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

519

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=(X-\mu)/(\sigma)`

For z = 2.53:

`0.253=(X-500)/(75)\\X=519`

The raw score to the 60th percentile is 519.