Question 436490
 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).
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a. If you have a body temperature of 99.00 °F, what is your percentile score?
z(99) = (99-98.2)/0.62 = 1.2903
P(z < 1.2903) = normalcdf(-100,1.2903) = 0.9015
So, 99.00 F is the 90%ile score
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b. Convert 99.00 °F to a standard score (or a z-score).
done above.
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c. Is a body temperature of 99.00 °F unusual? Why or why not?
Not unlikely as is is only 1.29 standard deviations above the mean.
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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?
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Note: std of all groups of size "50" = 0.62/sqrt(50) = 0.0877
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z(97.98) = (97.98-98.2)/0.0877 = -2.5086
P(x-bar < 97.98) = P(z < -2.5086) = 0.0198
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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?
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Find the z-value using std = 0.62.
If the temp is at least 2 std above the mean it is unusual.
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f. What body temperature is the 95th percentile?
Find the z-value with a left tail of 0.95
That value is 1.645
Solve for "x" where x = zs+u
x = 1.645*0.62+98.2 = 99.22 degrees
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g. What body temperature is the 5th percentile?
Same procedure but use z = -1.645
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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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If you understand a thru g you should be able to answer "h" yourself.
Cheers,
Stan H.