Statistics
posted by Lisa .
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?

99  98.2 = .8
.8 divided by .6 = 1.29...zscore
zscore of 1.29...on table
=.9015 = 90.15 %