Using 68-95-99.7% rule to find that $P(\vert Z \vert > 3)$ is approximately $0.003$; the exact value is $2 \cdot \Phi (-3)$.

Question

When reading an example of bounds on a Normal tail probability, I came across this:

Let $Z \sim N(0, 1)$. By the 68-95-99.7% rule, we know that $P(\vert Z \vert > 3)$ is approximately $0.003$; the exact value is $2 \cdot \Phi (-3)$.

In this Wikipedia article on the 68-95-99.7 rule, it gives the example of

$${\displaystyle \Pr(\mu -2\sigma \leq X\leq \mu +2\sigma )=\Phi (2)-\Phi (-2)\approx 0.9772-(1-0.9772)\approx 0.9545}$$

I'm presuming that this is $P(\vert Z \vert) \ge 2)$. So it has that $P(\vert Z \vert) \ge 2) = 2\Phi(2) - 1 \approx 0.9545$.

This Wikipedia article gives the standard normal CDF as

$${\displaystyle \Phi (x)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{x}e^{-t^{2}/2}\,dt}$$

This is the formula we would to calculate either of these two values.

Given this, how is it that $P(\vert Z \vert > 3) = 2 \Phi(-3)$? Shouldn't it be something like $P(\vert Z \vert > 3) = 1 - 2 \Phi(-3)$? I think I might be getting confused with the symmetry properties of the Normal distribution here?

Thank you.

Answer 1

We have

$= P (|Z|\le z)=P(-z\le Z\le z ) \\ = \Phi(z)-\Phi(-z)\\ = 1-\Phi(-z)-\Phi(-z), \ \ \textrm{due symmetry}\\ = 1-2\Phi(-z)$

That means that $P (|Z|\geq z)=1-P (|Z|\le z)=1-(1-2\Phi(-z))=2\Phi(-z)$