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:
[tex]z = \frac{X-\mu }{\sigma}[/tex]
For X = $15.00, the z-score is:
[tex]z = \frac{15.00-12.50 }{3.25}\\ z=0.7692[/tex]
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:
[tex]P(X>\$15.00) = 1-0.7692 = 0.2308[/tex]
The probability that a specific fare will exceed $15.00 is 0.2308.
