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# 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 $(i)\;V\bigg(\Large \frac{X}{2}\normalsize \bigg)\quad (ii)\;variance\;of\;\;X$

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A)
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