Answer:
16%
Explanation:
1) Mean, u = 100
standard deviation, o = 15
Score specified = x = 115
so, z-score = (x - u)/o
= (115 - 100)/15 = 1
From the probability standard normal distribution curve, 84.13% of the test scores falls within 1 SD (z score = 1), so, the percentage of test scores below 115 is 84.13%, so, for the percentage of test scores of 115 or higher = 100% - 84.13% = 16%.
Therefore, the correct answer is 16%.