Question

X is normally distributed with mean 250 and standard deviation 40. what value of x does only the top 15% exceed?

Answer 1

The top 15% (85th percentile) is the cutoff value $x$ such that

$\mathbb P(X\le x)=\mathbb P\left(\dfrac{X-250}{40}\le\dfrac{x-250}{40}\right)=\mathbb P(Z\le z)=0.85$

where $z$ is the corresponding cutoff for the standardized normal distribution. We have $z\approx1.0364$, and so

$\dfrac{x-250}{40}=z\implies x\approx291.456$