From 8cb689e661a76d4d70f03d453eaf6648694fdcd7 Mon Sep 17 00:00:00 2001 From: NunoSempere Date: Thu, 14 Apr 2022 16:38:12 -0400 Subject: [PATCH] fix: style --- .../website/docs/Internal/ProcessingConfidenceIntervals.md | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/packages/website/docs/Internal/ProcessingConfidenceIntervals.md b/packages/website/docs/Internal/ProcessingConfidenceIntervals.md index 202c291c..99e72e5a 100644 --- a/packages/website/docs/Internal/ProcessingConfidenceIntervals.md +++ b/packages/website/docs/Internal/ProcessingConfidenceIntervals.md @@ -18,14 +18,15 @@ module Normal = { We know that for a normal with mean $\mu$ and standard deviation $\sigma$, -$$ +$$ 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 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 $2 \cdot 1.6448536269514722$. +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. -## For lognormals \ No newline at end of file +## For lognormals