Assuming [tex]X[/tex] is normally distributed, you have
[tex]\mathbb P(X>x')=\mathbb P\left(\dfrac{X-12}4>\dfrac{x'-12}4\right)=\mathbb P\left(Z>\dfrac{x'-12}4\right)=0.24[/tex]
The right-tail probability of [tex]0.24[/tex] corresponds to a [tex]z[/tex] score of approximately [tex]0.7063[/tex].
In terms of [tex]X[/tex], this critical value is
[tex]0.7063=\dfrac{x'-12}4\implies x'\approx14.83[/tex]
