Using the t-distribution, as we have the standard deviation for the sample, it is found that the 99% confidence interval for the true mean number of visitors on summer days is given by:
c. 92.664 to 103.940 visitors
<h3>What is a t-distribution confidence interval?</h3>
The confidence interval is:

In which:
is the sample mean.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a<u> two-tailed 99% confidence interval</u>, with 30 - 1 = <em>29 df</em>, is t = 2.756.
The other parameters are given as follows:
.
Hence, the bounds of the interval are given by:


Hence option c is correct.
More can be learned about the t-distribution at brainly.com/question/16162795