Explanation:
The question is asking for the probability that a standard normal random variable Z is greater than 3.89. The standard normal distribution has a mean of 0 and a standard deviation of 1. The value 3.89 is quite far to the right on the standard normal distribution, so the probability that Z is greater than 3.89 is very small.
In practice, we would typically use a Z-table or a statistical software to find this probability. However, most Z-tables only go up to a Z-score of 3.4 because the probabilities beyond that point are so small. Therefore, for a Z-score of 3.89, the probability is so small that it is essentially zero.
Solution:
P(Z > 3.89) ≈ 0
