Suppose $X$ be the standard normal distriution. Its tail probability will be
$$Q(t)=\frac{1}{\sqrt{2\pi}}\int_{t}^{\infty}e^{-\frac {x^2}2}\,{\rm d}x$$
I need to find the upper bound of $Q$ function using Markov's inequality. It's easy to get the answer when $t>0$, but I have trouble working with $t<0$.
And I can't understand the hint.
The PDF of $|X|$ is $$ g(t)=\sqrt{\frac{2}{\pi}} e^{-t^2/2}, t\geq 0. $$
By integration, $E(|X|)=\sqrt{\frac{2}{\pi}}.$
$|X|$ is a positive distribution. Using Markov, $$ Q(t)=P(X\geq t)=\frac{1}{2}P(|X|\geq t)\\ \leq \frac{1}{2}\frac{E(|X|)}{t}=\frac{\sqrt{\frac{2}{\pi}}}{2t}=\frac{1}{\sqrt{2\pi }\space t} $$ for $t>0$ only. Of course, this upper bond is meaningless for $t<\frac{1}{\sqrt{2\pi }\space t}$, because we have $Q(t)< 1$ before anything.
For $t\leq 0$, the upper bond is $1$. You can find the lower bond by symmetry.