Question 769041: 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)
If you have a body temperature of 99.00 °F, what is your percentile score?
a) 90
b)92
c) 94
d)96
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)
If you have a body temperature of 99.00 °F, what is your percentile score?
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z(99) = (99-98.2)/0.62 = 0.8/0.62 = 1.2903
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P(z < 1.2903) = normalcdf(-100,1.2903) = 0.9015
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Ans: 90%ile
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Cheers,
Stan H.
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a) 90
b)92
c) 94
d)96
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