To find the required sample size for estimating the mean water usage with a specified maximum error and a given confidence level, we use the formula for the sample size in estimating a population mean:

$n = \left( \frac{Z \cdot \sigma}{E} \right)^2$

Where:

• $n$ is the required sample size.
• $Z$ is the Z-score corresponding to the desired confidence level.
• $\sigma$ is the population standard deviation.
• $E$ is the maximum allowable error (also known as the margin of error).

Step 1: Determine the Z-score

For a 95% confidence level, the Z-score corresponds to the critical value of the standard normal distribution. For 95% confidence, the Z-score is approximately $Z = 1.96$.

Step 2: Plug in the known values

From the problem:

• $\sigma = 2$ gallons (standard deviation),
• $E = 0.14$ gallons (maximum error),
• $Z = 1.96$ (Z-score for 95% confidence).

Now, substitute these values into the formula:

$n = \left( \frac{1.96 \cdot 2}{0.14} \right)^2$

$n = \left( \frac{3.92}{0.14} \right)^2$

$n = \left( 28 \right)^2$

$n = 784$

Step 3: Round up the sample size

Since the sample size must be a whole number, round up to the next integer, which gives:

$n = 785$

Thus, the required sample size is 785.

---

Would you like more details or clarification on any step? Here are five related questions for further study:

1. How does changing the confidence level affect the required sample size?
2. What happens if the standard deviation increases or decreases in the formula?
3. How would you calculate the sample size if the population mean was unknown?
4. How does the maximum error (E) influence the sample size?
5. What is the interpretation of the Z-score in this context?

Tip: Always round up when calculating sample sizes, because rounding down could lead to a sample size that's too small to meet the required precision.