Solution
= Solution
The preceding marginal calculation makes each $V_i$ <uniformly distributed> on $(0,1)$, with <expectation> $1/2$ and <variance> $1/12$. By <independence>,
$$
\mathbb E(S_n/n)=\frac12,\qquad
\operatorname{Var}(S_n/n)=\frac1{12n}.
$$
Apply part (a) to the centred <random variable> $Z=S_n/n-1/2$ to obtain
$$
\boxed{\mathbb P\left(\left|S_n/n-1/2\right|\geq a\right)\leq\frac1{12na^2}}.
$$
This is <Chebyshev's inequality> in this setting.