Answer:
519
Step-by-step explanation:
Let's assume that the scores for the national verbal proficiency test follow a normal distribution.
Mean score (μ) = 500
Standard deviation (σ) = 75
According to a Z-score table, the score for the 60th percentile is roughly z = 0.253
For any score "x", the z-score is given by:
[tex]Z=\frac{X-\mu}{\sigma}[/tex]
For z = 2.53:
[tex]0.253=\frac{X-500}{75}\\X=519[/tex]
The raw score to the 60th percentile is 519.
