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Suppose ACT Composite scores are normally distributed with a mean of 21.2 and a standard deviation of 5.4. A university plans to admit students whose scores are in the top 45%.

What is the minimum score required for admission?

Answer :

The minimum score required for admission to the university is approximately 30.117 on the ACT Composite scale.

To find the minimum score required for admission to the university, we need to determine the cutoff score that corresponds to the top 45% of ACT Composite scores.

First, we can convert the top 45% to a z-score, which represents the number of standard deviations above or below the mean. We can use a standard normal distribution table or a calculator to find the z-score associated with the cumulative probability of 0.45.

Using the standard normal distribution table, we find that the z-score associated with a cumulative probability of 0.45 is approximately 1.645.

Next, we can use the z-score formula to calculate the minimum score required for admission:

z = (x - μ) / σ

Where:

z is the z-score

x is the ACT Composite score

μ is the mean of the scores

σ is the standard deviation of the scores

Rearranging the formula to solve for x (the ACT Composite score):

x = z * σ + μ

Substituting the known values into the equation:

x = 1.645 * 5.4 + 21.2

Calculating this expression:

x ≈ 30.117

To know more about cumulative probability refer here:

https://brainly.com/question/30772963#

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