Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$? Suppose $X_1\sim N(\mu_1,\sigma^2)$ and $X_2\sim N(\mu_2,\sigma^2)$ where $\mu_2\ge \mu_1$.
Since $\mu_2\ge \mu_1$, based on a characterization of stochastic ordering, we can say that
$$P(X_1>c)\le P(X_2> c) \quad \text{ for any constant }c\,.$$
Now suppose $|\mu_2| \ge |\mu_1|$ instead of $\mu_2 \ge \mu_1$.
Can we also say that $$P(|X_1|>k)\le P(|X_2|> k)\quad \text{ for any constant }k\,?$$
This is certainly the case if $E|X_2|\ge E|X_1|$, but does this hold?
$|X_i|$ has a folded normal distribution and its mean is given by
$$E|X_i|=\sigma\sqrt{\frac 2{\pi}}\exp\left\{-\frac{\mu_i^2}{2\sigma^2}\right\}+\mu_i\left(1-2\Phi\left(-\frac{\mu_i}{\sigma}\right)\right)$$
But it is not immediately clear to me if $E|X_2|\ge E|X_1|$. Is there any simpler way to approach this?

A possible approach: wlog we may assume $\mu_1 \ge 0$, $k \ge 0$.
$$
P(|X_2|>k) - P(|X_1|>k) = \int_0^{\infty}(f_2(k+t) - f_1(k+t)) - (f_1(-k-t) - f_2(-k-t)) \,dt
$$
and it would suffice to show that the integrand is nonnegative for all $t\ge0$.
