Fix solution to problem 5.1

Author: RamonAraujo

src/chapters/5/sections/pdfs_and_cdfs/problems/1.tex

@@ -1,28 +1,35 @@
 The PDF is \[f(x) = x e^{-x^2 / 2}\]
 \begin{flalign}
-    P(x \leq a) & = \int_{0}^a f(x) dx \\
-    & = \int_{0}^a x e^{-x^2 / 2} dx \\
-    & = 1 - e^{-a^2 / 2}
+    P(X \leq a) & = \int_{0}^a f(x) dx \nonumber\\
+    & = \int_{0}^a x e^{-x^2 / 2} dx \nonumber\\
+    & = 1 - e^{-a^2 / 2} \nonumber
 \end{flalign}
 
+The quantile function $Q$, equal to the inverse of the CDF, is given by
+
+\[Q(a) = \sqrt{-2 \log(1-a)}\]
+
 (a) 
 \begin{flalign}
-    P(1 < X < 3) & = P(X \le 3) - P(X \le 1) \\
-    & = e^{-1/2} - e^{-9/2}
+    P(1 < X < 3) & = P(X \le 3) - P(X \le 1) \nonumber\\
+    & = e^{-1/2} - e^{-9/2} \nonumber
 \end{flalign}
 
 (b)
-For first quantile \(q_1\) \\
-\[P(X \le q_1) = \frac{1}{4}\]
-\[1 - e^{-q_1^2 / 2} = \frac{1}{4}\]
-\[q_1 = 0.54\]
-For second quantile \(q_2\) \\
-\[P(X \le q_2) = \frac{2}{4}\]
-\[1 - e^{-q_2^2 / 2} = \frac{2}{4}\]
-\[q_1 = 0.83\]
-For third quantile \(q_3\) \\
-\[P(X \le q_3) = \frac{1}{4}\]
-\[1 - e^{-q_3^2 / 2} = \frac{3}{4}\]
-\[q_3 = 1.17\]
+The general formula for calculating the quartiles is as follows
+
+\[P(X \le q_j) = \frac{j}{4}\]
+\[q_j = Q\left( \frac{j}{4} \right) = \sqrt{ -2 \log \left( \frac{3 j}{4} \right) } \text{ , for } j=1,2,3\]
+
+First quartile $q_1$
+
+\[q_1 = 0.759\]
+
+Second quartile $q_2$, also known as median
+
+\[q_2 = 1.18\]
+
+Third quartile $q_3$
 
+\[q_3 = 1.67\]