Question from the test:
Let $X_n$ (for $n=1,2,3,\ldots$) be a random variable with a binomial distribution with parameters $n$ and $p$, and let $\alpha$ be such that $P(Z \geq \alpha)=\frac{1}{2024}$, for $Z \sim N(0,1)$. Find a sequence $a_n$ such that $P(X_n > a_n)$ tends to $\frac{1}{2024}$. (In the formula, you can use $n, \alpha, p$).
So, we want to find a sequence $a_n$ such that
$$\lim_{n \rightarrow \infty} P(X_n > a_n) =P(Z \geq \alpha),$$ or alternatively (*) $$\lim_{n \rightarrow \infty}P(X_n \leq a_n) = P(Z \leq \alpha) = F_Z(\alpha) = \Phi(\alpha).$$ Therefore, we want $a_n$ such that $$\lim_{n \rightarrow \infty}P(X_n \leq a_n)= \Phi(\alpha).$$
From the Central Limit Theorem
$$ \lim_{n \rightarrow \infty} P\left(\frac{X_n - \mu n}{\sqrt{n}\sigma} \leq \alpha\right) = \Phi(\alpha), $$
$$ \lim_{n \rightarrow \infty} P\left(X_n \leq \alpha \cdot \sqrt{n} \cdot \sigma + \mu n\right) = \Phi(\alpha). $$
After comparison with $\lim_{n \rightarrow \infty}P(X_n \leq a_n)= \Phi(\alpha)$ we get that
$$ a_n = \alpha \cdot \sqrt{n} \cdot \sigma + \mu n = \alpha \cdot \sqrt{n} \cdot \sqrt{np(1-p)} + np\cdot n = \alpha n \sqrt{p(1-p)} + n^2p. $$
(*) - I wrote $P(Z \leq \alpha)$ instead of $P(Z < \alpha)$ to fit the formula of CDF. Is it possible to do it this way?
Is this the correct solution?