Normal distribution. Doubt

Question

If $Y$ is Normal($\mu$=0, $\sigma$=1), can it happen that, for a positive constant $\alpha$: $$\Pr[Y\gt2\alpha]=0 \text{ ?}$$

Answer 1

No. Note that $P(Y \gt 2\alpha) = 0 \iff P(Y \leq 2\alpha ) = 1$, and this is the standard normal distribution, so it's cdf is $ P(Y \leq 2\alpha ) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{2\alpha} e^{\frac{-t^2}{2}} dt$

One can show that for any $\alpha \in \mathbb{R}$ this cdf is smaller than one, although it $\it{converges}$ to one.

Meaning that $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{\frac{-t^2}{2}} = 1$, but for a specific $\alpha$, $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{2\alpha} e^{\frac{-t^2}{2}} \lt 1$

Answer 2

No, there is not.

For certain values of $\alpha$, this probability can get extremely small, but never zero.

The normal distribution has a PDF of $\displaystyle \frac{1}{\sqrt{2\pi}}e^{x^2/2}$, which has positive probability even for large $x$.