The prompt
There are 100 rooms in a hotel. Since the owner knows that 10% of the early reservations are canceled before the arrival, he ordered to accept reservations for more than 100 rooms. What is the probability that after accepting 104 reservations the hotel will run out of vacant rooms?
I am required to solve the following problem using Central Limit theorem. The way I attempted to solve the problem is by assuming $n = 100$, $$q = P(\text{reservation is cancelled}) = 0.1$$ $$p = P(\text{reservation is not cancelled}) = 0.9$$ So, $mean(np) = 90$, $var(\sqrt{npq}) = 3$
We are trying to find the probability that hotel doesn't have any more rooms to provide after 104 reservations, $$P(X \leq 104) = \phi(\frac{104-90}{3}) $$
Is this the correct way to attempt a problem like this? Hints appreciated.
The central limit theorem says that if $X_1,\dots, X_n$ are independent and iid with mean $\mu$ and variance $\sigma ^2$ and
$$S_n = \frac1n \sum_{i=1}^n X_i,$$
then we have
$$\lim_{n\to\infty}\mathbb P \Big(\sqrt n (S_n - \mu)\leqslant z\Big) = \Phi\left(\frac z\sigma\right),$$
where $\Phi$ is the cdf of the standard normal distribution.
In your case, it's easy to identify that $X_i$ is Bernoulli with $\mathbb P(X_i = 1) = 0.9$ and $\mathbb P(X_i = 0) = 0.1$, so that $\mu = 0.9$ and $\sigma ^2=0.09 \implies \sigma = 0.3$.
You wish to know $\mathbb P\left(\sum_{i=1}^{104}X_i > 100\right)$, the probability that there will not be enough rooms. Then:
\begin{align} \mathbb P\left(\sum_{i=1}^{104}X_i > 100\right) &= 1- \mathbb P\left(\sum_{i=1}^{104}X_i \leqslant 99\right) \\&= 1- \mathbb P\left(104\,S_{104} \leqslant 99\right) \\&= 1- \mathbb P\left(\sqrt{104}\,S_{104} \leqslant \frac{99}{\sqrt{104}}\right) \\&= 1- \mathbb P\left(\sqrt{104}\,\big(S_{104} - 0.9\big)\leqslant \frac{99}{\sqrt{104}} - \sqrt{104}\cdot0.9\right) \\&= 1- \mathbb P\left(\sqrt{104}\,\big(S_{104} - 0.9\big)\leqslant \frac{\sqrt{104}}{104}\big(99 - 104\cdot0.9\big)\right) \\&= 1- \mathbb P\left(\sqrt{104}\,\big(S_{104} - 0.9\big)\leqslant \frac{\sqrt{104}}{104}\,\frac{27}5\right) \\&\approx 1 - \Phi\left(\frac{\frac{\sqrt{104}}{104}\,\frac{27}5}{0.3}\right) \simeq 0.038778 \end{align}
Compare this with the actual result. $Y = \sum_{i=1}^{104}X_i$ has the binomial distribution $B(104, 0.9)$. Then
$$ 1 - \mathbb P(Y\leqslant 99) = 1 - \sum_{i=0}^{99}\,\binom{104}i {(0.9)}^i{(0.1)^{104 - i}} \simeq 0.017938. $$
Therefore, while they are both 'small', the percentage difference in value is not.
EDIT: Following Robert Israel's suggestion, if we apply the continuity correction we'll get
$$ p = \mathbb P\left(Z > 100.5\right), $$
where $Z\sim N(104\cdot 0.9, 104\cdot0.9\cdot0.1)$. This yields
$$p = \int_{100.5}^{\infty} \frac{1}{\sqrt{2\pi \cdot 9.36}}\,\exp\left(-\frac{{(x-93.6)}^2}{2\cdot9.36}\right)\,dx\approx 0.0120561,$$
which is much closer to our true result.