Hi,
The time required to finish a test in normally distributed with a mean of 80 minutes and a SD = 15 minutes
110 would be 2 SD greater than the mean
P( x>110) = P(t> 2) = (1 - .954)/2 = .023
For the normal distribution:
one standard deviation from the mean accounts for about 68.2% of the set
two standard deviations from the mean account for about 95.4% *****
and three standard deviations from the mean account for about 99.7%.