Question

Scores on an exam follow an approximately normal distribution with a mean of 76. 4 and a standard deviation of 6. 1 points. What is the minimum score you would need to be in the top 2%?.

Answer 1

88.9294

Explanation:

So for this problem we need to convert the top 2% into a z-score, which can be done using a calculator. You also first have to convert top 2% into 98% percentile. In doing so you'll approximately get:
`z\approx2.054`

This z-score just represents how many standard deviations away this is from the mean.

So this means that the minimum score would be represented as:
`76.4 + 2.054(6.1)\\88.9294`

This can be rounded as needed, but this would approximately be the minimum score to be in the top 2%