= Solution
The normalized <quantum harmonic oscillator> <ground state> has $|A|^2=\sqrt{m\omega/(\pi\hbar)}$ and <energy> $E_0=\hbar\omega/2$. Its <classical turning points> are $x=\pm\sqrt{\hbar/(m\omega)}$. With $y=x\sqrt{m\omega/\hbar}$, the <ground-state oscillator probability beyond the turning points> is
$$
\boxed{P=\frac2{\sqrt\pi}\int_1^\infty e^{-y^2}\,dy=1-\operatorname{erf}(1)\simeq0.16.}
$$
The stated rounded <error function> gives $0.16$; the more precise Gaussian integral is $0.15730$.
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