To calculate a 99% confidence interval for the population standard deviation ($\sigma$) at Bank B, we will use the sample data you provided. The steps involve:

1. Calculating the sample mean ($\bar{x}$) and sample variance ($s^2$).
2. Applying the chi-square distribution to find the confidence interval bounds for the variance ($\sigma^2$).
3. Deriving the confidence interval for the standard deviation.

Data for Bank B:

$4.2, 5.4, 5.9, 6.1, 6.6, 7.6, 7.6, 8.5, 9.4, 10.0$

Let's compute the necessary values:

1. Sample size ($n$) = 10 (since there are 10 values).
2. Degrees of freedom ($df$) = $n - 1 = 9$.
3. Sample mean $\bar{x}$ and sample variance $s^2$ for the Bank B data.

Next, I'll calculate these and use the chi-square distribution to construct the 99% confidence interval. Give me a moment.The sample variance for Bank B is approximately $3.35$, with a sample standard deviation of approximately $1.83$.

For a 99% confidence interval for the population standard deviation ($\sigma$) at Bank B, the interval is:

$(1.13, 4.17)$

This means we are 99% confident that the true population standard deviation lies between 1.13 and 4.17.

Do you want further details on the process or have any questions? Here are 5 related questions to expand your understanding:

1. How would the confidence interval change if the confidence level were 95%?
2. How is the chi-square distribution used in estimating the variance?
3. How would the confidence interval change if we had a larger sample size?
4. Can this method be applied to non-normally distributed data?
5. How do different levels of variance affect the reliability of service times at Bank B?

Tip: The wider your confidence interval, the less certain you are about the estimate, but you're more likely to capture the true parameter.