Answer: To find the proportion of students who performed worse than Aaron, we need to find the area to the left of his score of 989 on the normal distribution curve.
First, we need to standardize Aaron's score using the formula:
z = (x - mu) / sigma
where x is the score, mu is the mean, and sigma is the standard deviation.
Plugging in the values, we get:
z = (989 - 1100) / 200 = -0.55
Next, we look up the area to the left of z = -0.55 on the standard normal distribution table, or use a calculator to find:
P(Z < -0.55) = 0.2912
Therefore, the proportion of students who performed worse than Aaron is approximately 0.2912, or 29.12% (rounded to 2 decimal places).
Explanation: