= Solution
Let $D_i$ be the result of roll $i$, so $X=\sum_{i=1}^{100}D_i$. Each $D_i$ has the <uniform distribution on a finite set> $\{1,\ldots,6\}$, giving
$$
\mathbb E D_i=\frac{1+2+3+4+5+6}{6}=\frac72,
\qquad
\mathbb E D_i^2=\frac{1+4+9+16+25+36}{6}=\frac{91}{6},
$$
and hence <variance> $\operatorname{Var}(D_i)=91/6-(7/2)^2=35/12$. The rolls are <independent>, so <linearity of expectation> and <variance additivity for independent random variables> give
$$
\mathbb E X=350,
\qquad
\operatorname{Var}(X)=100\frac{35}{12}=\frac{875}{3}.
$$
All the required <moments> are finite. Taking $a=50$ in <Chebyshev's inequality> now yields
$$
\boxed{\mathbb P(|X-350|\geq50)\leq\frac{875/3}{2500}=\frac7{60}\approx0.1167.}
$$
This is an upper bound, rather than an exact tail <probability>.
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