Answer:
30.1
Step-by-step 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:
[tex]z=\frac{X-\mu}{\sigma} \\1.405=\frac{X-21.4}{6.2}\\ X=30.1[/tex]
The minimum score required for the scholarship is 30.1.
