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Answer :
Final answer:
In a normal distribution, about 2.5% of the score falls above a Z-score of approximately 1.96. By using the mean score and standard deviation provided, we can calculate the score corresponding to the top 2.5%, which is about 99.8.
Explanation:
To answer a question like this, you would typically use the concept of a Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean. In a normal distribution, 2.5% of the scores fall above the Z-score of approximately 1.96.
To find the score corresponding to the top 2.5% of students, we apply the formula which is Z = (X - μ) / σ. Here, μ is the mean, σ is the standard deviation (which is the square root of the given variance, 25, so σ = 5), and Z is the Z-score, which we know is approximately 1.96 for the top 2.5%.
So, we now adjust the formula to find X, which represents the top 2.5% score. So X = Z * σ + μ = 1.96 * 5 + 90 = 99.8. So, the score obtained by the top 2.5% of students is approximately 99.8. Therefore, the answer is a. 99.8.
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