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Question Number 42058 by Rio Michael last updated on 17/Aug/18

Given that  x^� = ((Σxi)/n_1 )   and y^� = ((Σyi)/n_2 )  show that    x_c ^� = ((n_1 (x^� ) + n_2 (y^� ))/(n_1 +n_2 ))

$${Given}\:{that}\:\:\bar {\mathrm{x}}=\:\frac{\Sigma\mathrm{x}{i}}{{n}_{\mathrm{1}} }\:\:\:{and}\:\bar {{y}}=\:\frac{\Sigma{yi}}{{n}_{\mathrm{2}} } \\ $$$${show}\:{that}\: \\ $$$$\:\bar {\mathrm{x}}_{{c}} =\:\frac{{n}_{\mathrm{1}} \left(\bar {\mathrm{x}}\right)\:+\:{n}_{\mathrm{2}} \left(\bar {{y}}\right)}{{n}_{\mathrm{1}} +{n}_{\mathrm{2}} } \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18

n_1 x^− =x_1 +x_2 +...+x_n_1    n_2 y^− =y_1 +y_2 +...+y_n_2    ((x_1 +x_2 +...+x_n_1  +y_1 +y_2 +...y_n_2  )/(n_1 +n_2 ))  =((n_1 x^− +n_2 y^− )/(n_1 +n_2 ))

$${n}_{\mathrm{1}} \overset{−} {{x}}={x}_{\mathrm{1}} +{x}_{\mathrm{2}} +...+{x}_{{n}_{\mathrm{1}} } \\ $$$${n}_{\mathrm{2}} \overset{−} {{y}}={y}_{\mathrm{1}} +{y}_{\mathrm{2}} +...+{y}_{{n}_{\mathrm{2}} } \\ $$$$\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +...+{x}_{{n}_{\mathrm{1}} } +{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +...{y}_{{n}_{\mathrm{2}} } }{{n}_{\mathrm{1}} +{n}_{\mathrm{2}} } \\ $$$$=\frac{{n}_{\mathrm{1}} \overset{−} {{x}}+{n}_{\mathrm{2}} \overset{−} {{y}}}{{n}_{\mathrm{1}} +{n}_{\mathrm{2}} } \\ $$

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