# continuity correction of normal approximation

In a game you win 10 with probability $$\frac{1}{20}$$ and lose 1 with probability $$\frac{19}{20}$$. Approximate the probability that you lost less than \$100 after the first 200 games.How will thisprobability change after 300 games?

I let $$S_n$$ be the number of winnings, then the total winning will be $$X_n=10S_n-1(n-S_n)$$. Then I want to calculate $$P(X_n>-100)=P(S_n>\frac{100}{11})$$ using normal approximation and continuity correction. I tried to calculate it 2 ways: $$1)P(S_n>\frac{100}{11})=P(\frac{S_n-np}{\sqrt{np(1-p)}}>\frac{\frac{100}{11}-\frac{1}{2}-np}{\sqrt{np(1-p)}})$$ and $$2) P(S_n>\frac{100}{11})=1-P(S_n\leq\frac{100}{11})=1-P(\frac{S_n-np}{\sqrt{np(1-p)}}\leq\frac{\frac{100}{11}+\frac{1}{2}-np}{\sqrt{np(1-p)}})$$

The answers I got are totally different and both times my answers didn't match the correct answer at the end of the book (which is 0,5636)? Is there any mistake in my computation?