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 $X=31.5+7.3Z$:

$P(X\le x)=P\left(\dfrac{X-31.5}{7.3}\le\dfrac{x-31.5}{7.3}\right)=P\left(Z\le\dfrac{x-31.5}{7.3}\right)=0.41$

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

$\dfrac{x-31.5}{7.3}\approx-0.2275\implies\boxed{x\approx29.8}$