Question

Scores on a test are normally distributed with a mean of 70 and a standard deviation of 11.5. find p81, which separates the bottom 81% from the top 19%.

Answer 1

Answer: The score p81 that separates the bottom 81% from the top 19% is 80.12

Step by step:

First, calculate the "z" value. Z is a normally distributed random variable with 0 mean and standard deviation 1. The score value corresponding to the desired percentile p81 can be determined from a z value as follows:

$z=\frac{s-\mu}{\sigma}\\z_{p81}=\frac{p_{81}-70}{11.5}\\\implies p_{81}=z_{p81}\cdot 11.5+70$

We use a z-table (check online) to find the z value for the 81-st percentile. I found $z_{p81}=0.88$ and so we use that value to calculate the score for the percentile:

$p_{81}=0.88\cdot 11.5+70=80.12$

The score p81 that separates the bottom 81% from the top 19% is 80.12