Final answer:
The percent of students who scored above 75 points on an exam with a mean of 76.4 and a standard deviation of 6.1 is approximately 59%.
Step-by-step explanation:
The student has asked what percent of students scored above 75 points on an exam if the scores are normally distributed with a mean of 76.4 and a standard deviation of 6.1 points.
To find this percentage, we first have to determine the z-score of a 75-point score. The z-score is calculated by subtracting the mean from the score and dividing by the standard deviation:
z = (score - mean) / standard deviation
z = (75 - 76.4) / 6.1
z = -1.4 / 6.1
z = -0.2295
A z-score of -0.2295 corresponds to a percentile rank just slightly lower than 50%, since the mean score has 50% of scores below it and 50% of scores above it. However, because we are interested in the percentage of students who scored above 75 points, and 75 points is below the mean, we know that more than 50% of the students scored above this mark.
Looking up this z-score on a normal distribution table or using standard statistical software, we find that the percentage of students who scored below 75 is about 41%. Therefore, the percentage of students who scored above 75 is 100% - 41% = 59%.
The correct answer to the student's question is 'D. 59%'