Posted by **anonymous** on Friday, April 2, 2010 at 10:44pm.

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).

Is a body temperature of 99.00 °F unusual?

Why or why not?

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

A person’s body temperature is found to be 101.00 °F.

Is the result unusual? Why or Why Not and What should you conclude?

- statistics -
**PsyDAG**, Saturday, April 3, 2010 at 11:26am
What is your criterion for being "unusual"? Statistically the most often used criterion is P ≤ .05, meaning that the results would occur solely by chance only 5% of the time. P ≤ .01 is also used. P can be determined by first finding the Z score.

Z = (x - μ)/SD, where μ = mean

For the first question, Z = (99 - 98.2)/.62

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

Use the same method for the last temperature (101).

The Z score for the difference between means (second is a little different.

Z = (μ1 - μ2)/SE (Standard Error of the mean)

SE = SD/√(n-1), where √ = square root

Use the same table for that Z score.

You will have to reach your own conclusions. I hope this helps.

- statistics -
**DQR**, Saturday, April 3, 2010 at 11:49am
First part of the question: no, 99°F isn't all that unusual. It is (99.0 - 98.2) / 0.62 = 1.29 standard deviations above the mean - and if you look that up in a set of Normal distribution tables, you'll find that the area to the left of 1.29 is 0.9015. That means that 90.15% of the population have a body temperature of 99°F or less, so fractionally under 10% have a body temperature higher than that.

The standard error of the mean (SEM) of 50 patients is S/sqrt(N), where S is the standard deviation of the population, and N is the sample size. In our case that's 0.62/sqrt(50) = 0.088, so a mean of 97.98 is (98.20 - 97.98) / 0.088 = 2.5 SEMs beneath the population mean. Look up -2.5 (minus because it's to the left of the mean) in a set of Normal tables to get the area to the left of that value, and you'll get the answer to part 2. (Bear in mind that the area to the LEFT of -2.5 is the same as the area to the RIGHT of +2.5)

Finally, you should be able to tackle the third part of the question the same way as the first part.

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