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For $G$ with the standard normal distribution and $t\ge 0$, $\mathbb{P}(|G|>t)=2(1-\Phi (t))\le e^{-t^2/2}$. The difference has derivative $e^{-t^2/2}(\sqrt{2/\pi }-t)$, starts at zero, and tends to zero, proving its nonnegativity.

For an independent normal sample with known standard deviation $\sigma$, the signed Wald statistic for testing mean zero is $W=\sqrt{n}\bar{Y}/\sigma$. It has the standard normal distribution under the null hypothesis and distribution $N(\sqrt{n}\delta /\sigma ,1)$ under mean $\delta$. Its square has a chi-squared distribution; retaining the sign permits a one-sided test.

If $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, then $(X-\mu )/\sigma$ has the standard normal distribution.

A standard normal random variable has the standard normal distribution, mean zero and variance one. If $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, the standardized variable $(X-\mu )/\sigma$ is standard normal.