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?
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.