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Solved typo in exercise 2.3
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alexsanchezpla committed Oct 24, 2024
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2 changes: 1 addition & 1 deletion 02-VariablesAleatorias_y_Distribuciones.Rmd
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Expand Up @@ -391,7 +391,7 @@ $$
1. Menos de seis meses ($x = 0.5$):

$$
P(X < 0.5) = F_X(0.5) = -0.25 \cdot 0.5^2 + 0.5 = 0.375
P(X < 0.5) = F_X(0.5) = -0.25 \cdot 0.5^2 + 0.5 = 0.4375
$$

2. Entre seis meses y un año ($x \in [0.5, 1]$):
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4 changes: 2 additions & 2 deletions Ejercicios-de-Inferencia-Estadistica.log
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Output written on Ejercicios-de-Inferencia-Estadistica.pdf (28 pages, 390553 by
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2 changes: 1 addition & 1 deletion docs/Ejercicios-de-Inferencia-Estadistica.tex
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Expand Up @@ -1100,7 +1100,7 @@ \subsubsection{Probabilidad de supervivencia}\label{probabilidad-de-supervivenci
\end{enumerate}

\[
P(X < 0.5) = F_X(0.5) = -0.25 \cdot 0.5^2 + 0.5 = 0.375
P(X < 0.5) = F_X(0.5) = -0.25 \cdot 0.5^2 + 0.5 = 0.4375
\]

\begin{enumerate}
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2 changes: 1 addition & 1 deletion docs/search_index.json

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Expand Up @@ -596,7 +596,7 @@ <h3><span class="header-section-number">2.3.5</span> Probabilidad de supervivenc
<li>Menos de seis meses (<span class="math inline">\(x = 0.5\)</span>):</li>
</ol>
<p><span class="math display">\[
P(X &lt; 0.5) = F_X(0.5) = -0.25 \cdot 0.5^2 + 0.5 = 0.375
P(X &lt; 0.5) = F_X(0.5) = -0.25 \cdot 0.5^2 + 0.5 = 0.4375
\]</span></p>
<ol start="2" style="list-style-type: decimal">
<li>Entre seis meses y un año (<span class="math inline">\(x \in [0.5, 1]\)</span>):</li>
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