The top 15% (85th percentile) is the cutoff value [tex]x[/tex] such that
[tex]\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[/tex]
where [tex]z[/tex] is the corresponding cutoff for the standardized normal distribution. We have [tex]z\approx1.0364[/tex], and so
[tex]\dfrac{x-250}{40}=z\implies x\approx291.456[/tex]
