If \(E(X)\gt\text{Var}(X)\) then \(X\) follows Binomial distribution.

Answer: False

For a binomial distribution \(E(X)=np\) and \(\text{Var}(X)=npq\); since \(0\lt q\lt 1\), \(E(X)\gt\text{Var}(X)\) always holds. But the converse is not true: many other distributions also have mean greater than variance, so \(E(X)\gt\text{Var}(X)\) alone does not mean that \(X\) is binomial.

For example, let \(X\) take the values 1, 2, 3 with probability \(\frac{1}{3}\) each. Then \(E(X)=2\) and \(\text{Var}(X)=\frac{14}{3}-4=\frac{2}{3}\), so \(E(X)\gt\text{Var}(X)\), but \(X\) is not binomial (a binomial variable takes the values \(0,1,\ldots,n\)).