var-x-2-then-var-2x-3-E-x-2-then-E-2x-3-

Question Number 76726 by Rio Michael last updated on 29/Dec/19

$$\:\mathrm{var}\left(\mathrm{x}\right)\:=\:\mathrm{2}\:\mathrm{then}\:\mathrm{var}\left(\mathrm{2x}\:−\mathrm{3}\right)=? \\ $$$$\mathrm{E}\left(\mathrm{x}\right)\:=\:\mathrm{2}\:\mathrm{then}\:\mathrm{E}\left(\mathrm{2x}\:−\mathrm{3}\right)\:=\:? \\ $$

Answered by john santu last updated on 30/Dec/19

E(x)=Σ_x xf(x)=2 so E(2x−3) = Σ_x (2x−3)f(x) = 2Σ_x xf(x)−Σ_x 3f(x) = 4−3Σ_x f(x)

$${E}\left({x}\right)=\underset{{x}} {\sum}{xf}\left({x}\right)=\mathrm{2}\: \\ $$$${so}\:{E}\left(\mathrm{2}{x}−\mathrm{3}\right)\:=\:\underset{{x}} {\sum}\left(\mathrm{2}{x}−\mathrm{3}\right){f}\left({x}\right) \\ $$$$=\:\mathrm{2}\underset{{x}} {\sum}{xf}\left({x}\right)−\underset{{x}} {\sum}\mathrm{3}{f}\left({x}\right) \\ $$$$=\:\mathrm{4}−\mathrm{3}\underset{{x}} {\sum}{f}\left({x}\right) \\ $$

Commented by john santu last updated on 30/Dec/19

$${note}\::\:\underset{{x}} {\sum}{f}\left({x}\right)=\mathrm{1} \\ $$

Answered by benjo 1/2 santuyy last updated on 30/Dec/19

using by theorem E(2x−3)=2E(x)−3 so we get E(2x−3)=2×2−3=1

$${using}\:{by}\:{theorem}\: \\ $$$${E}\left(\mathrm{2}{x}−\mathrm{3}\right)=\mathrm{2}{E}\left({x}\right)−\mathrm{3}\: \\ $$$${so}\:{we}\:{get}\:{E}\left(\mathrm{2}{x}−\mathrm{3}\right)=\mathrm{2}×\mathrm{2}−\mathrm{3}=\mathrm{1} \\ $$

Answered by benjo 1/2 santuyy last updated on 30/Dec/19