From 5d095644e17a1d7e869b7447bbd39dcd5c1a3e3c Mon Sep 17 00:00:00 2001
From: Ramon Araujo <ramoncfaraujo@gmail.com>
Date: Tue, 14 Oct 2025 20:39:14 -0300
Subject: [PATCH] add solution to problem 5.19

---
 src/chapters/5/sections/normal/index.tex      |  2 ++
 .../5/sections/normal/problems/19.tex         | 35 +++++++++++++++++++
 2 files changed, 37 insertions(+)
 create mode 100644 src/chapters/5/sections/normal/problems/19.tex

diff --git a/src/chapters/5/sections/normal/index.tex b/src/chapters/5/sections/normal/index.tex
index f355c40e..8e5abf0d 100644
--- a/src/chapters/5/sections/normal/index.tex
+++ b/src/chapters/5/sections/normal/index.tex
@@ -1,5 +1,7 @@
 \section{Normal}
 
+\subsection{problem 19}
+\input{problems/19}
 \subsection{problem 26}
 \input{problems/26}
 \subsection{problem 35}
diff --git a/src/chapters/5/sections/normal/problems/19.tex b/src/chapters/5/sections/normal/problems/19.tex
new file mode 100644
index 00000000..9f8d5731
--- /dev/null
+++ b/src/chapters/5/sections/normal/problems/19.tex
@@ -0,0 +1,35 @@
+From the statement, $Y$ is a normal-distributed r.v. with mean $\mu=1$ and variance $\sigma^2=4$, which implies standard deviation $\sigma=2$.
+
+
+We can obtain $Y$ as a function of $Z$ using the location-scale transform, applicable to Normal distributions
+
+$$
+Z = \frac{Y-\mu}{\sigma} = \frac{Y-1}{2}
+$$
+
+$$
+Y = 1 + 2 Z
+$$
+
+
+Let's check that the mean of $Y$ is correct by taking the expectation operator in both sides of the above equation
+
+$$
+E(Y) = E(1 + 2Z) = 1 + 2 E(Z)
+$$
+
+$$
+E(Y) = 1
+$$
+
+Doing the same with variance
+
+$$
+\mathrm{Var}(Y) = \mathrm{Var}(1 + 2Z) = 0 + 2^2 \mathrm{Var}(Z)
+$$
+
+$$
+\mathrm{Var}(Y) = 4
+$$
+
+The mean and variance matched, therefore the simple-looking function between $Y$ and $Z$ is correct.