There were 49.7 million people with some type of long-lasting condition or disability living in the United States in 2000. This represented 19.3 % of the majority of civilians aged five and above. A sample of 1000 persons is selected at random. What is the approximate the probability that more than 200 persons in the sample have a disability?

This is one of the questions that I'm still confuse which is the right way to solve it.... As far as I know, if we need to find the probability that is more than n, then we're going to use the continuity correction factor of P(X > n - 0.5)

Thus, the formula and solution for this problem should be:

P(X>x0.5)=P(Z>np(1p)x0.5np)

P(X>2000.5)=P(Z>(1000)(0.193)(10.193)2000.5(1000)(0.193))

However, after I check this solution from some websites, their formula and solution for this problem is

P(X>x+0.5)=P(Z>np(1p)x+0.5np)

P(X>200+0.5)=P(Z>(1000)(0.193)(10.193)200+0.5(1000)(0.193))

Which formula and solution is correct?