Standard normal Distribution

    The times taken by applicants for data entry positions to type a standard passage were normally distributed with a mean of 90 seconds and a standard deviation of 12 seconds. If it is company policy to consider only the fastest 20% of the applicants, what is the cutoff point between those considered by the company and those not considered?  (Hint: fastest applicants will use the least amount of time)

1-.2 = .80 = -.84

-.84=(X-90)/12= -.84 ×12+90=79.92 <---this is correct but how do people get the -.84??

My question is how are people getting the -.84???