Answer to Question #322902 in Discrete Mathematics for Ketcha Jim Tabot

Question

2. A random variable X can have values -4. - 1, 2, 3 and 4, each with probability 1/5 Find: the density function f_{Y} . the distribution function F_{y} , the mean E(Y) . and the variance sigma_{y} ^ 2 , of the random Y = 3X ^ 3

Answer 1

$P\left( Y=-192 \right) =P\left( X=-4 \right) =\frac{1}{5}\\P\left( Y=-3 \right) =P\left( X=-1 \right) =\frac{1}{5}\\P\left( Y=24 \right) =P\left( X=2 \right) =\frac{1}{5}\\P\left( Y=81 \right) =P\left( X=3 \right) =\frac{1}{5}\\P\left( Y=192 \right) =P\left( X=4 \right) =\frac{1}{5}\\f_Y\left( y \right) =\left\{ \begin{array}{c} 0.2,y\in \left\{ -192,-3,24,81,192 \right\}\\ 0,otherwise\\\end{array} \right. \\F_Y\left( y \right) =\left\{ \begin{array}{c} 0,y<-192\\ 0.2,-192\leqslant y<-3\\ 0.4,-3\leqslant y<24\\ 0.6,24\leqslant y<81\\ 0.8,81\leqslant y<192\\ 1,y\geqslant 192\\\end{array} \right. \\EY=\frac{-192-3+24+81+192}{5}=20.4\\EY^2=\frac{\left( -192-20.4 \right) ^2+\left( -3-20.4 \right) ^2+\left( 24-20.4 \right) ^2+\left( 81-20.4 \right) ^2+\left( 192-20.4 \right) ^2}{5}=15758.6\\var\left( Y \right) =15758.6-20.4^2=15342.5$