Contradiction between me and my professor solution in normal app. to binomial 
%$20$ of people smoke , if we choose a sample space consisting of $225$ people , what is the probability that more than $40$ people are smokers

This is an probability question given by my professor.As you see ,it can be solved by normal approximation to binomial distribution.However , there is problem here. The problem is the limits of z value. Me and my professor have different solutions , i want you guys decide which of us is right..
My professors'approach: If it is wanted more than $40$ , then find $P(x=41)+P(x=42)+...+P(x=225)$ , so make use of normal approach such that $P(X >39.5)$
My approach : If it is wanted more than $40$ , then find $P(x=41)+P(x=42)+...+P(x=225)$ , so make use of normal approach such that $P(X >40.5)$
The rest is classical solution and not our problem here.. I thought that it must be $P(X >40.5)$ instead of $P(X >39.5)$ , because $P(X >39.5)$ include the probability of $P(x=40)$.However , he continues to solve other problems in this way. For example , another binomial question asked for $P(10<x<18)$ and he solved it such that $P(9.5<x<18.5)$ , but i think that because of it is binomial ,i.e discrete, it must be $P(11 \leq x \leq 17)$ ,so $P(10.5<X<17.5)$
When you look at these $2$ example , which of us right ? Thanks in advance..
NOTE = He gives this course in many years  , if i am missing something please explain clearly
 A: When using continuity correction (see the link below) when estimating $P(X \leq 40)$ where $X\sim \text{Binomial}(n,p)$ , you use the probability $P(Y \leq 40.5)$ where $Y \sim N(np,\sqrt{np(1-p)})$. Hence for estimating $P(X > 40)$ you would naturally have to use $P(Y>40.5)$. So this would mean your approach is correct.
However, your professor might be taking into account that for certain values of $p$ and $n$, the Normal approximation underestimates the Binomial distribution and hence choosing to correct for this by looking at $P(X > 39.5)$ instead. I am not an expert on this topic so I won't comment on whether or not this is reasonable or not. I just wanted to point out that this could be an explanation.
As a final remark, note that $P(X > 39.5) \approx P(X > 40.5)$ as long as $np$ is not too close to 40. Hence in a some situations it does not matter which approach you use (as you are using approximations anyway).

Sources:
Continuity correction: https://en.wikipedia.org/wiki/Continuity_correction

Edit: It just occured to me that your professor might intepret it as greater or equal to 40 in which case the probability of interest is $P(X \geq 40)=P(X>39)$. Then you would both agree and pick $39.5$.
