Assume $10\%$ of the population are left-handed. We consider a stadium with $5000$ people. What is the probability that there are between $750$ and $800$ left-handers in the stadium? So we have $p = 0.1$ and $n = 5000$.
Now, we can see that the probability for a single person is given by $$X_k \sim \text{Bernoulli(0.1)}$$
For $5000$ people we know that the sum of the sequence is binomial distributed $$S_n = \sum_{k=1}^n X_k \sim \text{Binomial}(5000, 0.1)$$
I know that the de Moivre-Laplace Theorem states if $np(1-p) > 9$: $$\mathbb{P}(S_n \leq x) \approx \Phi\left(\frac{x-np}{\sqrt{np(1-p)}}\right)$$
Which gives us now $$\mathbb{P}(750 \leq S_n \leq 800) \approx \Phi\left(\frac{800-5000 \cdot 0.1}{\sqrt{5000 \cdot 0.1(1-0.1)}}\right) - \Phi\left(\frac{750-5000 \cdot 0.1}{\sqrt{5000 \cdot 0.1(1-0.1)}}\right)$$
Which results in $0$. Did I solve this wrong and where is my mistake? Glad for any help.