Hi there,

Assume that body temperatures of healthy adults are normally distributed
with a mean of 98.20°F and a standard deviation of 0.62°F (based on data from
University of Maryland researchers).

a. If you have a body temperature of 99.00°F, your percentile score is

z = (x-mu)/s

= (99-98.2)/0.62

= 1.2903

prob(z < 1.2903) from a z table is:

0.9015

That's 90.15%

b. Convert 99.00°F to a standard score (or z score).

z = (x-mu)/s

= (99-98.2)/0.62

= 1.2903

c. Is a body temperature of 99.00°F "unusual"? Why or why not?

No, this is not very unusual, since nearly 10% of people will have higher temperatures.

d. Fifty adults are randomly selected. What is the likelihood
that the mean of their body temperatures is 97.98°F or lower?
Answer:

z = (x-mu)/(s/sqrt(N))

z = (97.98-98.2)/(0.62/sqrt(50))

z = -2.509

prob(z < -2.509) from the z table is:

0.0061

Let me know if you have any questions,

Scott