SOLUTION: Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 degrees F and a standard deviation of 0.62 degrees F If you have a body temperature

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Question 398216: Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 degrees F and a standard deviation of 0.62 degrees F
If you have a body temperature of 99 degrees F what is your percentile score
Convert 99.00 degrees F to a standard score
What body temperature is the 95th percentile
What body percent is the 5th percentile
Is a body temperature of 99.00 degrees F unusual? Why or Why not
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 degrees F and a standard deviation of 0.62 degrees F
If you have a body temperature of 99 degrees F what is your percentile score
Convert 99.00 degrees F to a standard score
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z(99) = (99-98.2)/0.62 = 1.29
P(z < 1.29) = normalcdf(-100,1.29) = 90%ile
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What body temperature is the 95th percentile
invNorm(0.95) = 1.645
temp = zs+u
temp = 1.645*0.62+98.2 = 99.22 degree
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What body percent is the 5th percentile
invNorm(0.05) = -1.645
temp = -1.645*0.62+98.2 = 97.18 degrees
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Is a body temperature of 99.00 degrees F unusual?
Why or Why not
Not really. It's only 1.29 standard deviations above the mean.
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Cheers,
Stan H.