SOLUTION: 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 Maryla

Question 309818: 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?

This is one question of many on a Statistics quiz due monday night for me. I would greatly appreciate any and all help provided. Thank you in advance!
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).
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.

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