Final answer:
To find the probability of the mean time spent with customs officers for a sample of 50 travelers, we need to standardize the values using the z-score formula. For a) the probability of being over 30 seconds is 59.1%, for b) the probability of being under 35 seconds is 55.9%, and for c) the probability of being under 30 seconds or over 35 seconds is 96.8%.
Step-by-step explanation:
To determine the probability that the mean time spent with customs officers will be:
a) Over 30 seconds:
First, we need to standardize the value of 30 seconds using the z-score formula:
z = (x - μ) / σ
where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.
Plugging in the values:
z = (30 - 33) / 13 = -0.23
Now, looking up the z-value in the z-table, we find that the probability of getting a value less than -0.23 is 0.409.
Therefore, the probability of getting a value greater than 30 seconds is 1 - 0.409 = 0.591, or 59.1%.
b) Under 35 seconds:
Using the same process as above, we find that the z-value for 35 seconds is (35 - 33) / 13 = 0.15.
Looking up the z-value in the z-table, we find that the probability of getting a value less than 0.15 is 0.559.
Therefore, the probability of getting a value less than 35 seconds is 0.559, or 55.9%.
c) Under 30 seconds or over 35 seconds:
To find the probability of either event happening, we add the probabilities from parts a) and b). So the probability is 0.409 + 0.559 = 0.968, or 96.8%.