Answer: 0.9996
Step-by-step explanation:
Given : The body temperatures of adults are normally distributed with a mean of 98.6° F and a standard deviation of 0.60° F.
Sample size : n=25
Let x be the random variable that represents the body temperatures of adults.
z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 99° F
[tex]z=\dfrac{99-98.6}{\dfrac{0.60}{\sqrt{25}}}\approx3.33[/tex]
Now, the probability that their mean body temperature is less than 99° F will be :-
[tex]P(X<99)=P(z<3.33)=0.9995658\approx0.9996[/tex]
Hence, the probability that their mean body temperature is less than 99° F = 0.9996
