Question

To receive a scholarship, a student needs to be in the top 10% of test scores on a mandatory competency test with a mean of 400 and a standard deviation of 100. What is the minimum score needed to receive this award?

Answer 1

A student needs a minimum score of 528 to be in the top 10% of the competency test and be eligible for the scholarship, assuming a mean of 400 and a standard deviation of 100.

Step-by-step explanation:

To determine the minimum test score needed to be in the top 10% of test scores for a scholarship, we employ certain principles of statistics, specifically dealing with the normal distribution characterized by a mean and a standard deviation. Given that the mean score is 400 and the standard deviation is 100, and recognizing that the 90th percentile is the cutoff for the top 10%, we can use a z-score table or a calculator function to find the z-score that corresponds to the 90th percentile.

From standard z-score tables or calculator functions, we find that the z-score corresponding to the 90th percentile is approximately 1.28. Using the formula for converting a z-score to an actual score (X = µ + zσ), we calculate the minimum score needed as follows:

Minimum Score (X) = Mean (µ) + Z-score (z) * Standard Deviation (σ)
Minimum Score (X) = 400 + 1.28 * 100
Minimum Score (X) = 400 + 128 = 528

Therefore, a student would need to score at least 528 to be in the top 10% and eligible for the scholarship.