How to work out the top 5% when given a mean and standard deviation?

Question

A Company's human resources department administers a test of "executive aptitude". Test scores are known to be normally distributed with a mean of 100 and a standard deviation of 15.

What score do you need to be in the top 5% of applicants?

What is the 15th Percentile?

Answer 1

Top 5% score = (1.645)(15) + (100) = 124.675 = 125 to nearest whole number

Answer: You need a score of 125 (to nearest whole number) to be in the top 5%.

**1.645 is the z-score associated with the top 5% or 95th percentile of the normal distribution.

** Formula for a score is x = z*o + u or in words score = (z-score)(standard deviation) + (mean)

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15th percentile = (-1.0364338)(15) + (100) = 84.4534993 = 85 to nearest whole number

** z-score = z-value at 15% of the total area of the normal distribution = -1.04 to two decimals.

** Repeat same formula as above x = z*o + u