Question 436490: 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. 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?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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.
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