A study was commissioned to find the mean weight of the residents in a certain town. The study examined a random sample of 67 residents and found the mean weight to be 187 pounds with a standard deviation of 26 pounds.

Use the normal distribution/empirical rule to estimate a 95% confidence interval for the mean, rounding all values to the nearest tenth.

To find a 95% confidence interval for the mean weight of the residents in the town, we need to follow these steps:

1. Understand the given information:
- Sample size ([tex]\(n\)[/tex]) = 67
- Sample mean ([tex]\(\overline{x}\)[/tex]) = 187 pounds
- Sample standard deviation ([tex]\(s\)[/tex]) = 26 pounds
- Confidence level = 95%

2. Determine the z-score for the confidence level:
For a 95% confidence level in a normal distribution, the z-score is approximately 1.96.

3. Calculate the standard error of the mean:
The standard error (SE) is calculated using the formula:
[tex]\[
SE = \frac{s}{\sqrt{n}}
\][/tex]
Plugging in the numbers:
[tex]\[
SE = \frac{26}{\sqrt{67}} \approx 3.2
\][/tex]

4. Calculate the margin of error:
The margin of error (ME) is determined by multiplying the z-score by the standard error:
[tex]\[
ME = z \cdot SE
\][/tex]
[tex]\[
ME = 1.96 \cdot 3.2 \approx 6.2
\][/tex]

5. Calculate the confidence interval:
To find the 95% confidence interval, use the margin of error to determine the lower and upper bounds:
- Lower bound = [tex]\(\overline{x} - ME = 187 - 6.2 = 180.8\)[/tex]
- Upper bound = [tex]\(\overline{x} + ME = 187 + 6.2 = 193.2\)[/tex]

So, the 95% confidence interval for the mean weight of the residents is approximately [tex]\(180.8\)[/tex] to [tex]\(193.2\)[/tex] pounds.