Answer :
To find the z-score corresponding to the 55th percentile. This z-score is approximately 0.13. The minimum score required for admission is approximately 22.0.
To determine the minimum score required for admission, we need to consider that ACT composite scores are normally distributed with a mean (µ) of 21.3 and a standard deviation (σ) of 5.3. The university plans to admit students in the top 45%, which means that we need to find the cutoff score corresponding to the 55th percentile (since 100% - 45% = 55%).
Using a standard normal distribution table or a calculator with a built-in function, we can find the z-score corresponding to the 55th percentile. This z-score is approximately 0.13.
Now, we'll use the z-score formula to find the minimum score required for admission:
X = µ + (z * σ)
Where X is the minimum score, µ is the mean, z is the z-score, and σ is the standard deviation. Plugging in the values:
X = 21.3 + (0.13 * 5.3)
X ≈ 21.3 + 0.689 = 21.989
Rounding the score to the nearest tenth, the minimum score required for admission is approximately 22.0.
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