Thank you for visiting Scores on a certain exam are normally distributed with a mean of 90 and a variance of 25 What is the score obtained by the. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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.
Learn more about Z-Score Calculation here:
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