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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

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pedrocarratore

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.