The combined math and verbal scores for students taking a national standardized examination for college admission are normally distributed with a mean of 550 and a standard deviation of 290. If a college requires a student to be in the top 40% of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission to the college?

The minimum score a student can obtain to be in the top 40% of students taking a national standardized exam, with scores normally distributed with a mean of 550 and a standard deviation of 290, is approximately 573.

Explanation:

In a normal distribution, the area under the curve to the left of the cut-off point signifies the percentile rank. Here the question states that the student should be in the top 40% of all students taking the test, which implies that the student should score more than 60% of the population i.e. in the 60th percentile.

Now, using the z-score formula which is (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation, we can now find the z-score for the 60th percentile. Note that in our case the mean (μ) is 550 and the standard deviation (σ) is 290. Checking from the Z table, the z-score for the 60th percentile is approximately 0.25.

Substituting this into the formula, X = Zσ + μ, we get X = 0.25 * 290 + 550 = 572.5. So for a student to be in the top 40% of students taking the test, he/she would need to score a minimum of approximately 573.

Learn more about Normal Distribution and Z-Scores here:

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