Assume the body temperature of healthy adults are normally distributed with a mean of 98.2 degrees F and a standard deviation of .62 degrees F. If you have a temperature of 99.0 what is your percentile score?

Answer 1

99 - 98.2 = .8

.8 divided by .6 = 1.29...z-score

z-score of 1.29...on table
=.9015 = 90.15 %