= Solution
{c}
Assuming $g$ is <measurable>, applying it separately to <independent and identically distributed random variables> preserves those properties. Thus $W_i=g(X_i)$ have common <mean> $\mu_g$ and <variance> $\sigma_g^2$. For $\sigma_g^2>0$, the <central limit theorem> gives
$$
\frac{\sqrt n}{\sigma_g}\left(\frac1n\sum_{i=1}^ng(X_i)-\mu_g\right)\xrightarrow{d}N(0,1).
$$
Hence, for large $n$, the approximate <probability distribution> is
$$
\boxed{\frac1n\sum_{i=1}^ng(X_i)\ \approx\ N\left(\mu_g,\frac{\sigma_g^2}{n}\right).}
$$
The second argument is the <variance>. If $\sigma_g^2=0$, this <sample mean> equals $\mu_g$ <almost surely>, so the appropriate description is a <point mass> at $\mu_g$.
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