Final answer:
Using the properties of the normal distribution, the z-score for 70.5 is calculated to be 3. The area to the right of a z-score of 3 represents approximately 0.13%, thus around 0.13% of students will score above 70.5.
Step-by-step explanation:
To determine the percentage of students who scored over 70.5 with a mean of 66 and a standard deviation of 1.5, we can use the properties of the normal distribution. First, we calculate the z-score for 70.5:
Z = (X - µ) / σ
Where X is the value (70.5), µ is the mean (66), and σ is the standard deviation (1.5).
Plugging in the numbers, we get:
Z = (70.5 - 66) / 1.5
Z = 4.5 / 1.5
Z = 3
Using standard normal distribution tables or a calculator with normal distribution functions, we find the area to the right of a z-score of 3. The area to the left of Z=3 is approximately 0.9987, so the area to the right, which represents the percentage of students scoring above 70.5, is 1 - 0.9987 = 0.0013, or 0.13%.
Thus, approximately 0.13% of students are expected to score over 70.5.