= Solution
Let $X$ be one score. It has the <discrete uniform distribution> on $1,\ldots,6$, so its <expected value> and <second moment> are
$$
\mathbb EX=\frac{1+2+3+4+5+6}{6}=\frac72,\qquad
\mathbb EX^2=\frac{1^2+2^2+3^2+4^2+5^2+6^2}{6}=\frac{91}{6}.
$$
Therefore
$$
\boxed{\mathbb EX=\frac72,\qquad\operatorname{Var}(X)=\frac{91}{6}-\frac{49}{4}=\frac{35}{12}.}
$$
For <independent> throws, <linearity of expectation> and additivity of <variance> give $\mathbb EY_n=7n/2$ and $\operatorname{Var}(Y_n)=35n/12$. The sample average consequently has <expectation> $7/2$ and <variance> $35/(12n)$.
The <Chebyshev inequality> follows by applying the <Markov inequality> to the nonnegative square of a centered <random variable>. It gives
$$
\mathbb P\left(\left|\frac{Y_n}{n}-\frac72\right|>\frac32\right)
\leq\frac{35/(12n)}{(3/2)^2}=\frac{35}{27n}.
$$
This is at most $1/10$ if $n\geq350/27$. Thus \b[$n=13$ suffices], as does every larger integer. This is the smallest integer guaranteed by this particular <Chebyshev inequality> bound; it need not be the smallest integer satisfying the actual <probability> requirement.
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