The average ACT score submitted to a certain university is 26. To be eligible for admission, a student must score in the top 6% of applicants. If the scores are normally distributed with a standard deviation of 2.4, what is the minimum score \( x \) needed to be considered for admission?

To find the minimum score needed to be considered for admission, we need to determine the score that corresponds to the top 6% of applicants.

First, we know that the average ACT score submitted to the university is 26.

Since the scores are normally distributed with a standard deviation of 2.4, we can use the Z-score formula to find the minimum score.

The Z-score formula is given by:

Z = (X - μ) / σ

where X is the score we want to find, μ is the mean, and σ is the standard deviation.

To find the Z-score corresponding to the top 6% of applicants, we need to find the Z-score that corresponds to the cumulative probability of 0.94 (1 - 0.06).

Using a standard normal distribution table or calculator, we find that the Z-score corresponding to a cumulative probability of 0.94 is approximately 1.55.

Now, we can solve for X:

1.55 = (X - 26) / 2.4

Multiply both sides by 2.4:

3.72 = X - 26

Add 26 to both sides:

X = 29.72

Therefore, the minimum score needed to be considered for admission is 29.72.

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