Final answer:
For the first exam, approximately 15.87% of the students would have scored above 91%. For the second exam, approximately 97.72% of the students would have scored above 65%.
Step-by-step explanation:
a. For the first exam:
To find the percentage of students who would have scored above 91%, we need to calculate the z-score for 91. The formula for calculating the z-score is:
z = (x - μ) / σ
where x is the value we want to find the percentage for, μ is the mean, and σ is the standard deviation. Plugging in the values for the first exam:
z = (91 - 82) / 9
z ≈ 1
Looking up the z-score in a z-table, we find that approximately 84.13% of the students would have scored below 91. Therefore, approximately 15.87% of the students would have scored above 91%.
b. For the second exam:
To find the percentage of students who would have scored below 65%, we can use the same formula for calculating the z-score. Plugging in the values for the second exam:
z = (65 - 81) / 8
z ≈ -2
Looking up the z-score in the z-table, we find that approximately 2.28% of the students would have scored below 65%. Therefore, approximately 97.72% of the students would have scored above 65%.