Final answer:
15.87% of students scored above 90.
84.13% scored below 90.
85.31% scored between 80 and 95.
The Z-score of a student who scored 72 is -2.6.
Step-by-step explanation:
1) To find the percentage of students that scored above 90, we need to find the area under the normal distribution curve to the right of 90.
Using a standard normal distribution table or a calculator, we can find that this area is 0.1587, which is equivalent to 15.87%.
Therefore, approximately 15.87% of the students scored above 90.
2) To find the percentage of students that scored below 90, we subtract the percentage above 90 from 100%.
Therefore, approximately 84.13% of the students scored below 90.
3) To find the number of students that scored between 80 and 95, we need to find the area under the normal distribution curve between these two scores.
Using a standard normal distribution table or a calculator, we can find that this area is 0.8531, which is equivalent to 85.31%.
Therefore, approximately 85.31% of the students scored between 80 and 95.
4) To find the Z-score of a student who scored 72, we use the formula Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation.
Plugging in the values, we get Z = (72 - 85) / 5
= -2.6.
Therefore, the Z-score of a student who scored 72 is -2.6.