\documentclass[12pt]{article}
\usepackage{amsmath, amssymb, graphicx, array, fancyhdr, unicode-math, multirow}

\DeclareMathOperator{\Var}{Var}
\DeclareMathOperator{\sd}{SD}
\DeclareMathOperator{\Pois}{Pois}
\DeclareMathOperator{\Bin}{Bin}
\DeclareMathOperator{\NegBin}{NegBin}
\DeclareMathOperator{\sech}{sech}


% Text: Calibri, all bold
\setmainfont{Calibri}
\renewcommand{\seriesdefault}{b}

% Math: choose a math font and visually embolden it
\setmathfont{TeX Gyre Termes Math}                  % base math font
\setmathfont[FakeBold=1.9]{TeX Gyre Termes Math}    % embolden; adjust 1.6–2.4

% header
\pagestyle{fancy}
\fancyhf{}
\renewcommand{\headrulewidth}{0pt}   % <-- removes first line

\lhead{\resizebox{!}{1.5cm}{E}} 
\chead{%
  \raisebox{0.2cm}{%
    \shortstack{\Huge Cumulative \\ \Huge Distribution Function}%
  }%
}
\rhead{\resizebox{!}{1.5cm}{E}}

\setlength{\headheight}{1cm}
\setlength{\headsep}{12pt}

\begin{document}
\par\noindent\rule{\linewidth}{1.2pt}\par
\thispagestyle{fancy}

% -------- PAGE 1 --------
\begin{table}[h!]
  \renewcommand{\arraystretch}{1.2}

% -------- Single -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{4}{*}{\fbox{\includegraphics[width=1in]{images/single.png}}} 
&$N=5$\\
&$K=3$\\
&$n=2$\\
&\\
\end{tabular}
\par\noindent\rule{\linewidth}{1.2pt}\par
% -------- Abstract -------------------
\begin{tabular}{m{1.2in} m{4in}}
\\
\multirow{4}{*}{\fbox{\includegraphics[width=1in]{images/abstract.png}}} 
&$P(X=k)={\displaystyle\binom{3}{k}\displaystyle\binom{2}{2-k}}/{\displaystyle\binom{5}{2}}={\displaystyle\binom{3}{k}\displaystyle\binom{2}{2-k}}/10$\\
&\\
&$P(X=0)={\displaystyle\binom{3}{0}\displaystyle\binom{2}{2}}/{10}={1}/{10}=0.10$\\
&\\
&$P(X=1)={\displaystyle\binom{3}{1}\displaystyle\binom{2}{1}}/{10}={6}/{10}=0.60$\\
&\\
&$P(X=2)={\displaystyle\binom{3}{2}\displaystyle\binom{2}{0}}/{10}={3}/{10}=0.30$\\
&$F(0)=0.10$\\
&$F(1)=0.10+0.60=0.70$\\
&$F(2)=1$\\
&\\
\end{tabular}
\par\noindent\rule{\linewidth}{1.2pt}\par
% -------- Question -------------------
\begin{tabular}{m{1.2in} m{4in}}
\\
\multirow{4}{*}{\fbox{\includegraphics[width=1in]{images/question.png}}} 
&$PMF$\\
&$P(X=-4)=0.3$\\
&$P(X=1)=0.4$\\
&$P(X=4)=0.3$\\
&$E[X]=?$\\
&$E[X^2]=?$\\
&$\mathrm{Var}(X)=?$\\
&$\sigma_X=?$\\
\end{tabular}
\end{table}


% second page
\newpage
\thispagestyle{fancy}
\par\noindent\rule{\linewidth}{1.2pt}\par
\begin{table}[h!]
% -------- Answer -------------------
\begin{tabular}{m{1.2in} m{4in}}
\\
\multirow{4}{*}{\fbox{\includegraphics[width=1in]{images/key.png}}} 
&$PMF$\\
&$P(X=-4)=0.3$\\
&$P(X=1)=0.4$\\
&$P(X=4)=0.3$\\
&$E[X]=?$\\
&$E[X^2]=?$\\
&$\mathrm{Var}(X)=?$\\
&$\sigma_X=?$\\
&\hrule\\
&\\
&$E[X]= (-4)\cdot0.3 \;+\; 1\cdot0.4 \;+\; 4\cdot0.3$\\
&$E[X]= -1.2 + 0.4 + 1.2$\\
&$E[X]= \boxed{0.4}$\\
&\hrule\\
&\\
&$E[X^2]= (-4)^2\cdot0.3 + 1^2\cdot0.4 + 4^2\cdot0.3$\\
&$E[X^2]= 16\cdot0.3 + 1\cdot0.4 + 16\cdot0.3$\\
&$E[X^2]= 4.8 + 0.4 + 4.8$\\
&$E[X^2]= \boxed{10.0}$\\
&\hrule\\
&\\
&$\mathrm{Var}(X)= E[X^2] - \bigl(E[X]\bigr)^2$\\
&$\mathrm{Var}(X)= 10.0 \;-\; (0.4)^2$\\
&$\mathrm{Var}(X)= 10.0 - 0.16$\\
&$\mathrm{Var}(X)= \boxed{9.84}$\\
&\hrule\\
&\\
&$\sigma_X = \sqrt{\mathrm{Var}(X)} = \sqrt{9.84}\approx \boxed{3.136}$\\
\end{tabular}
\end{table}
\end{document}