Posted by **sweet** on Monday, August 16, 2010 at 12:58pm.

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

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

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

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

e. A person’s body temperature is found to be 101.00 °F. Is the result unusual? Why or why not? What should you conclude?

f. What body temperature is the 95th percentile?

g. What body temperature is the 5th percentile?

h. Bellevue Hospital in New York City uses 100.6 °F as the lowest temperature considered to indicate a fever. What percentage of normal and healthy adults would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6 °F is appropriate?

- statistics -
**PsyDAG**, Tuesday, August 17, 2010 at 12:10pm
Use the processes from your previous post.

However for d, you are using a distribution of means rather than Individual scores.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.

That table can be used for your other problems too, finding Z scores to convert from percentile to score.

- statistics -
**JSL**, Sunday, February 6, 2011 at 1:52pm
Am I close? a- 99%, b- 1.29, c- yes, unusual because only 1% of the healthy adult population would be higher,d- 24.5 of the 50 are to be at or below 98.98, e- 4.52, an unusual result and the adult is probably not healthy because the 101 is outside the niormal range by a large percentage. f- 98.97

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