Suppose the scores $ x $ on a college entrance examination are normally distributed with a mean of 550 and a standard deviation of 100. A certain prestigious university will consider for admission only those applicants whose scores exceed the 90th percentile of the distribution.

Find the minimum score an applicant must achieve in order to receive consideration for admission to the university.

To solve this problem, we need to find the score x such that the area to the right of x under the standard normal curve is 0.1 (since the 90th percentile corresponds to the top 10% of scores).

Using a standard normal table (also known as a Z-table), we can find the z-score that corresponds to the area of 0.1. The closest value we can find in the table is 1.28. This means that 10% of the scores fall above a z-score of 1.28.

Now we can use the formula for converting a z-score to an x-score:

z = (x - mu) / sigma

where mu is the mean and sigma is the standard deviation of the distribution. Substituting the given values, we have:

1.28 = (x - 550) / 100

Solving for x, we get:

x = 100(1.28) + 550 = 678

Therefore, the minimum score an applicant must achieve in order to receive consideration for admission to the university is 678.

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