Question

Scores on an English test are normally distributed with a mean of 31.5 and a standard deviation of 7.3. Find the score that separates the top 59% from the bottom 41%. Round your answer to the nearest tenth.

Answer 1

If
`X` is a random variable representing test scores, then the 41st percentile is the score
`x` such that
`P(X\le x)=0.41`. Transform
`X` to the random variable
`Z` that follows the standard normal distribution via
`X=31.5+7.3Z`:

`P(X\le x)=P\left((X-31.5)/(7.3)\le(x-31.5)/(7.3)\right)=P\left(Z\le(x-31.5)/(7.3)\right)=0.41`

The
`z`-score for the 41st percentile is about -0.2275, so

`(x-31.5)/(7.3)\approx-0.2275\implies\boxed{x\approx29.8}`