(i) We know that:
(ii) Now, the distribution becomes
\[E({{X}^{2}})\text{ }=\text{ }1\text{ }\times \text{ 1/2 }+\text{ }4\text{ }\times \text{ }1/5\text{ }+\text{ }16\text{ }\times 3/25\text{ }+\text{ }36\text{ }\times 1/10\text{ }+\text{ }81\text{ }\times 1/25\text{ }+\text{ }225\text{ }\times \text{ }1/25\]
= \[1/2+\text{ }4/5\text{ }+\text{ }48/25\text{ }+\text{ }36/10\text{ }+\text{ }81/25\text{ }+\text{ }225/25\]
= \[0.5\text{ }+\text{ }0.8\text{ }+\text{ }1.92\text{ }+\text{ }3.6\text{ }+\text{ }3.24\text{ }+\text{ }9\text{ }=\text{ }19.06\]
Variance (X) = \[E({{X}^{2}})\text{ }\text{ }{{\left[ E\left( X \right) \right]}^{2}}\]
\[=\text{ }19.06\text{ }\text{ }{{\left( 2.94 \right)}^{2}}~=\text{ }19.06\text{ }\text{ }8.64\text{ }=\text{ }10.42\]