$\mathrm{erf}(\eta) = 0.95$, what is a method to get $\eta$ value from the expression?

Question

I have an error function $\mathrm{erf}(\eta) = 0.95$.

How can I calculate the value of $\eta$ from this expression?

I know that $\eta \approx 1.4$. And I can get the value of $0.95$ by using Excel's ERF.PRECISE(1.4) function. However, I would like to know the reverse process, i.e. how to find $\eta$ from having only $\mathrm{erf}(\eta) = 0.95$.

Answer 1

Here you can find a method for the calculation of the inverse of the $erf(\eta)$ http://www.ams.org/journals/mcom/1968-22-101/S0025-5718-1968-0223070-2/S0025-5718-1968-0223070-2.pdf For small values of $x$, you can write: $$erf(x)^{-1}=\sum_{n=1}^\infty C_nx^{2n-1}$$ and you obtaine the following expression for $C_n$: $$1+\left(\sum_{m=1}^\infty(2m-1)C_mx^{2m-2}\right)\left(\sum_{n=1}^\infty n^{-1}C_nx^{2n}-\dfrac{2}{\sqrt\pi}\right)$$ if you put $x=erf(y)$ and so $y'=\dfrac{\sqrt\pi}{2}\exp(y^2)$, $y''=2yy'y'$ and then $$-\dfrac{1}{y'}=2\int_0^xy(t)dt-\dfrac{2}{\sqrt\pi}$$

Answer 2

Hint: =NORM.INV(prob,mean,sd)

For your case =NORM.INV(0.95,0,1)