A set of 10 numbers has a mean of 10 and a standard deviation of 2.0 another set of 10 numbers have a mean of 4 and a standard deviation of 3.0 find the standard deviation of the 20 numbers
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suppose first set of number is {$x_1,x_2,x_3,......,x_{10}$}.since mean of this set is $10$. so$$\dfrac{x_1+x_2+......+x_{10}}{10}=10\implies \mathbf {{x_1+x_2+......+x_{10}}=100}$$ To calculate Standard deviation we need deviation: $$S.D.=\sqrt{\dfrac 1{N-1}\sum_ {i=1}^N(x_i-x)^2}$$ S.D.=$2$ $$4=\dfrac{{(x_1-10)}^2+{(x_2-10)}^2+...........{(x_{10}-10)}^2}{10-1}$$ $$4=\dfrac{(x_1^2+x_2^2+.......+x_{10}^2)-20({x_1+x_2+......+x_{10}})+1000}{9}$$ $$36={(x_1^2+x_2^2+.......+x_{10}^2)-(20\times 100)+1000}$$ $$\mathbf {1036={x_1^2+x_2^2+.......+x_{10}^2}}$$
similarly for second set {$x_{11},x_{12},.......x_{20}$}
Mean=$4$$${\mathbf{x_{11}+x_{12}+......+x_{20}}=40}$$ and from S.D.=$3$ $$\mathbf {x_{11}^2+x_{12}^2+......+x_{20}^2=241}$$
similarly we can calculate S.D. for these 20 numbers where those mean = 7
Now new set is {$x_1,x_2,...............,x_{20}$} $$\mathbf{x_1+x_2+...............+x_{20}=140}$$ $$S.D.^2=\dfrac{(x_1-7)^2+(x_2-7)^2+........+(x_{20}-7)^2}{20-1}$$ $$S.D.^2=\dfrac{(x_1^2+x_2^2+....x_{10}^2+x_{11}^2.......+x_{20}^2)-14(x_1+x_2+....+x_{20})+20\times49}{20-1}$$ $$S.D.^2=\dfrac{(1036+241)-14(140)+20\times49}{20-1}$$ $$S.D.^2=\dfrac{(1036+241)-14(140)+20\times49}{20-1}$$ $$S.D.^2=\dfrac{1277-1960+980}{20-1}$$ $$S.D.^2=\dfrac{1277-1960+980}{20-1}$$ $$S.D.=3.954$$
iostream007
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \angles{x} = {1 \over 20}\sum_{n = 1}^{20}x_{n} = {1 \over 20}\pars{10\,{\sum_{n = 1}^{10}x_{n} \over 10} + 10\,{\sum_{n = 11}^{20}x_{n} \over 10}} = \half\pars{\angles{x}_{1} + \angles{x}_{2}} $$ Similarly, $$ \angles{x^{2}} = \half\pars{\angles{x^{2}}_{1} + \angles{x^{2}}_{2}} $$ Then, \begin{align} &\angles{x^{2}} - \angles{x}^{2} = \half\pars{\angles{x^{2}}_{1} + \angles{x^{2}}_{2}} - {1 \over 4}\pars{\angles{x}_{1}^{2} + 2 \angles{x}_{1}\angles{x}_{2} + \angles{x}_{2}^{2}} \\[3mm]&= \half\pars{\angles{x^{2}}_{1} - \angles{x}_{1}^{2}} + \half\pars{\angles{x^{2}}_{2} - \angles{x}_{2}^{2}} + {1 \over 4}\pars{\angles{x}_{1}^{2} - 2 \angles{x}_{1}\angles{x}_{2} + \angles{x}_{2}^{2}} \end{align} $$\color{#0000ff}{\large% \sigma ={\root{2} \over 2}\root{% \sigma_{1}^{2} + \sigma_{2}^{2} + \half\pars{\angles{x}_{1} - \angles{x}_{2}}^{2}}} $$ $$ \sigma = {\root{2} \over 2}\root{2^{2} + 3^{2} + \half\pars{10 - 4}^{2}} = {\root{62} \over 2} \approx 3.9370 $$
Felix Marin
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