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the _correct_ 90 vs. 95 thing

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Quinn Dougherty 2022-04-27 16:56:30 -04:00
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packages/website/docs/Internal/Processing-Confidence-Intervals.md
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@ -25,7 +25,7 @@ $$
a \cdot Normal(\mu, \sigma) = Normal(a \cdot \mu, |a| \cdot \sigma)
$$
We can now look at the inverse cdf of a $Normal(0,1)$. We find that the 90% 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 $2 \cdot 1.6448536269514722$.
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 $2 \cdot 1.6448536269514722$.
So then, if we take a $Normal(0,1)$ and we multiply it by $\frac{(high -. low)}{(2. *. 1.6448536269514722)}$, it's 90% confidence interval will be multiplied by the same amount. Then we just have to shift it by the mean to get our target normal.