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

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following