Step 1 :Given a random sample of 19 mortgage institutions, the mean interest rate was $3.41 \%$ and the standard deviation was $0.36 \%$. The interest rates are assumed to be normally distributed.
Step 2 :We need to decide which distribution to use to construct a $99 \%$ confidence interval for the population mean.
Step 3 :Option A suggests using a t-distribution because it is a random sample, the population standard deviation ($\sigma$) is unknown, and the interest rates are normally distributed.
Step 4 :Option B suggests using a normal distribution because the interest rates are normally distributed and $\sigma$ is known. However, this is incorrect because $\sigma$ is not known.
Step 5 :Option C suggests using a normal distribution because the sample size is less than 30 and the interest rates are normally distributed. However, this is incorrect because when the sample size is less than 30 and $\sigma$ is unknown, we should use a t-distribution.
Step 6 :Option D suggests using a t-distribution because the interest rates are normally distributed and $\sigma$ is known. However, this is incorrect because $\sigma$ is not known.
Step 7 :Option E suggests that neither the standard normal distribution nor the t-distribution can be used because $\sigma$ is unknown, the sample size is less than 30, and the interest rates are not normally distributed. However, this is incorrect because the interest rates are assumed to be normally distributed.
Step 8 :Therefore, the correct answer is \(\boxed{\text{A. Use a t-distribution because it is a random sample, } \sigma \text{ is unknown, and the interest rates are normally distributed.}}\)