# Determine $x,x∈R^+$ such that $φ(x)=0,9505$

I tried to use the definition: $$\displaystyle φ(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-{s^2}/{2}}\,\mathrm ds$$

So, according to this site: $$\int \:e^{-{x^2}/{2}}\mathrm dx=\frac{\sqrt{\pi }}{\sqrt{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+C$$

But by definition: $${\displaystyle \operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-s^{2}}\,\mathrm {d} s}$$

I do not know how to follow after the function $erf (...)$

Maybe the value is only possible to get it through tables?

How to determine $x,x∈R^+$ such that $φ(x)=0,9505$?

Thank you very much.

• Reread the problem you were given. You're using $\varphi$ to refer to the standard normal distribution function, but this is strictly between 0 and 1 on all of $\mathbb{R}$, much less equal to 0 or 9505 for some positive real.
– user135912
Feb 16, 2017 at 4:02
• I really do not understand your contribution. But thanks anyway. They already answered me. Feb 16, 2017 at 15:01
• @51413 In many non-English-speaking countries, a comma is used in place of what is called the decimal point to separate the integer part $\lfloor x \rfloor$ of a real number $x$ from its non-integer part $x - \lfloor x \rfloor$. Thus, $0,9505$ does not necessarily mean $0$ or $9505$ as you have interpreted it. Feb 17, 2017 at 16:00

Tables of $\Phi(x)$ can be found in many textbooks, on-line (e.g. here), etc, and you simply look in the table for the value of $x$ for which $\Phi(x)$ equals $0.9505$. Alternatively, there are various on-line calculators (e.g. this one) that can find the value of $x$ for you. If you want to know how these calculators find the answer, well, one possibility is that they use a formula such as 26.2.22 in the well-known reference book Handbook of Mathematical Functions by Abramowitz and Stegun.
• Yes, I actually have the table printed. I would like to know if the only way is to look at the table, or if you can also take the numbers of the tables to have a $x$ more approximate. Feb 16, 2017 at 5:11
• @GeoMatt22 I don't understand. The second link in my answer above connects to a page on which one can enter $\mu=0$, $\sigma=1$ and the value of $\Phi(x)$ (0.9505) in three boxes and it will calculate the fourth box (value of $x$ is 1.65) for you. Or does "this answer" refer to a different (unspecified) answer on stats.SE or elsewhere? Feb 16, 2017 at 14:02