SOLUTION: Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 degrees fahrenheit and a standard deviation of 0.62 degrees fahrenheit
a) what is your
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Question 280833: Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 degrees fahrenheit and a standard deviation of 0.62 degrees fahrenheit
a) what is your percentile score if your body temp is 99.0 degrees fahrenheit?
b) convert 99.00 degrees fahrenheit to a standart score (z-score)
c) Is a body temperature of 99.0 degrees fahrenheit unusual? Why or why not?
d)50 adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 degrees fahrenheit or lower?
e) A person's body temperature is found to be 101.00 degrees fahrenheit. 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?
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 fahrenheit and a standard deviation of 0.62 degrees fahrenheit
a) what is your percentile score if your body temp is 99.0 degrees fahrenheit?
z(99) = (99-98.2)/0.62 = 1.2903
P(z < 1.2903) = 90.15
99 degrees is at the 90th %ile
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b) convert 99.00 degrees fahrenheit to a standart score (z-score)
Done above
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c) Is a body temperature of 99.0 degrees fahrenheit unusual? Why or why not?
It's 1.29 standard deviations above the mean: not too unusual.
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d)50 adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 degrees fahrenheit or lower?
z(97.98) = (97.98-98.2)/[0.62/sqrt(50)] = -2.5091
P(xbar<97.98) = P(z<-2.509) = 0.0061
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e) A person's body temperature is found to be 101.00 degrees fahrenheit.
Is the result unusual? Why or why not? What should you conclude?
Find the z-value of 101 and draw your conclusion.
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f)What body temperature is the 95th percentile?
Find the z-value that has a 95% left tail: invNorm(0.95) = 1.645
Find the corresponding x-value:
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?
Follow the same pattern used for "f".
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Cheers,
Stan H.
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