\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
For some \(1 \leq i \leq n\), we have \begin {align*} \Prob (Y = x_i) & = \Prob (Y = X_i, Y = X_1) + \Prob (Y = X_i, Y = X_2) \\ & = \Prob (Y = x_i \mid Y = X_1) \cdot \Prob (Y = X_1) + \Prob (Y = x_i \mid Y = X_2) \cdot \Prob (Y = X_2) \\ & = \Prob (X_1 = x_i) \cdot \Prob (Y = X_1) + \Prob (X_2 = X_i) \cdot \Prob (Y = x_2) \\ & = p a_i + q b_i. \end {align*}
Hence, \begin {align*} \Expt (Y) & = \sum _{i = 1}^{n} x_i \Prob (Y = x_i) \\ & = \sum _{i = 1}^{n} x_i (p a_i + q b_i) \\ & = p \sum _{i = 1}^{n} x_i a_i + q \sum _{i = 1}^{n} x_i b_i \\ & = p \Expt (X_1) + q \Expt (X_2) \\ & = p \mu _1 + q \mu _2. \end {align*}
For the variance, we have \begin {align*} \Expt (Y^2) & = \sum _{i = 1}^{n} x_i^2 \Prob (Y = x_i) \\ & = \sum _{i = 1}^{n} x_i^2 (p a_i + q b_i) \\ & = p \sum _{i = 1}^{n} x_i^2 a_i + q \sum _{i = 1}^{n} x_i^2 b_i \\ & = p \Expt (X_1^2) + q \Expt (X_2^2) \\ & = p \left (\Expt (X_1)^2 + \Var (X_1)\right ) + q \left (\Expt (X_2)^2 + \Var (X_2)\right ) \\ & = p \left (\mu _1^2 + \sigma _1^2\right ) + q \left (\mu _2^2 + \sigma _2^2\right ), \end {align*}
and hence \begin {align*} \Var (Y) & = \Expt (Y^2) - \Expt (Y)^2 \\ & = p \left (\mu _1^2 + \sigma _1^2\right ) + q \left (\mu _2^2 + \sigma _2^2\right ) - \left (p \mu _1 + q \mu _2\right )^2 \\ & = p \sigma _1^2 + q \sigma _2^2 + p \mu _1^2 + q \mu _2^2 - p^2 \mu _1^2 - q^2 \mu _2^2 - 2pq \mu _1 \mu _2 \\ & = p \sigma _1^2 + q \sigma _2^2 + p (1 - p) \mu _1^2 + q (1 - q) \mu _2^2 - 2pq \mu _1 \mu _2 \\ & = p \sigma _1^2 + q \sigma _2^2 + pq \mu _1^2 + pq \mu _2^2 - 2pq \mu _1 \mu _2 \\ & = p \sigma _1^2 + q \sigma _2^2 + pq \left (\mu _1 - \mu _2\right )^2, \end {align*}
as desired.
We have \[ \Prob (B = 1) = \frac {1}{2} \cdot \frac {1}{6} + \frac {1}{2} \cdot \frac {5}{6} = \frac {1}{2}. \]
\(Z_1\) is the sum of \(n\) independent values of \(B\), and counts the number of times when \(B = 1\).
Hence, \(Z_1 \sim \Binomial \left (n, \frac {1}{2}\right )\).
Since \(n \gg 1\), we have \[ Z_1 \sim \Binomial \left (n, \frac {1}{2}\right ) \dot {\sim } \Normal \left (\frac {n}{2}, \frac {n}{4}\right ). \]
The probability of \(Z_1\) being within \(10\) percent of its mean is given by \begin {align*} \Prob \left (\frac {n}{2} - \frac {n}{20} \leq Z_1 \leq \frac {n}{2} + \frac {n}{20}\right ) & = \Prob \left (- \frac {\frac {n}{20}}{\frac {\sqrt {n}}{2}} \leq Z \leq \frac {\frac {n}{20}}{\frac {\sqrt {n}}{2}}\right ) \\ & = \Prob \left (- \frac {\sqrt {n}}{20} \leq Z \leq \frac {\sqrt {n}}{20}\right ) \end {align*}
where \(Z \sim \Normal (0, 1)\) is the standard normal.
As \(n \to \infty \), \(- \frac {\sqrt {n}}{20} \to -\infty \), and \(\frac {\sqrt {n}}{20} \to \infty \), and so the probability approaches \(\Prob (-\infty < Z < \infty )\) which is \(1\).
Let \(X_1 \sim \Binomial \left (n, \frac {1}{6}\right )\), and \(X_2 \sim \Binomial \left (n, \frac {5}{6}\right )\). \(Z_2\) has \(\frac {1}{2}\) chance of taking \(X_1\) and \(\frac {1}{2}\) chance of taking \(X_2\).
We have \(\mu _1 = \frac {n}{6}, \mu _2 = \frac {5n}{6}, \sigma ^2_1 = \sigma ^2_2 = \frac {5n}{36}\).
Hence, \[ \Expt (Z_2) = \frac {1}{2} \cdot \frac {n}{6} + \frac {1}{2} \cdot \frac {5n}{6} = \frac {n}{2}, \] and \[ \Var (Z_2) = \frac {1}{2} \cdot \frac {5n}{36} + \frac {1}{2} \cdot \frac {5n}{36} + \frac {1}{4} \left (\frac {n}{6} - \frac {5n}{6}\right )^2 = \frac {n^2}{9} + \frac {5n}{36}. \]
A normal approximation will not be a good approximation since in this case, \(Z_2\) is bimodal – it is likely to take values close to \(\frac {n}{6}\) or \(\frac {5n}{6}\), but not near the mean \(\frac {n}{2}\).
The bounds within \(10\) percent of the mean is \(\frac {n}{2} \pm \frac {n}{20}\). We have \begin {align*} \Prob \left (\frac {n}{2} - \frac {n}{20} \leq Z_2 \leq \frac {n}{2} + \frac {n}{20}\right ) & = \frac {1}{2} \Prob \left (\frac {n}{2} - \frac {n}{20} \leq X_1 \leq \frac {n}{2} + \frac {n}{20} \right ) + \frac {1}{2} \Prob \left (\frac {n}{2} - \frac {n}{20} \leq X_2 \leq \frac {n}{2} + \frac {n}{20}\right ) \\ & = \frac {1}{2} \Prob \left (\frac {n}{2} - \frac {n}{20} \leq X_1\right ) + \frac {1}{2} \Prob \left (X_2 \leq \frac {n}{2} + \frac {n}{20}\right ) \\ & = \Prob \left (\frac {n}{2} - \frac {n}{20} \leq X_1\right ). \end {align*}
Since \(n\) is large, we have \(X_1 \sim \Binomial \left (n, \frac {1}{6}\right ) \dot {\sim } \Normal \left (\frac {n}{6}, \frac {5n}{36}\right )\), and hence \begin {align*} \Prob \left (\frac {n}{2} - \frac {n}{20} \leq X_1\right ) & = \Prob \left (Z \geq \frac {\frac {n}{2} - \frac {n}{20} - \frac {n}{6}}{\frac {\sqrt {5n}}{6}}\right ) \\ & = \Prob \left (Z \geq \frac {30n - 3n - 10n}{10 \sqrt {5n}}\right ) \\ & = \Prob \left (Z \geq \frac {17 \sqrt {n}}{10 \sqrt {5}}\right ), \end {align*}
and as \(n \to \infty \), \(\frac {17 \sqrt {n}}{10 \sqrt {5}} \to \infty \), and hence the probability tends to \(0\), as desired.