Question 309818
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?
Find the percent of scores that are below 99.
z(99) = (99-98.2)/0.62 = 1.2903
P(x < 99) = P(z < 1.2903) = 0.9015 = 90.15%ile
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b. Convert 99.00 °F to a standard score (or a z-score). = 1.2903
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c. Is a body temperature of 99.00 °F unusual? Why or why not?
No; z >= 2 is usually considered to be the cut-off for "unusual".
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d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower?
z(97.98) = (97.98-98.2)/[0.62/sqrt(50)] = -2.5091
P(x-bar <= 97.98) = P(z <= -2.5001) = 0.0061
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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?
The z-value is 4.5161
What do you think?
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f. What body temperature is the 95th percentile?
Find the z-value that has a left-tail of 0.95.
invNorm(0.95) = 1.645
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Solve for "x":
x = zs+u = 1.645*0.62+98.2 = 99.22 degrees
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g. What body temperature is the 5th percentile?
x = -1.645*0.62+98.2 = 97.18 degrees

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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?
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Can you handle "h" now?
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Cheers,
Stan H.