We are given $X_1, X_2, ...$ independent random variables with uniform distribution on $[80,100]$.
What is the probability that for $n=100$ at least half of the variables has value $X_i \ge 95$?
I think this is what we are looking for here $P(S_{50} \ge 50 \cdot 95)$. This would give us $P(Z_{50} \ge 0,8659)$ so something is wrong, because that is very close to $1$.
Could you tell me where I'm making a mistake?
Summarising, if we let $N=$the number of $i \in {1,...,100}$ for which $X_i \ge 95$, we have:
$$P(N=k) = \binom{100}{k} P(X_1 \ge 95)^k P(X_1 < 95)^{100-k} = \binom{100}{k}(1/4)^k(3/4)^{100-k}$$ so $$P(N\ge 50) = \sum_{k=50}^{100}\binom{100}{k}(1/4)^k(3/4)^{100-k}$$
If we want to evaluate the probability with $0,001$ accuracy, we use the central limit theorem for $N \in \mathcal{B}(100, \frac{1}{4})$ and standardize the variable $N$.
Then $N$ has approximately distribution $N(25, 4,33)$ and therefore $P(N \ge 50) \approx 1- \phi(\frac{50-25}{4,33}) \approx 1- \phi(5,77) \approx 0$