Suppose the scores on a college entrance examination are normally distributed with a mean of 525 and a standard deviation of 110. A certain prestigious university will consider admission only for those students whose scores are in the top 10%.

What score must a student achieve to be considered for admission?

To determine the cut-off score for the top 10% on a college entrance exam with a normal distribution, we use the Z-score formula and a Z-score table. Given a mean of 525 and a standard deviation of 110, the necessary Z-score for the top 10% is approximately 1.28 resulting in a required exam score of about 666.

Explanation:

This question is concerned with the applications of the concept of a normal distribution and Z-scores within the context of college entrance exams. In order to find the score needed to fall into the top 10%, we symbolize the importance of the Z-score in relation to normal distribution.

We begin with the formula for a Z-score which is Z = (X - μ)/σ, where X represents a specific value, μ represents the mean and σ is the standard deviation. In your scenario, the mean is 525 and the standard deviation is 110. The Z-score needed to be in the top 10% (or the 90th percentile) is approximately 1.28 according to the standard Normal Distribution table.

Plugging 1.28 into the Z-score formula, we get X = μ + Z*σ = 525 + 1.28 * 110 ≈ 665.8. So the score needed for the top 10% is approximately 666, rounding up to the whole number.

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