\begin{figure}
  \begin{align*}
    \sigma_{low} &= S_{n-1}^2 - z_{1-\alpha/2} \frac{\sqrt{2}S_{n-1}^2}{\sqrt{n}} \\
    \sigma_{high} &= S_{n-1}^2 + z_{1-\alpha/2} \frac{\sqrt{2}S_{n-1}^2}{\sqrt{n}} \\
    \mathcal{P}(\sigma_{low} \le \sigma \le \sigma_{high}) &\approx 1 - \alpha \\
    z_a \;\text{est tel que} &\int_{-\infty}^{z_a}{\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}}dx=a \\
  \end{align*}
  \begin{align*}
    &I = [\mu^{\star} \pm \sigma_{high} \cdot z_{1 - \beta/2}] \\
    &\text{Avec un taux d'erreur de $\alpha$,} \\
    X \hookrightarrow \mathcal{N}(\mu, \sigma^2)
    &\Rightarrow \forall i, \mathcal{P}(\omega_i \in I) = 1 - \beta' \ge 1 - \beta \\
    &\Rightarrow S = \sum_{i=1}^{n}{\mathds{1}_{I}(\omega_i)} \hookrightarrow \mathcal{B}(1 - \beta', n) \\
    &\Rightarrow \mathbb{E}(S) = (1 - \beta') n \ge (1 - \beta) n
  \end{align*}
  \caption{Calcul du test de normalité par valeurs anormales}
  \label{normalitytest}
\end{figure}
