A normal distribution uses a z-score:
z=(x-μ)/σ where the mean is μ and the standard deviation is σ. Calculating this would give you (400-500)/100=-1.0 as a z-score.
You can look up how to calculate percentage from a z-score from http://www.z-table.com/. The z-score of 1.0 yields a p(z=-1.0)=15.87% as % of people who scored a 400 or less. Because you want the population who scored 400 or higher, you would calculate this by 100% - p(z=-1.0).
Because your question states "approximately", beware of significant digits.