normal approximation of binomial distribution ("overbooking")

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for the following example of an "Overbooking" I have to calculate the probability by using the Central Limit Theorem:

  • An airline books 52 seats whilst there are only 50 seats available. A guest appears with a probability of 90%. What's the probablity that maximum 50 guests are coming?

I thought about normal approximation:

  • $P(0 \leq S_n \leq 50) \approx \phi (\frac{50+0.5-µ}{\sigma}) - \phi (\frac{0-0.5-µ}{\sigma})$.
  • $µ = np = 52\cdot 0.1 = 5.2$.
  • $\sigma = \sqrt{npq} = \sqrt{4.68}$

Finally I get $0.99978$, whereas the probability by using binomial distribution gives me $0.9717$.

Have I done it correctly or are there some mistakes?