# How Do I show that $\sqrt{2\pi}<\Phi^{-1}\left( \frac{23}{\exp(\pi)}\right)$ without calculators?

Question: How might I show without calculators that
$$\sqrt{2\pi}<\Phi^{-1}\left( \frac{23}{\exp(\pi)}\right); \label{a}\tag{1}$$ where $$\Phi^{-1}()$$ is the inverse of the cumulative distribution function of the standard normal distribution, sometimes called the probit function.

Indeed numerical calculations show that $$\color{blue}{2.506628274631\ldots}=\sqrt{2\pi}<\Phi^{-1}\left( \frac{23}{\exp(\pi)}\right)=\color{blue}{2.50747379179\ldots}$$ So $$\ref{a}$$ is justified numerically. But attempting to show it is true without resorting to calculators has proven difficult. Of course $$\Phi\sqrt{2\pi}=\frac{1}{2}\bigg(1+\text{Erf}\sqrt{2\pi}\bigg)$$ and so we might want to show that $$\frac{1}{2}\bigg(1+\text{Erf}\sqrt{2\pi}\bigg)>\frac{23}{\exp(\pi)}$$ is false drawing the contradiction and so establishing the inequality in $$\ref{a}.$$ In particular we might leverage our knowledge that $$\ln23<\pi$$ and so $$23<\exp(\pi)$$, all of which can be shown without calculators. But this route has proven illusive. I also tried reasoning about $$\ref{a}$$ geometrically by looking at the summated volume of unit balls over $$\mathbb{R}^{n}.$$

I should mention that I am a new learner to math and statistics and probability. Just to be pedantic all calculation can be found here on Wolfram Alpha.

• This begs an awful lot of questions about how you know various things (e.g., that $\ln 23 < \pi$) without using a "calculator". – Ben Feb 22 at 22:25
• Wow that's a tiny difference. Why do you want to do this? – Benjamin Wang Feb 23 at 14:17