Solution:

$M_0(t) = E(e^{tx})$

=$\sum pi e^{tx_i}$

= $\sum (\frac{1}{2})^x.e^{tx}$

=$\sum (\frac{1}{2^x}).e^{tx}$

=$\sum (\frac{e^t}{2})^x$

since =1,2,3,...

=$\frac {e^t}{2}$+$(\frac {e^t}{2})^2$+($\frac {e^t}{2})^3$

$\frac{e^t}{2}\{1+(\frac{e^t}{2})+(\frac{e^t}{2})^2$

=$\frac {e^t}{2}[\frac{1}{(1-\frac {e^t}{2})}]$

= $\frac {e^t}{2}[\frac{1}{\frac{2-e^t}{2}}]$

= $\frac {e^t}{2}[\frac{2}{2-e^t}]$

=$m.g.f.= M_0(t)=[\frac{e^t}{2-e^t}]$-----(1)

mean , $\mu_1'=[\frac{d}{dt} M_0(t)]_{t=0}$

=$\mu_1'=[\frac{(2-e^t)(e^t)-e^t(0-e^t)}{(2-e^t)^2}]_{t=0}$

=$\mu_1'=[\frac{(2-e^0)(e^0)-e^0(0-e^0)}{(2-e^0)^2}]$

=$\mu_1'=[\frac{(2-1)(1)-1(0-1)}{(2-1)^2}]$

=$\mu_1' = mean =2 $

variance = $\mu_2'-\mu_1'^2$-----(2)

to find $\mu_2'$,

$\mu_2'=[\frac {d^2}{dt^2} M_0(t)]_{t=0}$

$\mu_2'=\frac {d^2}{dt^2}[\frac {e^t}{2-e^t}]_{t=0}$

=$\mu_2'=\frac{d}{dt}[\frac{(2-e^t)(e^t)-e^t(0-e^t)}{(2-e^t)^2}]_{t=0}$

=$\mu_2'=\frac{d}{dt}[\frac{(2e^t-e^{2t}+e^{2t}}{(2-e^t)^2}]_{t=0}$

=$\mu_2'=\frac{d}{dt}[\frac{2e^t}{(2-e^t)^2}]_{t=0}$

==$\mu_2'=[\frac{(2-e^t)^2(2e^t)-(2e^t.2(2-e^t)(0-e^t)}{(2-e^t)^4}]_{t=0}$

=$(2-e^t)[\frac {2(2-e^t)e^t-4e^t(-e^t)}{(2-e^t)^4}]_{t=0}$

=$[\frac {2(2-e^t)e^t+4e^{2t}}{(2-e^t)^3}]$

=$[\frac {2(2-1)1+4{1}}{(2-1)^3}]$

$\mu_2'=6$

variance = $\mu_2'-\mu_1'^2$

=$6-2^2$

Variance=2