\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{}
\lhead{\resizebox{!}{2cm}{C}}
\chead{\raisebox{0.5cm}{\Huge Variance / SD}} % (add title text if you want)
\rhead{\resizebox{!}{2cm}{C}}
\setlength{\headheight}{2.5cm}
\setlength{\headsep}{12pt} % space between header and body

\begin{document}
\thispagestyle{fancy}

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

% -------- Single -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/single.png}}} 
&$X = \{1,2,3,4,5,6\}$\\
&$P(X = x) = \frac{1}{6}$\\
&\\
&\\
&\\
&\\
\end{tabular}

% -------- Abstract -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/abstract.png}}} 
&$E(X) = \sum_{x=1}^6 x \cdot \frac{1}{6} = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$\\
&\hrule\\
&\\
&$E(X^2) = \sum_{x=1}^6 x^2 \cdot \frac{1}{6} = \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6} = \frac{91}{6}$\\
&\hrule\\
&\\
&$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{91}{6} - (3.5)^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12} = \frac{35}{12}$\\
&\hrule\\
&\\
&$\operatorname{SD}(X) = \sqrt{\operatorname{Var}(X)} = \sqrt{\frac{35}{12}} = \sqrt{\frac{35}{12}} \approx 1.708$\\
&\\
&$\mathrm{Var}(4X + 2) = 16\,\mathrm{Var}(X)$\\
&\\
\end{tabular}

% -------- Question -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/question.png}}} 
&$\mu = E[X] = ?$\\
&$E[X^2] = ?$\\
&$Var(X) = ?$\\
\end{tabular}
\end{table}


% second page
\newpage
\thispagestyle{fancy}
\begin{table}[h!]
% -------- Answer -------------------
\begin{tabular}{m{1.2in} m{4in}}
\multirow{5}{*}{\fbox{\includegraphics[width=1in]{images/key.png}}} 
&$x = \{-2,-1,0,1,2\}$\\
&$p(x)=1/55\,(x-3)^2$\\
&\\ 
&$P(X=-2) = 25 \times 1/55 = 25/55$\\
&$P(X=-1) = 16 \times 1/55 = 16/55$\\
&$P(X=0) = 9 \times 1/55 = 16/55$\\
&$P(X=1) = 4 \times 1/55 = 16/55$\\
&$P(X=2) = 1 \times 1/55 = 16/55$\\
&\\ 
&$\mu = E[X] = ?$\\
&$E[X^2] = ?$\\
&$Var(X) = ?$\\
&\hrule\\
&\\
&$-2\cdot25 = -50$\\
&$-1\cdot16 = -16$\\
&$0\cdot9   = 0$\\
&$1\cdot4   = 4$\\
&$2\cdot1   = 2$\\
&\\
&$\sum x\,(x-3)^2 = -50 -16 + 0 + 4 + 2 = -60$\\
&\\
&$\mu = \sum_{x=-2}^{2} x\,p(x) = \frac1{55}\sum_{x=-2}^{2} x\,(x-3)^2  -\frac{60}{55} = -\frac{12}{11}\approx -1.0909$\\
&\hrule\\
&\\
&$E[X^2] = \sum_{x=-2}^{2} x^2\,p(x) = \frac1{55}\sum_{x=-2}^{2} x^2\,(x-3)^2$\\
&$E[X^2] = \sum x^2\,(x-3)^2 = 4\cdot25 +1\cdot16 +0\cdot9+1\cdot4+4\cdot1$\\
&$E[X^2] = \sum x^2\,(x-3)^2 = 100+16+0+4+4 = 124$\\
&$E[X^2] = \frac{124}{55}\approx 2.2545$\\
&\hrule\\
&\\
&$\mathrm{Var}(X) = E[X^2] - \mu^2$\\
&$\mathrm{Var}(X) = \frac{124}{55} - \Bigl(-\tfrac{12}{11}\Bigr)^2$\\
&$\mathrm{Var}(X) = \frac{124}{55} - \frac{144}{121}$\\
&$\mathrm{Var}(X) = \frac{1364 - 720}{605}$\\
&$\mathrm{Var}(X) = \frac{644}{605}$\\
&$\mathrm{Var}(X) = \approx 1.0645$\\
\end{tabular}
\end{table}

\end{document}