bound on normal distribution $P\left\{ \left|Z\right|\le\epsilon\right\} $

Question

I have seen a statement that assume $Z\sim N\left(0,1\right) $, for any $\epsilon>0$, we have $$P\left\{ \left|Z\right|\le\epsilon\right\} \le\epsilon $$

I think this statement is true, but I do not know how to prove it. Can anyone help me?

Answer 1

$\frac 1 {\sqrt {2\pi}}\int_{-\epsilon}^{\epsilon} e^{-x^{2}/2}dx < \frac 1 {\sqrt {2\pi}}\int_{-\epsilon}^{\epsilon} dx=\frac {2\epsilon} {\sqrt {2\pi}}<\epsilon $ since $\pi >2$.

Answer 2

Hint: Try using the fact that $$P(|Z|<\epsilon)=\frac{1}{\sqrt{2\pi}}\int_{-\epsilon}^{\epsilon}e^{-t^2/2} \ dt$$ in combination with the integral inequality $$\int_a^b f(t) \ dt\leq (b-a)M$$ with $f$ a non negative function in $[a,b]$ satisfying $f(t)\leq M$ for every $t\in [a,b]$.