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.

  • $\begingroup$ 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. $\endgroup$
    – user135912
    Feb 16, 2017 at 4:02
  • $\begingroup$ I really do not understand your contribution. But thanks anyway. They already answered me. $\endgroup$
    – cfrostte
    Feb 16, 2017 at 15:01
  • 1
    $\begingroup$ @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. $\endgroup$ Feb 17, 2017 at 16:00

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – cfrostte
    Feb 16, 2017 at 5:11
  • $\begingroup$ Did you check the second link in this answer? $\endgroup$
    – GeoMatt22
    Feb 16, 2017 at 5:57
  • $\begingroup$ @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? $\endgroup$ Feb 16, 2017 at 14:02
  • $\begingroup$ Dilip: Sorry, my comment was intended as a response to that of @emi, but forgot to cc him! $\endgroup$
    – GeoMatt22
    Feb 16, 2017 at 14:29
  • $\begingroup$ @GeoMatt22 I want to avoid the calculation using software since I am preparing for an exam. The idea of my previous comment was to know if perhaps with the same information in the table it was possible to use some additional process to provide more accuracy. In fact I'm looking at the formulas of the third link. $\endgroup$
    – cfrostte
    Feb 16, 2017 at 14:58

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