Probability of no more than?

You would expect 54.5 headaches per 500 people (10.9% x 500) on average. That is the mean.

You would expect the standard deviation of a distribution of 500 samples to be normal in shape with a sigma of sqrt(500 * 10.9% * 89.1%) = 6.97

The z score required for 50 is (50 - 54.5)/6.97 = -.646

Looking up that z score you get 25.9%