Convert z-scores to percentile rank Does anyone know the formula to convert a z-score to a percentile rank instead of using the z-score table? I searched the website, but did not find the formula.
thank you
 A: Like in the comments, there's a known function that describes the relationship. The reason you use a table is that the function is ugly. It can be calculated very easily by a computer but not at all by hand. So before everyone had R in front of them it was the best way to "calculate" a value from $F(x)$ when you needed one -- just look it up in the table.
If you want to know more about the theory, I'd refer you to any intro probability textbook... which imo you should already be using if you're going to be using z-scores.
This is done in R with function pnorm. This is the standard set of functions for a distribution in R:


*

*r:  input $N$, get $N$ random values

*p:  input $X$, get $F(X)$

*d:  input $X$, get $f(x)$

*q:  input $p$, get $F^{-1}(p)$


Try it out:
dt(1.65,df=30)
df(2,df1=2,df2=30)
df(-2,df1=2,df2=30) # F distribution is for positive random variables
qnorm(0.95)
pnorm(1.65)
dnorm(1.65)
runif(15)

And if you want to know more about how exactly the computer calculates $\frac{1}{2\pi}\int_{-\infty}^x e^{\frac{-t^2}{2}}dt$ when you give it an $x$, you'll be wanting to know about numerical integration.
