Hypothesis test: Type I error I've been trying to understand the solution to this exercise:
It is estimated that the proportion of adults living in a small town who are college graduates is $p = 0.6$. To test this hypothesis, we selected a random sample of $200$ adults. If the number of graduates in our sample is any number between $110\leq x\leq 130$, accept the null hypothesis that $p = 0.6$, otherwise, we conclude that $ p\neq 0.6$.
Evaluate $\alpha$ (Type I error) with the assumption that $p = 0.6$. Use the normal distribution.
My anwers is: 
\begin{align*}
\alpha&= P(\text{Type I error})\\
&=P\left(z\leq \frac{109,5-200(0,6)}{\sqrt{200(0,6)(0,4)}}\right)+P\left(z\geq \frac{130,0-200(0,6)}{\sqrt{200(0,6)(0,4)}}\right)\\
&\approx2\cdot(0,0655)\\
&=0,131
\end{align*}
Is this correct? I do not know why (and when) I have to use the values $​​109.5$ and $130.5$, because the theorem of normal approximation to the binomial does not say anything about it. Can anyone help?
Thank you very much.

Probably there is currently a typo, you meant $130.5$, not $130.0$. For one thing, our expression should be symmetric about $120$. 
We reject if we get a result of $131$ or bigger. The probability of this is $1$ minus the probability that the result is $130$ or less. Let $X$ be the number of college graduates. Then $X$ takes on consecutive integer values $0,1,2,\dots,200$.  When we approximate $\Pr(X \le k)$ by the normal, one usually gets a better approximation by calculating the probability that the normal is $\le k+\frac{1}{2}$.
So we want $1$ minus the probability that the normal is $\le 130.5$. That is exactly the probability that the normal is $\ge 130.5$.
Remark: Minor tip for the continuity correction: It is best to only try to remember the method for $\Pr(X\le k)$. and use this to figure out each time what to do when we have a situation like $X \lt k$, or $X\ge k$, or $X\gt k$. 
