The 87th percentile is the value [tex]x_{0.87}[/tex] such that 87% of the distribution lies below [tex]X=x_{0.87}[/tex], i.e.
[tex]P(X \le x_{0.87}) = 0.87[/tex]
Transform [tex]X[/tex] to the standard normal random variable [tex]Z[/tex], then solve for [tex]x_{0.87}[/tex] using the inverse CDF of [tex]Z[/tex].
[tex]P(X\le x_{0.87}) = P\left(\dfrac{X-60}{12} \le \dfrac{x_{0.87}-60}{12}\right) = P\left(Z \le \dfrac{x_{0.87}-60}{12}\right) = 0.87[/tex]
[tex]\implies \dfrac{x_{0.87}-60}{12} = F_Z^{-1}(0.87) \approx 1.12369[/tex]
[tex]\implies \boxed{x_{0.87} \approx 73.5167}[/tex]
