Browse Questions

# Consider the probability distribution of a random variable X.

X 0 1 2 3 4 p(X) 0.1 0.25 0.3 0.2 0.15   Calculate $(i)\;V\bigg(\Large \frac{X}{2}\normalsize \bigg)\quad (ii)\;variance\;of\;\;X$

Toolbox:
• define y=$\frac{x}{2}$
• x=$(0\;1\;2\;3\;4)$
• y=$\frac{x}{2}$=$(0\;0.5\;1\;1.5\;2)$
• P(X)=$(.1\;.25\;.3\;.2\;.15)$
• $\vee$($\large\frac{x}{2}$)=$\vee$(y)=$\sum$$P_i$$Y^2_i$_($\sum$$P_i$$Y_i$)$^2$
• $\vee$(X)=$\sum$$P_i$$X^2_i$_($\sum$$P_i$$X_i$)$^2$
$\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

edited Jun 6, 2013