Limit of $P(100(X_1+\cdots+X_n)>97n)$ as $n \to \infty$ for $(X_k)$ i.i.d.

Question

Let $X_1, X_2, \ldots, X_n$ be iid random variables with the following distribution: $P(X_1=0)=\frac{1}{18}$, $P(X_1=1)=\frac{8}{9}$, and $P(X_1=2)=\frac{1}{18}$.

Let $T_n=100(X_1+\cdots+X_n)$. Determine the limit of $P(T_n>97n)$ as $n \to \infty$.

Attempt: $P(T_n>97n)=P\left(\frac{T_n-n\mu}{\sigma\sqrt{n}}>\frac{97n-n\mu}{\sigma\sqrt{n}}\right)\cong\Phi(-0.09\sqrt{n})$ by the CLT. I am not sure whether my application is correct though, and I'm stuck here.

Answer 1

Setting $\mu:=\mathbb EX_1$ it is not difficult to find $\mu=1$.

Observe that: $$|\overline{X}_n-1|<0.03\implies\overline{X}_n-1>-0.03$$ and:$$\overline{X}_n-1>-0.03\iff T_n>97n$$

So we have: $$P(T_n>97n)\geq P(|\overline{X}_n-\mu|<0.03)=1-P(|\overline{X}_n-\mu|\geq 0.03)$$ Now apply the WLLN.