Previous in Probability and Statistics Next in Probability and Statistics
Question Number 180190 by Ar Brandon last updated on 08/Nov/22
Answered by Acem last updated on 09/Nov/22
![X(Ω)= {5, 2, 1, 0} ; P(X=0)= 0.4 x_i 5 2 1 0 P(X =x_i ) 0.1 0.2 0.3 0.4 V(x)= Σ_(i= 1) ^n x_i ^2 p_i − [E(x)]^2 ...(1) E(x)=Σ_(i= 1) ^n x_i p_i = 1.2 V(x)= 3.6−1.44= 2.16](Q180199.png)
$$\:{X}\left(\Omega\right)=\:\left\{\mathrm{5},\:\mathrm{2},\:\mathrm{1},\:\mathrm{0}\right\}\:;\:{P}\left({X}=\mathrm{0}\right)=\:\mathrm{0}.\mathrm{4} \\ $$$$ \\ $$$$\:{x}_{{i}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{0} \\ $$$${P}\left({X}\:={x}_{{i}} \right)\:\:\:\:\:\mathrm{0}.\mathrm{1}\:\:\:\:\:\mathrm{0}.\mathrm{2}\:\:\:\:\:\mathrm{0}.\mathrm{3}\:\:\:\:\:\mathrm{0}.\mathrm{4} \\ $$$$ \\ $$$$\:{V}\left({x}\right)=\:\underset{{i}=\:\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{i}} ^{\mathrm{2}} \:{p}_{{i}} \:−\:\left[{E}\left({x}\right)\right]^{\mathrm{2}} \:...\left(\mathrm{1}\right) \\ $$$$\:{E}\left({x}\right)=\underset{{i}=\:\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{i}} \:{p}_{{i}} \:=\:\mathrm{1}.\mathrm{2} \\ $$$$\:{V}\left({x}\right)=\:\mathrm{3}.\mathrm{6}−\mathrm{1}.\mathrm{44}=\:\mathrm{2}.\mathrm{16} \\ $$$$\: \\ $$
Commented by Ar Brandon last updated on 09/Nov/22
Thank you, Sir!