Suppose ACT Composite scores are normally distributed with a mean of 20.6 and a standard deviation of 5.2. A university plans to admit students whose scores are in the top 40%. What is the minimum score required for admission? Round your answer to the nearest tenth, if necessary.

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Answer:

The minimum score required for admission is 21.9.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 20.6, \sigma = 5.2[/tex]

A university plans to admit students whose scores are in the top 40%. What is the minimum score required for admission?

Top 40%, so at least 100-40 = 60th percentile. The 60th percentile is the value of X when Z has a pvalue of 0.6. So it is X when Z = 0.255. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.255 = \frac{X - 20.6}{5.2}[/tex]

[tex]X - 20.6 = 0.255*5.2[/tex]

[tex]X = 21.9[/tex]

The minimum score required for admission is 21.9.