Answer:
The travel time that separates the top 2.5% of the travel times from the rest is of 91.76 seconds.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 80 seconds and a standard deviation of 6 seconds.
This means that
What travel time separates the top 2.5% of the travel times from the rest?
This is the 100 - 2.5 = 97.5th percentile, which is X when Z has a p-value of 0.975, so X when Z = 1.96.
The travel time that separates the top 2.5% of the travel times from the rest is of 91.76 seconds.
