The combined math and verbal scores for students taking a national standardized examination for college admission are normally distributed with a mean of 650 and a standard deviation of 300.

If a college requires a student to be in the top 20% of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission at the college?

Remember that scores on this test are whole numbers.

Percentiles in a normal distribution can be calculated using z-scores which are a measure of how many standard deviations an element is from the mean. Here, we'll need to find the z-score that corresponds to the 80th percentile. Using a z-table or a statistical calculator, the z-score that corresponds to the 80th percentile is approximately 0.84.

Knowing this z-score, we can use the formula for a z-score in a normal distribution, which is:

z = (X - μ) / σ

Where:
- z is the z-score,
- X is the value in the dataset,
- μ is the mean of the dataset,
- σ is the standard deviation of the dataset.

Rearranging this formula to solve for X gives us:

X = z×σ + μ

Now we can substitute the values we have:

X = 0.84 × 300 + 650

= 902.

However, since the scores must be whole numbers, and we want to ensure that the student is in the top 20%, we need to round up to the next whole number, which gives us a final score of 903.

In conclusion, to be within the top 20% of test-takers, a student would need to score at least 903 on the test.

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