From 7e9a62e02f4960e6a39273cec71c0c37fb1c2dc3 Mon Sep 17 00:00:00 2001
From: Louis-Nicolas Douce <52084354+sayu92@users.noreply.github.com>
Date: Sat, 1 Apr 2023 22:23:25 +0200
Subject: [PATCH] Typo in chapter 3

The mean of the set Y was not 0
---
 03-Gaussians.ipynb | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/03-Gaussians.ipynb b/03-Gaussians.ipynb
index 47b761ae..2e205701 100644
--- a/03-Gaussians.ipynb
+++ b/03-Gaussians.ipynb
@@ -748,7 +748,7 @@
     "\n",
     "This is clearly incorrect, as there is more than 0 variance in the data. \n",
     "\n",
-    "Maybe we can use the absolute value? We can see by inspection that the result is $12/4=3$ which is certainly correct — each value varies by 3 from the mean. But what if we have $Y=[6, -2, -3, 1]$? In this case we get $12/4=3$. $Y$ is clearly more spread out than $X$, but the computation yields the same variance. If we use the formula using squares we get a variance of 3.5 for $Y$, which reflects its larger variation.\n",
+    "Maybe we can use the absolute value? We can see by inspection that the result is $12/4=3$ which is certainly correct — each value varies by 3 from the mean. But what if we have $Y=[6, -2, -3, -1]$? In this case we get $12/4=3$. $Y$ is clearly more spread out than $X$, but the computation yields the same variance. If we use the formula using squares we get a variance of 3.5 for $Y$, which reflects its larger variation.\n",
     "\n",
     "This is not a proof of correctness. Indeed, Carl Friedrich Gauss, the inventor of the technique, recognized that it is somewhat arbitrary. If there are outliers then squaring the difference gives disproportionate weight to that term. For example, let's see what happens if we have:"
    ]