Binomial distribution (working included)

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I've been working on this question but I'm not too sure if I am correct... Any feedback or advice would be much appreciated.

Question: Suppose an aircraft has a capacity of 500 passengers. The airline that operates it 'overbooks' the plane by accepting reservations for up to 575 passengers, because it knows from experience that 5% of passengers do not check in for the flight.

a) Assuming that X, the number of passengers who check in for the flight, can be modelled by a binomial distribution, calculate the probability that one or more passengers will be denied the chance to board.

My working: We can safely use the normal approximation to the binomial with such large numbers.

mean = np = 575*0.95 = 546.25

sd = √(npq) = √(575*0.95*0.05) = 5.226

with continuity correction, > 500 becomes > 500.5

z-score = (500.5-546.25)/5.226 = -8.75

P(z> -8.75) ≈ 1

This means that for all practical persons, it is certain that one or more persons will be denied the chance to board.