Question

We gave a national verbal proficiency test that has a mean score of 500 with a standard deviation of 75. What raw score would correspond to the 60th percentile

Answer 1

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:

$Z=\frac{X-\mu}{\sigma}$

For z = 2.53:

$0.253=\frac{X-500}{75}\\X=519$

The raw score to the 60th percentile is 519.