Answer:
97.72% probability that their mean body temperature is greater than 98.4degrees° F.
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation 
In this problem, we have that:

If 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4degrees° F.
This is 1 subtracted by the pvalue of Z when X = 98.4. So

By the Central Limit Theorem



has a pvalue of 0.0228
1 - 0.0228 = 0.9772
97.72% probability that their mean body temperature is greater than 98.4degrees° F.