Question

The fares received by taxi drivers working for the City Taxi line are normally distributed with a mean of $12.50 and a standard deviation of $3.25. Based on this information, what is the probability that a specific fare will exceed $15.00?

Answer 1

Answer:

0.2308 or 23.08%

Explanation:

Mean (μ) = $12.50

Standard deviation (σ) = $3.25

Assuming a normal distribution, for any given fare X, the z-score is calculated as:

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

For X = $15.00, the z-score is:

$z = \frac{15.00-12.50 }{3.25}\\ z=0.7692$

A z-score of 0.7692 corresponds to the 77.91-th percentile of a normal distribution. Therefore, the probability that a fare exceeds $15.00 is:

$P(X>\ 15.00) = 1-0.7692 = 0.2308$

The probability that a specific fare will exceed $15.00 is 0.2308.