Let \(Y = Z^{2}\), \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is a normal random variable with mean \(\mu\) and variance \(\sigma^{2}\). The variance of \(Y\) is
Answer
(B)
Explanation
Standardising a normal variable gives \(Z \sim N(0, 1)\), so \(E[Z^{2}] = \operatorname{Var}(Z) = 1\). The fourth moment of a standard normal is \(E[Z^{4}] = 3\): its moment generating function \(e^{t^{2}/2}\) has \(t^{4}\) coefficient \(\frac{1}{8}\), and \(4! \cdot \frac{1}{8} = 3\). Therefore
\[\operatorname{Var}(Y) = E[Z^{4}] - \left(E[Z^{2}]\right)^{2} = 3 - 1 = 2.\]
Equivalently, \(Y = Z^{2}\) is chi-squared with 1 degree of freedom, whose variance is twice the degrees of freedom. Option C is \(E[Z^{4}]\) alone, not the variance.
Answer: B