Final answer:
The percentage of students that scored above 90 is approximately 15.87%, while the percentage scoring below 90 is roughly 84.13%. About 4093 students will have scores between 80 and 95. The z-score of a student who scored a 72 is -2.6.
Step-by-step explanation:
The question involves normal distribution and z-scores. Given a normal distribution with a mean score of 85 and a standard deviation of five for a student's test scores, we can compute the following:
- To find the percentage of students that scored above 90, we look for the z-score corresponding to 90 and then consult the standard normal distribution table. A score of 90 would have a z-score of (90-85)/5 = 1. From the standard normal distribution table, we can find the proportion of scores above a z-score of 1. The percentage of students scoring above 90 is approximately 15.87%.
- The percentage of students that scored below a 90 is the complement of the above percentage, which is approximately 84.13%.
- To find how many students scored between 80 and 95, we calculate the z-scores for both values. The z-score for 80 is (80-85)/5 = -1, and for 95 it's (95-85)/5 = 2. We then find the area between these two z-scores using the standard normal distribution table. This area corresponds to approximately 81.85%. Multiplying this percentage by the number of students (5000), we get approximately 4092.5 students, which can be rounded accordingly based on context.
- The z-score of a student who scored a 72 is calculated as (72-85)/5 = -2.6.
Therefore, we've determined the percentages of students scoring below and above 90, the number of students who score between 80 and 95, and the z-score for a score of 72 using the principles of normal distribution and z-scores.