The combined math and verbal scores for students taking a national standardized examination for college admission are normally distributed with a mean of 880 and a standard deviation of 180. If a college requires a student to be in the top 10% of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission at the college?

To qualify for the top 10% of students taking a national standardized test for college admission, with a mean of 880 and a standard deviation of 180, a student needs to score approximately 1104. This is calculated using z-score for the 90% percentile (1.28), and the formula for z-score: X = Z * σ + μ, where Z is the z-score, μ is the mean, σ is the standard deviation, and X is the score.

Explanation:

The question is asking for the minimum score a student needs to be in the top 10% of students taking a national standardized test for college admission, which follows normal distribution. Here, we need to find a z-score associated with the top 10% (or 90% percentile) and then convert this z-score into a test score.

For normally distributed data, a z-score for the 90% percentile is approximately 1.28. This value can be found in a standard normal distribution table..

The z-score represents how many standard deviations an element is from the mean. It is calculated by subtracting the mean from the individual score and then dividing by the standard deviation:

Z = (X - μ) / σ

We know that Z = 1.28, μ = 880 (the mean score), and σ = 180 (the standard deviation). All we need to do is to find the score X. By rearranging the formula we get:

X = Z * σ + μ

By substituting the values we get: X = 1.28 * 180 + 880, therefore, X = 1104 approximately.

So, the minimum score a student can have and still be in the top 10% is approximately 1104.

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