Let's solve the problem by finding the 95% confidence interval for the true mean waiting time.

Given:

• Sample mean ($\bar{x}$) = 9.3 minutes
• Standard deviation ($s$) = 6.2 minutes
• Sample size ($n$) = 200 passengers
• Confidence level = 95%

Steps:

1. Identify the z-score for a 95% confidence level. The critical value $z_{\alpha/2}$ for 95% confidence is approximately 1.96 (from the standard normal distribution table).

2. Calculate the standard error (SE): $SE = \frac{s}{\sqrt{n}} = \frac{6.2}{\sqrt{200}} \approx \frac{6.2}{14.14} \approx 0.4383$

3. Compute the margin of error (ME): $ME = z_{\alpha/2} \times SE = 1.96 \times 0.4383 \approx 0.858$

4. Determine the confidence interval:

   • Lower limit = $\bar{x} - ME = 9.3 - 0.858 = 8.442$
   • Upper limit = $\bar{x} + ME = 9.3 + 0.858 = 10.158$

Thus, the 95% confidence interval is approximately (8.4, 10.2).

The correct answer is B: 8.4 and 10.2.

Would you like more details or explanations on any part of this calculation?

Related Questions:

1. What is the formula for calculating the confidence interval for a population mean?
2. How do you interpret a 95% confidence interval?
3. How does the sample size affect the width of the confidence interval?
4. What assumptions must be satisfied to use the z-score in confidence intervals?
5. What is the impact of changing the confidence level from 95% to 99%?

Tip:

Always ensure your sample size is large enough to assume a normal distribution when calculating confidence intervals, especially if the population standard deviation is unknown.