math
posted by rick .
The average middledistance runner at a local high school runs the mile in 4.5 minutes, with a standard deviation of 0.3 minute. The percentage of a runners that will run the mile in less than 4 minutes is %.

.5/.3 = 1.67 standard deviations below mean
Look at table for normal distribution
My quick table is very rough
for z = 1.6 F(z) = .055
for z = 1.7 F(z) = .045
so about F(1.67) = about .05 or 5 %
By the way, I do not believe it. Suspect sigma much lower than 0.3 min or possibly not a normal distribution at all.