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.

  • 1
    $\begingroup$ This begs an awful lot of questions about how you know various things (e.g., that $\ln 23 < \pi$) without using a "calculator". $\endgroup$ – Ben Feb 22 at 22:25
  • $\begingroup$ Wow that's a tiny difference. Why do you want to do this? $\endgroup$ – Benjamin Wang Feb 23 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.