To construct the 99% confidence interval for the population standard deviation (σ) of Bank A's waiting times, we can follow these steps:

1. Identify the sample data for Bank A: The waiting times for Bank A are: 6.4, 6.6, 6.7, 6.8, 7.1, 7.3, 7.5, 7.9, 7.9, 7.9.

2. Calculate the sample variance and standard deviation: The formula for sample variance ($s^2$) is: $s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}$ where $x_i$ are the data points, $\bar{x}$ is the sample mean, and $n$ is the number of data points.

3. Use the Chi-Square distribution: To construct a confidence interval for the population standard deviation, we use the chi-square distribution: $\left( \frac{(n-1) s^2}{\chi^2_{\alpha/2}} \right)^{1/2} < \sigma < \left( \frac{(n-1) s^2}{\chi^2_{1-\alpha/2}} \right)^{1/2}$ where $\chi^2_{\alpha/2}$ and $\chi^2_{1-\alpha/2}$ are the chi-square critical values for the 99% confidence level, and $\alpha = 0.01$.

Let me calculate this interval for you.The sample standard deviation for Bank A's waiting times is approximately 0.58 minutes.

The 99% confidence interval for the population standard deviation ($\sigma$) is: $0.36 \, \text{min} < \sigma_{Bank \, A} < 1.31 \, \text{min}$

Would you like further clarification on any part of this process?

Here are some related questions to consider:

1. How would the confidence interval change if the confidence level was reduced to 95%?
2. What does the width of the confidence interval tell us about the variability of Bank A's waiting times?
3. How would the interval change if the sample size for Bank A increased?
4. How can we interpret the chi-square distribution in this context?
5. What are the implications of a wider versus narrower confidence interval?

Tip: A larger sample size leads to a more precise estimate of the population standard deviation, thus narrowing the confidence interval.