Here the population variance is $\sigma^{2}=\mu^{2}$ .

(A), (C) and (D) are true. By the weak law of large numbers $S_n^{2}\overset{P}{\longrightarrow}\sigma^{2}=\mu^{2}$ , giving (A). Also $\overline{X}\overset{P}{\longrightarrow}\mu$ , giving (C). And the normal law is symmetric about $\mu$ , so its population median is $\mu$ and the sample median is consistent for it, giving (D).

(B) is the statement that fails. Applying the continuous mapping theorem to (A),

$S_n=\sqrt{S_n^{2}} \overset{P}{\longrightarrow} \sqrt{\mu^{2}}=|\mu| .$

So whenever $\mu>0$ we have $S_n\overset{P}{\longrightarrow}\mu$ and $S_n$ is a perfectly good consistent estimator of $\mu$ . The blanket word "cannot" is therefore too strong, which is what makes (B) the false statement.

The genuine subtlety. $S_n$ recovers only the magnitude of $\mu$ , never its sign: for $\mu<0$ it converges to $-\mu\neq\mu$ and is inconsistent. So $S_n$ is consistent on the half $\mu>0$ and inconsistent on $\mu<0$ ; read as a claim about the whole parameter space $\mu\in\mathbb{R}$ , (B) has a defensible reading too. Since (A), (C) and (D) are unambiguously true, (B) is the intended key.