The body temperatures of adults are normally distributed with a mean of 98.6° f and a standard deviation of 0.60° f. if 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° f. 0.0228 0.8188 0.9360 0.9772

Here the population standard deviation is 0.60 degree F.  If a sample of 36 adults is randomly selected, that results in a sample standard deviation of 0.60 degree F divided by the square root of 36:  0.10 degree F.

The probability in question is the area under the standard normal probability distribution between 98.4 degree F and infinity, and intuitively you can detect that this will be more than 0.5 (corresponding to 50%).

Convert 98.4 degrees F to a z-score, using the sample standard deviation (0.10 degree F).  That z score is 
       98.4-98.6
z =   =  -0.20/0.10 = -2
           0.10

We need to determine the area under the standard normal curve to the right of z=-2.  Use a table of z-scores to do this, or use your calculator's built-in probability functions.  My result is 98.21% (corresponding to an area of 0.9821).

With my calculator I can find this probability using the following command:

normalcdf(-2,100000,0.10).