\(\sum\)\(P_i\)\(Y_i\)=\(0\times.1+\large\frac{0.25}{2}\)+1\(\times\).3+\(\large\frac{3}{2}\times.2+2\times.15\)
=\(0+0.125+.3+0.3+0.3\)
=\(1.025\)
=\(\sum\)\(P_i\)\(Y_i\)\(^2\)=\((0\times.1+(\large\frac{1}{2})^2\)\(\times\).25+1\(^2\)\(\times\).3+\((\large\frac{3}{2})^2\)\(\times\).2=2\(^2\)\(\times0.15\)
=\(0.0625+0.3+0.45+.6\)
=\(1.976\)
var(Y)=var(\(\large\frac{x}{2}\))=\(1.976\)-(\(1.025\))\(^2\)
=\(0.1118\)
\(\sum\)\(P_i\)\(x_i\)=\(0\times.1+1\times2.5+2\times.3+3\times.2+4\times1.5\)
=\(2.05\)
=\(\sum\)\(P_i\)\(x_i\)\(^2\)=\(0\times.1+1{^2}\times.25+2{^2}\times.3+3{^2}\times.2+4{^2}\times.15\)
=.25+1.2+1.8+2.4
=5.65
var(X)=5.65-(2.05)\(^2\)
=5.65-4.1025
=0.4475