Final answer:
To find the z-score for P(Z>?)=0.6892 in a standard normal distribution, we use the complement area 0.3108 and look it up in a z-table or use statistical software to find the corresponding z-value, which is approximately -0.49.
Step-by-step explanation:
You are asking how to find the z-score on a standard normal distribution given that P(Z>?)=0.6892. To find this z-score, we must look up the complementary probability in the z-table, which is 1 - 0.6892 = 0.3108. This represents the area to the left of the z-score.
Looking at the z-table, we need to find the value corresponding to an area as close to 0.3108 as possible. You may find the exact value using statistical software or a calculator that provides inverse normal distribution functions. However, using the standard normal distribution table, we observe that the z-score with an area to the left of approximately 0.3108 is around -0.49. Remember that most tables give you the area to the left of the z-score. Therefore, if you're working with a table that gives the area to the right, you would directly use the given probability of 0.6892 to find the corresponding z-score.