College

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 to receive consideration for admission to the university.

Answer :

Application acceptance requires a minimum score of 392, which is required.

How to calculated the minimal score for admission?

Make "x" the required minimum score.

Given:

The average score is 500.

Standard deviation () is equal to 100

Admission percentage, P > 86 percent, or 0.86

Thus, the area under the normal distribution curve to the right of the z-score, which is 86%, is provided.

The portion of the score left over is displayed in the z-score table. As a result, we will calculate the z-score value for area as 100 - 86 = 14% or 0.14.

The z-score is therefore equal to -1.08 for value of 0.1401.

P(x>x0) = P(z > -1.08)

= -1.08 = x0 - 500 /100

x0 - 500 = -1.08 × 100

x0 = -1.08 + 500 = 392.

As a result, 392 is the minimal score needed for admission.

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