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?

Statistics 
qwerty123
99  98.2 = .8
.8 divided by .6 = 1.29...zscore
zscore of 1.29...on table
=.9015 = 90.15 %
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