\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}{C}} 
\chead{%
  \raisebox{0.2cm}{%
    \shortstack{\Huge Expected Value \& Variance\\ \Huge Binomial Distribution}%
  }%
}
\rhead{\resizebox{!}{1.5cm}{C}}


\setlength{\headheight}{2.5cm}
\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}}} 
&$n=6$\\
&$p=\tfrac12$\\
&\\
&\\
&\\
\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}}} 
&$E[X]=np=6\cdot\frac12=\boxed{3}$\\
&\hrule\\
&\\
&$\Var(X)=np(1-p)=3\cdot\frac12=\boxed{\frac32}$\\
&\hrule\\
&\\
&$E[X^2]=\Var(X)+\bigl(E[X]\bigr)^2$\\
&$E[X^2]=\frac32 + 3^2$\\
&$E[X^2]=\frac32 + 9$\\
&$E[X^2]=\boxed{\frac{21}{2}}$\\
\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}}} 
&$n=5$\\
&$p=0.4$\\
&$PMF=?$\\
\end{tabular}
\end{table}


% second page
\newpage
\thispagestyle{fancy}
\begin{table}[h!]
% -------- Answer -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{4}{*}{\fbox{\includegraphics[width=1in]{images/key.png}}} 
&$n=5$\\
&$p=0.4$\\
&$PMF=?$\\
&\hrule\\
&\\
&$PMF$\\
&$P(X = x) = \binom{5}{x}(0.4)^x(0.6)^{5-x}$\\
&$P(X = 0)= \binom{5}{0}(0.4)^0(0.6)^5= 0.6^5 \approx 0.0778$\\
&$P(X = 1)=\binom{5}{1}(0.4)^1(0.6)^4 = 5·0.4·0.6^4 \approx  0.2592$\\
&$P(X = 2)= \binom{5}{2}(0.4)^2(0.6)^3 = 10·0.16·0.216  \approx 0.3456$\\
&$P(X = 3)= \binom{5}{3}(0.4)^3(0.6)^2 = 10·0.064·0.36  \approx   0.2304$\\
&$P(X = 4)= \binom{5}{4}(0.4)^4(0.6)^1 = 5·0.0256·0.6  \approx  0.0768$\\
&$P(X = 5)= \binom{5}{5}(0.4)^5(0.6)^0 = 0.4^5 = 0.0102$\\
&\hrule\\
&\\
&$\mu = E[X] = np = 5 \times 0.4 = \boxed{2.0}$\\
&\hrule\\
&\\
&$\Var(X) = np(1-p) = 5 \times 0.4 \times 0.6 = \boxed{1.2}$\\
&\hrule\\
&\\
&$\sigma = \sqrt{\Var(X)} = \sqrt{1.2} \approx \boxed{1.0954}$\\
\end{tabular}
\end{table}
\end{document}