Posted by Maile on .
The body temperatures of adults are normally distributed with a mean of 98.6 degree F and a standard deviation of 0.60 degree F. If 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4 degree F.
z = (98.4 - 98.6)/(0.60/√36) = ?
Once you calculate the z-score, check a z-table for your probability. (Remember that the problem is asking to find the probability "greater than" 98.4)
-2 z=0.0228 1-0.0228 answer 0.9772