

3. **Find the percentage greater than $260**: 
   The percentage greater than $260 is simply the complement of the percentage less than $260.
   \[
   \text{Percentage} = 1 - 0.1357 = 0.8643 = 86.43\%
   \]

Thus, approximately **86.43%** of the hotels have per day expenses greater than $260.

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Would you like further clarification on any of these steps?

### Related Questions:
1. What percentage of hotels have expenses less than $260?
2. What percentage of hotels have expenses greater than $400?
3. How would the percentage change if the mean increased to $400 but the standard deviation stayed the same?
4. What percentage of hotels have expenses between $260 and $348?
5. How do we calculate the expenses for the top 5% of hotels?

#### Tip: 
The z-score helps you relate raw data to the standard normal distribution, making it easier to calculate probabilities.

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 $260?

This problem involves calculating the percentage of hotels with daily expenses greater than $260, given a normal distribution with a mean of $348 and a standard deviation of $80. Using the Z-score formula and standard normal distribution, we find that approximately 86.43% of hotels have expenses higher than $260.

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