Healthy people have body temperatures that are normally distributed with a mean of 98.20∘f and a standard deviation of 0.62∘f . (a) if a healthy person is randomly selected, what is the probability that he or she has a temperature above 98.9∘f?

0.098525

Step-by-step explanation:

According to the given description, we have

Let x be a random variable that represents the  body temperatures.

Also,Healthy people have body temperatures that are normally distributed.

Then , the z-score corresponds to x= 98.9 on normal curve ,

              ()

P-value = P(x> 98.9)= P(z>1.129)=1-P(z≤1.129)

=1-0.901475=0.098525 ≈ 0.098525  (using z-value table)

Hence, the probability that he or she has a temperature above 98.9°F = 0.098525