XS="100%" YS="100%"/> 2 <ART .205 FILE="02_56eq03.eps" W="88pt" H="20.809pt" XS="100%" YS="100%"/> 3 <ART .051 FILE="02_56eq04.eps" W="86.559pt" H="20.578pt" XS="100%" YS="100%"/> 4 <ART .0064 FILE="02_56eq06.eps" W="94pt" H="20.117pt" XS="100%" YS="100%"/> 5 <ART .00032 FILE="02_56eq05.eps" W="98.248pt" H="20.809pt" XS="100%" YS="100%"/> 1.0 (X (X XP(X) X E(X) E(X))2 E(X))2P(X) – – – 0.0 .997 .326 –.9985 .41 .0015 0 0 .41 1.0015 1.003 .2056 .153 2.0015 4.006 .2043 .024 3.0015 9.009 .0541 .0015 4.0015 16.012 .0048 .9985 .7948 = .9985 1.0 2 = (X E(X) E(X))2P(X) – = .7948 0.80 Using the formulas for the binomial: = E(X) np = (5)(.2) = 1.0 2 = np(1 p) – = 5(.2)(.8)
XS="100%" YS="100%"/> 2 <ART FILE="02_56eq03.eps" W="88pt" H="20.809pt" XS="100%" YS="100%"/> .205 3 <ART FILE="02_56eq04.eps" W="86.559pt" H="20.578pt" XS="100%" YS="100%"/> .051 4 <ART FILE="02_56eq06.eps" W="94pt" H="20.117pt" XS="100%" YS="100%"/> .0064 5 <ART FILE="02_56eq05.eps" W="98.248pt" H="20.809pt" XS="100%" YS="100%"/> .00032 1.0 XP(X) X – E(X) (X – E(X))2 (X – E(X))2P(X) 0.0 –.9985 .997 .326 .41 .0015 0 0 .41 1.0015 1.003 .2056 .153 2.0015 4.006 .2043 .024 3.0015 9.009 .0541 .0015 4.0015 16.012 .0048 .9985 .7948 E(X) = .9985 1.0 2 = (X – E(X))2P(X) = .7948 0.80 Using the formulas for the binomial: E(X) = np = (5)(.2) = 1.0 2 = np(1 – p) = 5(.2)(.8)
solution-manual-quantitative-analysis-for-management-12th-edition-render.docx