Final answer:
To be in the top 10% of exam scores with a mean of 75 and a standard deviation of 12, a student must score approximately 90. The closest answer choice to this is C. 92.
Step-by-step explanation:
The question asks to determine the grade required to be among the top 10% of exam scores which are normally distributed with a mean of 75 and a standard deviation of 12. To find the score that corresponds to the top 10%, we look for the z-score that marks the 90th percentile. From the z-score table, or using a calculator, we find that the z-score for the 90th percentile is approximately 1.28. Using the z-score formula (z = (X - μ) / σ), where X is the exam score, μ is the mean (75), and σ is the standard deviation (12), we can solve for X:
1.28 = (X - 75) / 12
X - 75 = 15.36
X = 90.36
Since exam scores are typically rounded to the nearest whole number, the grade required to be in the top 10% of scores is approximately 90.
Out of the options provided, none exactly matches 90. However, option C (92) is the closest and the only score that would be within the top 10%, as it is higher than 90. Therefore, the correct answer is C. 92.