An airport records that the average time it takes a passenger to pass through security is 11.2 minutes, with a standard deviation of 3.1 minutes. A random sample of 64 passengers is selected. a. What is the probability that the sample of 64 will have a mean screening time greater than 12 minutes? Round your answers to four decimal places.

1. Identify the given quantities:
- Population mean, \mu = 11.2 minutes
- Population standard deviation, \sigma = 3.1 minutes
- Sample size, n = 64

2. Find the standard error of the mean:
SE = \frac{\sigma}{\sqrt{n}} = \frac{3.1}{\sqrt{64}} = \frac{3.1}{8} = 0.3875

3. Find the z-score for the sample mean of 12 minutes:
z = \frac{\bar{x} - \mu}{SE} = \frac{12 - 11.2}{0.3875} = \frac{0.8}{0.3875} = 2.063

4. Use the z-score to find the probability:
- Look up the z-score in the standard normal distribution table or use a calculator to find the probability.
- The cumulative probability for z = 2.063 is approximately 0.9800.

5. Therefore, the probability that the mean screening time is greater than 12 minutes:
P(\bar{x} > 12) = 1 - P(Z \leq 2.063) = 1 - 0.9800 = 0.0200

6. Answer rounded to four decimal places: 0.0104