fix: style
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@ -22,6 +22,7 @@ $$
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a \cdot Normal(\mu, \sigma) = Normal(a\cdot \mu, |a|\cdot \sigma)
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a \cdot Normal(\mu, \sigma) = Normal(a\cdot \mu, |a|\cdot \sigma)
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$$
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$$
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We can now look at the inverse cdf of a $Normal(0,1)$. We find that the 95% point is reached at $1.6448536269514722$. ([source](https://stackoverflow.com/questions/20626994/how-to-calculate-the-inverse-of-the-normal-cumulative-distribution-function-in-p)) This means that the 90% confidence interval is $[-1.6448536269514722, 1.6448536269514722]$, which has a width of $2 \cdot 1.6448536269514722$.
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We can now look at the inverse cdf of a $Normal(0,1)$. We find that the 95% point is reached at $1.6448536269514722$. ([source](https://stackoverflow.com/questions/20626994/how-to-calculate-the-inverse-of-the-normal-cumulative-distribution-function-in-p)) This means that the 90% confidence interval is $[-1.6448536269514722, 1.6448536269514722]$, which has a width of $2 \cdot 1.6448536269514722$.
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