To find the number of gallons of bottled water consumed by 99.5% of the population, we need to determine the 99.5th percentile of the normal distribution.

Given:

• Mean ($\mu$) = 33.1 gallons
• Standard deviation ($\sigma$) = 13 gallons

We need to find the value $X$ such that 99.5% of the population consumes less than $X$ gallons. This corresponds to the 99.5th percentile of the normal distribution.

Steps to solve:

1. Determine the z-score corresponding to the 99.5th percentile:

   • The z-score for the 99.5th percentile can be found using a z-table or a statistical calculator. The z-score corresponding to 99.5% is approximately $z = 2.575$.

2. Use the z-score formula to find the value of $X$: $X = \mu + z \cdot \sigma$ Substituting the known values: $X = 33.1 + 2.575 \times 13$ $X = 33.1 + 33.475 = 66.575 \text{ gallons}$

Thus, 99.5% of people consumed less than approximately 66.58 gallons of bottled water.

Would you like more details on this solution or have any other questions?

Here are five related questions you might find useful:

1. How would the answer change if the standard deviation was smaller?
2. What does the 0.5th percentile represent in this context?
3. How do you calculate the mean and standard deviation of a dataset?
4. What is the difference between a z-score and a percentile?
5. How would this calculation change if the distribution were not normal?

Tip: Percentiles are useful in understanding how data is distributed within a population, helping you to compare individual data points with the overall population.