the admissions policy at a certain university requires that incoming students score in the upper 20% on a standardized test. if the mean score on the test is 510 and the standard deviation of the scores is 80, what is the minimum score that a student can earn on the test to meet the admissions requiremnt? Scores on the test are normally distributed and are reported in intervals of 10.

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To find the minimum score that a student can earn on the test to meet the admissions requirement, we need to determine the cutoff point for the upper 20% of scores.

Step 1: Calculate the Z-score
The Z-score measures the number of standard deviations a score is from the mean. We can calculate the Z-score using the formula: Z = (X - μ) / σ
Where:
- X is the score we want to find,
- μ is the mean score on the test (given as 510),
- σ is the standard deviation of the scores (given as 80).

Z = (X - 510) / 80

Step 2: Find the Z-score for the 80th percentile
Since we are looking for the upper 20% of scores, we need to find the Z-score that corresponds to the 80th percentile. The 80th percentile represents the cutoff point for the upper 20% of scores.

To find the Z-score for the 80th percentile, we can use a standard normal distribution table or a calculator. Look for the closest value to 0.80 or 80% in the body of the table (z column) and find the corresponding Z-score. In this case, the Z-score for the 80th percentile is approximately 0.84.

Step 3: Solve for X
Now that we have the Z-score, we can substitute it back into the formula and solve for X.

0.84 = (X - 510) / 80

Rearrange the equation to solve for X:

0.84 * 80 = X - 510
67.2 = X - 510
X = 510 + 67.2
X ≈ 577.2

Therefore, the minimum score that a student can earn on the test to meet the admissions requirement is approximately 577.2. However, since scores are reported in intervals of 10, the minimum score rounded down to the nearest interval of 10 would be 570.