Final answer:
To determine the probability of a sample mean being over 30 seconds, we use the Central Limit Theorem and calculate the standard error and the associated z-score. We then find the probability corresponding to the z-score from the standard normal distribution table, given the sample size and the population mean and standard deviation.
Step-by-step explanation:
To find the probability that the mean time spent with customs officers for a random sample of 50 travelers is over 30 seconds, we apply the Central Limit Theorem. Since we have a large sample size (n = 50), the sampling distribution of the sample mean will be approximately normally distributed with the population mean μ = 33 seconds and the standard error σ√(n) = 13√(50).
Steps to calculate:
Calculate the standard error (SE) of the mean: SE = σ / √(n) = 13 / √(50).
Find the z-score for 30 seconds: z = (X - μ) / SE = (30 - 33) / (13 / √(50)).
Use the z-score to find the corresponding probability from the standard normal distribution table.
Since the mean is greater than 30 seconds and the population is normally distributed, the probability of the sample mean being over 30 seconds is quite high.