Question

The travel time on a section of a Long Island Expressway (LIE) is normally distributed with a mean of 80 seconds and a standard deviation of 6 seconds. What travel time separates the top 2.5% of the travel times from the rest

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 80, \sigma = 6$

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.

$Z = \frac{X - \mu}{\sigma}$

$1.96 = \frac{X - 80}{6}$

$X - 80 = 6*1.96$

$X = 91.76$

The travel time that separates the top 2.5% of the travel times from the rest is of 91.76 seconds.