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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:
For X = $15.00, the z-score is:
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:
The probability that a specific fare will exceed $15.00 is 0.2308.