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  "body": "\n## Statement\n\nLet $\\mu \\in \\mathbb{R}$, let $\\sigma^2 \u003e 0$ and let $X$ have the [[normal distribution|07RXBVTVZB]] $\\mathcal{N}(\\mu, \\sigma^2)$. Then\n$$\\operatorname{Var}(X) = \\sigma^2.$$\n\n## Remarks\n\nProof: $Z = (X - \\mu)/\\sigma$ has the law $\\mathcal{N}(0, 1)$, where $\\sigma = \\sqrt{\\sigma^2}$. Integration by parts gives $E[Z^2] = \\int z^2 \\varphi(z) \\, dz = \\int \\varphi(z) \\, dz = 1$.\nSince $E[Z] = 0$ by the [[mean of the normal distribution|09PR1V6RWR]], $\\operatorname{Var}(Z) = 1$. The [[scaling|0KVQ9D86BM]] and [[translation|04CYN5G6RY]] rules then give $\\operatorname{Var}(X) = \\sigma^2 \\operatorname{Var}(Z) = \\sigma^2$.\nSo the two parameters of $\\mathcal{N}(\\mu, \\sigma^2)$ are its mean and its variance.\n\n## Explanations\n\n### L1 human\n\nThe second parameter of the normal distribution is exactly its variance. So $\\sigma$ is the standard deviation, the typical distance of a value from the center.\n",
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