Question on z-score

Question

I saw this somewhere. Is this right?

$ P(z< -4.3333)=0.00003 $

Most z-score table stops at 3.4. If this is right, kindly explain how it was gotten. Thanks

Answer 1

This $z$-score is calculated as $$P(z<-4.3333) = \int_{-\infty}^{4.333} f(z)\mathrm{d}z$$

where $f(z)$ is the pdf for the standard normal distribution.Every $z$-score is calculated in this way.

Answer 2

It was most likely obtained by the integral representation:

$$ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{t} e^{-\frac{z^2}{2}} \ dz $$

Of course any computer can do this. Many tables stop at a score of 3 because it is less than anything practical.