Задача 29140 5. Найти внутригрупповую, межгрупповую и...
Условие
Решение
N_(1)=30+15+5=50
vector{x_(1)}=(x_(1)*n_(1)+x_(2)*n_(2)+x_(3)*n_(3))/N_(1)=
=(1*30+2*15+8*5)/50=2
N_(2)=10+15=25
vector{x_(2)}=(x_(1)*n_(1)+x_(2)*n_(2)+x_(3)*n_(3))/N_(2)=
=(1*10+6*15)/25=4
N_(3)=20+5=25
vector{x_(3)}=(x_(1)*n_(1)+x_(2)*n_(2)+x_(3)*n_(3))/N_(3)=
=(3*20+8*5)/25=4
Найдем групповые дисперсии:
D_(1 гр)=((x^2_(1)*n_(1)+x^2_(2)*n_(2)+x^2_(3)*n_(3))/N_(1)) - (vector{x_(1)})^2=
=(1/50)*(1^2*30+2^2*15+8^2*5)-2^2=4,2
D_(2 гр)=((x^2_(1)*n_(1)+x^2_(2)*n_(2)+x^2_(3)*n_(3))/N_(2)) - (vector{x_(2)})^2=
=(1/25)*(10*1^2+15*6^2)-2^2=22-16=6
D_(3 гр)=((x^2_(1)*n_(1)+x^2_(2)*n_(2)+x^2_(3)*n_(3))/N_(3)) - (vector{x_(3)})^2=
=(1/25)*(20*3^2+5*8^2)-4^2=20-16=4
n=N_(1)+N_(2)+N_(3)=50+25+25=100
D_(внгр)=(N_(1)*D_(1гр)+N_(2)*D_(2гр)+N_(3)*D_(3гр))/n=
=(1/100)*(50*4,2+25*6+25*4)=4,6
Найдем общую среднюю
vector{x}=
(1/100)*(1*30+2*15+8*5+1*10+6*15+3*20+8*5)=3
D_(мжгр)=(1/100)*(50*(2-3)^2+25*(4-3)^2+25*(4-3)^2)=1
D_(общ)=(1/100)*(30*(1-3)^2+15*(2-3)^2+5*(8-3)^2+
+10*(1-3)^2+15*(6-3)^2+20*(3-3)^2+5*(8-3)^2)=5,6
О т в е т. D_(внгр)=4,6; D_(мжгр)=1; D_(общ)=5,6