Question 314444
a.{{{mu=98.20}}}
{{{sigma=0.62}}}
{{{x=99}}}
{{{z=(x-mu)/sigma=(99-98.2)/0.62=1.33}}}
{{{P(1.33)=0.9087}}}
91st percentile.
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b. {{{z=1.33}}}
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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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d. Set up a t test.
The null hypothesis is that the means are the same. 
{{{N=50}}}
{{{X=97.98}}}
{{{S=sigma/sqrt(N-1)=0.62/sqrt(50-1)=0.62/7=0.00886}}}
{{{t=(X-mu)/S=(97.98-98.2)/0.00886=-2.484}}}
Now the challenge is in working out the critical t value. 
t(0.05,49)=1.647
t(0.01,49)=2.405
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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e. {{{x=101}}}
{{{z=(x-mu)/sigma=(101-98.2)/0.62=4.52}}}
{{{P(4.52)=0.999998}}}
Yes, that would be odd because very few people would normally have that temperature. 
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f. {{{P(z)=0.95}}}
{{{z=1.645=(x-98.2)/0.62}}}
{{{x=99.2}}}
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g. What body temperature is the 5th percentile?
{{{P(z)=0.05}}}
{{{z=-1.645=(x-98.2)/0.62}}}
{{{x=97.2}}}
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h. {{{z=(x-mu)/sigma=(100.6-98.2)/0.62=1.33}}}
{{{P(z)=0.999968}}}
Normally and healthy would be ones who have higher than 100.6 but are not sick.
{{{P=1-0.999968=0.000032}}} 
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.