# finding percent of people that fall bellow a certain value in a list of data

I have a file containing the times taken to do something. The data is fits a normal distribution. I am trying to find the percent of people that take longer than 1800 seconds. So I find the z-score $z=\frac{X-\mu}{\sigma}=\frac{1800-1369.53}{257.504}\approx 1.67$ By symmetry I take -1.67 and from a table I look up -1.67 and find 0.0475 which gives me 4.75%