document.write( "Question 314444: 2. 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).
\n" ); document.write( "a. If you have a body temperature of 99.00 °F, what is your percentile score?
\n" ); document.write( "b. Convert 99.00 °F to a standard score (or a z-score).
\n" ); document.write( "c. Is a body temperature of 99.00 °F unusual? Why or why not?
\n" ); document.write( "d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower?
\n" ); document.write( "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?
\n" ); document.write( "f. What body temperature is the 95th percentile?
\n" ); document.write( "g. What body temperature is the 5th percentile?
\n" ); document.write( "h. Bellevue Hospital in New York City uses 100.6 °F as the lowest temperature considered to indicate a fever. What percentage of normal and healthy adults would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6 °F is appropriate?\r
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Algebra.Com's Answer #224974 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
a.\"mu=98.20\"
\n" ); document.write( "\"sigma=0.62\"
\n" ); document.write( "\"x=99\"
\n" ); document.write( "\"z=%28x-mu%29%2Fsigma=%2899-98.2%29%2F0.62=1.33\"
\n" ); document.write( "\"P%281.33%29=0.9087\"
\n" ); document.write( "91st percentile.
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\n" ); document.write( "b. \"z=1.33\"
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\n" ); document.write( "c. Depends on what you consider normal, unusual, etc., you should make that determination and be able to defend your position.
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\n" ); document.write( "d. Set up a t test.
\n" ); document.write( "The null hypothesis is that the means are the same.
\n" ); document.write( "\"N=50\"
\n" ); document.write( "\"X=97.98\"
\n" ); document.write( "\"S=sigma%2Fsqrt%28N-1%29=0.62%2Fsqrt%2850-1%29=0.62%2F7=0.00886\"
\n" ); document.write( "\"t=%28X-mu%29%2FS=%2897.98-98.2%29%2F0.00886=-2.484\"
\n" ); document.write( "Now the challenge is in working out the critical t value.
\n" ); document.write( "t(0.05,49)=1.647
\n" ); document.write( "t(0.01,49)=2.405
\n" ); document.write( "Depending on the critical t value you choose, you could reject the null hypothesis, in which case, you would say the means are not the same, there is a statistically significant difference in the findings.
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\n" ); document.write( "e. \"x=101\"
\n" ); document.write( "\"z=%28x-mu%29%2Fsigma=%28101-98.2%29%2F0.62=4.52\"
\n" ); document.write( "\"P%284.52%29=0.999998\"
\n" ); document.write( "Yes, that would be odd because very few people would normally have that temperature.
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\n" ); document.write( "f. \"P%28z%29=0.95\"
\n" ); document.write( "\"z=1.645=%28x-98.2%29%2F0.62\"
\n" ); document.write( "\"x=99.2\"
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\n" ); document.write( "g. What body temperature is the 5th percentile?
\n" ); document.write( "\"P%28z%29=0.05\"
\n" ); document.write( "\"z=-1.645=%28x-98.2%29%2F0.62\"
\n" ); document.write( "\"x=97.2\"
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\n" ); document.write( "h. \"z=%28x-mu%29%2Fsigma=%28100.6-98.2%29%2F0.62=1.33\"
\n" ); document.write( "\"P%28z%29=0.999968\"
\n" ); document.write( "Normally and healthy would be ones who have higher than 100.6 but are not sick.
\n" ); document.write( "\"P=1-0.999968=0.000032\"
\n" ); document.write( "That's pretty low, so you would think it's pretty unlikely that someone coming in off the street would have a 100.6 temperature and be normal. The cutoff is probably appropriate. Again check the caveats in section c.
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