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???

From here:
1-.2 = .80 = -.84
.80 = -.84
-.84

>>1
The -0.84 is the standard normal score (the Z-score) associated with the leftmost 20% of the distribution.  (It is wrong to write "0.80 = -0.84" which makes no sense.)  People don't try to calculate it directly, but instead use an electronic function or a printed table of Z-scores versus probabilities.

20% of a normal distribution is .84 standard deviations away from the mean (use one z-table and look up 1-tailed, 20% critical value). The mean is 90. One standard deviation is 12. Thus we have 90 - (.84) 12 = 79.92.

20% of applicants will be below 79.92.