Calculate the z score. The well-known formula is z = (score - mean) / (standard deviation). Then use the standard normal table.
P(score above 629) = ?
First, z = (629 - 515) / 114 = 1.
Second, look at the standard normal table, something like http://www.sjsu.edu/faculty/gerstman...fo/z-table.htm: the table tells me that P(z<1)=.5398.
Lastly, some straightforward logic: we wanted score above, so we want to find P(z>1). It is 1 - .5398.