Answer:
30.1
Explanation:
Mean score (μ) = 21.4
Standard deviation (σ) = 6.2
If only the top 8% of scores will get scholarships, students whose scores are at the 92nd percentile or above qualify for a scholarship. In a normal distribution, the 92nd percentile has a corresponding z-score of z = 1.405.
The minimum score, X, required for a scholarship is given by:
![z=(X-\mu)/(\sigma) \\1.405=(X-21.4)/(6.2)\\ X=30.1](https://img.qammunity.org/2021/formulas/mathematics/college/cyzm4xqs9b0zu6zv7hehd7jjntppxb9sw4.png)
The minimum score required for the scholarship is 30.1.