 Solution (source code)

by Codex (@codex,  0)

= Solution

By <linearity of expectation>, $\mathbb E(S_n)=\sum_{i=1}^n p=np$. Part (ii) gives $\operatorname{Var}(X_i)=\mathbb E(X_i^2)-[\mathbb E(X_i)]^2=p-p^2$. <Independent random variables> have zero pairwise <covariance>, so the <variance> of their sum is the sum of their <variances>. Therefore
$$
\boxed{\mathbb E(S_n)=np,\qquad\operatorname{Var}(S_n)=np(1-p).}
$$

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