Assume the body temperture of healthy adults are normally distributed with a mean of 98.20 degrees F and a standard deviation of 0.62 degrees F...

Question

a. If you have a body temperture of 99.00 degrees F, what is your percentile score?

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

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

d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperature is 97.98 degrees F or lower?

e. A person's body temperature is found to 101.00 degrees F. Is the result unusual? Why or why not?

f. What body temperature is the 95th percentile?

g. What body temperature is the 5th percentile?

Answer 1

Z = (x - mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions corresponding to your Z scores.

d. Z = (mean1 - mean2)/Standard Error (SE) of the mean

SE = SD/√n

This information should lead you to find your own answers. We do not do your work for you.

Answer 2

Assume the 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 the University of Maryland researchers).

a. (0.1 point) If you have a body temperature of 99.00 °F, what is your percentile score?

b. (0.1 point) Convert 99.00 °F to a standard score (or a z-score).



c. 1. (0.1 point) Is a body temperature of 99.00 °F unusual?



2. (0.1 point) Why or why not?



d. (0.1 point) Fifty adults are randomly selected. What is the likelihood that the mean

of their body temperatures is 97.98 °F or lower?



e. (0.1 point) A person’s body temperature is found to be 101.00 °F. Is the result unusual?



(0.1 point) Why or Why Not and What should you conclude?









f. (0.1 point) What body temperature is the 95th percentile? :



g. (0.1 point) What body temperature is the 5th percentile?

Answer 3

a 101 body temperature is unusual