Question 316002: 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?
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?
z(99) = (99-98.2)/0.62 = 1.2903
P(z < 1.2903) = normalcdf(-100,1.2903) = 0.9015
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99 degrees is at the 91%ile
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b. Convert 99.00 °F to a standard score (or a z-score).
See the problem above.
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c. Is a body temperature of 99.00 °F unusual? Why or why not?
Not too unusual. It is 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?
z(97.98-98.2)/[0.62/sqrt(50)] = -2.5091
P(z< -2.5091) = 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?
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Find the z-value of 101 degrees.
Based on the z-value decide if the result is "unusual".
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Cheers,
Stan H.
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