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Suppose the scores on a college entrance examination are normally distributed with a mean of 525 and a standard deviation of 110. A certain prestigious university will consider admission only for those applicants whose scores exceed the 95th percentile of the distribution.

Find the minimum score an applicant must achieve in order to receive consideration for admission to the university. Round your answer up to the nearest integer.

Answer :

To be considered for admission at the university in question, an applicant must score at least 706 on the college entrance exam, which is the score exceeding the 95th percentile for a normal distribution with a mean of 525 and standard deviation of 110.

To find the minimum score an applicant must achieve to receive consideration for admission to a university that only accepts applicants above the 95th percentile, based on scores that are normally distributed with a mean of 525 and a standard deviation of 110, we need to use the concept of z-scores in a normal distribution.

The z-score corresponding to the 95th percentile is first determined. This value is typically around 1.645, but you may need to use a z-score table or calculator for more precise values. Once we have the z-score, we use the formula:

Score = mean + (z-score times standard deviation)

Plugging in the values, we have:

Score = 525 + (1.645 times 110) = 525 + 180.95 = 705.95

Since we are asked to round up to the nearest integer, the minimum score required for consideration is 706.

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