Download Here is the whole problem. Assume body temperatures are normally

Here is the whole problem.
Assume body temperatures are normally distributed with mean of 98.2 and a standard
dev. of 0.62
Part A) Assuming a hospital defines a fever as 100 or over, determine the percentage of a
normal healthy person considered to have a fever using Z table
Answer: 100-98.2/.62=2.90 =.19% rounded.
Solution. Denote the body temperature by X. Then by hypothesis, X follows normal
distribution with mean of 98.2 and a standard dev. of 0.62, i.e., X ~ N (98.2,0.62 2 ) .
Now we need to find the probability
.2
10098.2
X 98.2
Pr( X  100)  Pr( X0.98
62 
0.62 )  Pr( 0.62  2.9032)
.2
Note 1: If X ~ N (98.2,0.62 2 ) , then X0.98
62 ~ N (0,1)
Using a website at
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html or
looking up Z table, we get
.2
10098.2
X 98.2
Pr( X  100)  Pr( X0.98
62 
0.62 )  Pr( 0.62  2.9032)  0.00185  0.19%
Note 2: Using the above website, you can see a window which is a probability calculator.
You only need to input 2.9032 in the lower bound of X and nothing in the upper bound,
then click ”calculate probability”, you will get 0.001847
Part B) I have the procedure right but I can't get the right answer. z=x-u/o Determine the
minimum fever if doctors at the hospital want only 3.0% of healthy persons to be
diagnosed as having a fever. Use closest z score. Answer is 99.37.
Solution. By part A), we know that if a hospital defines a fever as 100 or over, then the
percentage of a normal healthy person considered to have a fever is 0.19%
Now we need to re-define a fever, for instance, as x 0 or over. We need to find
this x 0 . How to find this x 0 ?
We know that if we use this new definition, then the percentage of a normal
healthy person considered to have a fever is 3%, namely, we have
x0 98.2
.2
Pr( X  x0 )  Pr( X0.98
62  0.62 )  3%
By looking up Z table or using these two websites
http://duke.usask.ca/~rbaker/Tables.html and
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html, we can
get
Pr( Z  1.88)  3%
(*)
where Z~N(0,1).
.2
We know that if X ~ N (98.2,0.62 2 ) , then X0.98
62 ~ N (0,1) . So, in order to find x 0 such
that
x0 98.2
.2
(**)
Pr( X  x0 )  Pr( X0.98
62  0.62 )  3%
By the above fact (*) and note that (**), we get
x0 98.2
0.62  1.88
So,
x0  98.2  1.88 * 0.62  99.3656  99.37
Another part B, different problem. How does 100.1-98.2/.62=.11 in part A and part B =
99.55 wanting 1.5% of healthy to be diagnosed?
Solution.
(I) Denote the body temperature by X. Then by hypothesis, X follows normal distribution
with mean of 98.2 and a standard dev. of 0.62, i.e., X ~ N (98.2,0.62 2 ) .
Now we need to find the probability
.2
100.198.2
X 98.2
Pr( X  100.1)  Pr( X0.98
62 
0.62 )  Pr( 0.62  3.0645)
.2
Note 1: If X ~ N (98.2,0.62 2 ) , then X0.98
62 ~ N (0,1)
Using a website at
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html or
looking up Z table, we get
.2
100.198.2
X 98.2
Pr( X  100.1)  Pr( X0.98
62 
0.62 )  Pr( 0.62  3.0645)  0.00109  0.11%
Note 2: Using the above website, you can see a window which is a probability calculator.
You only need to input 3.0645 in the lower bound of X and nothing in the upper bound,
then click ”calculate probability”, you will get 0.00109
x0 98.2
.2
(II) We want to find x 0 such that Pr( X  x0 )  Pr( X0.98
.
62  0.62 )  1.5%
By looking up Z table or using these two websites
http://duke.usask.ca/~rbaker/Tables.html and
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html, we can
get
Pr( Z  2.17)  1.5%
where Z~N(0,1).
So, we get
x0 98.2
0.62  1.88
So,
x0  98.2  2.17 * 0.62  99.545  99.55
A different B: 99.2-98.2/.62 =5.37 for part A, and B wanting 2.5% = 99.42.?? I have
been taking 1-the %s (3.0%, 2.5% et cetera as above) but I'm still not getting the right
answers.
Solution.
(I) Denote the body temperature by X. Then by hypothesis, X follows normal distribution
with mean of 98.2 and a standard dev. of 0.62, i.e., X ~ N (98.2,0.62 2 ) .
Now we need to find the probability
.2
99.298.2
X 98.2
Pr( X  99.2)  Pr( X0.98
62 
0.62 )  Pr( 0.62  1.61)
.2
Note 1: If X ~ N (98.2,0.62 2 ) , then X0.98
62 ~ N (0,1)
Using a website at
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html or
looking up Z table, we get
.2
99.298.2
X 98.2
Pr( X  99.2)  Pr( X0.98
62 
0.62 )  Pr( 0.62  1.61)  0.053699  5.37%
Note 2: Using the above website, you can see a window which is a probability calculator.
You only need to input 1.61 in the lower bound of X and nothing in the upper bound,
then click ”calculate probability”, you will get 0.053699
x0 98.2
.2
(II) We want to find x 0 such that Pr( X  x0 )  Pr( X0.98
.
62  0.62 )  2.5%
By looking up Z table or using these two websites
http://duke.usask.ca/~rbaker/Tables.html and
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html, we can
get
Pr( Z  1.96)  2.5%
where Z~N(0,1).
So, we get
x0 98.2
0.62  1.96
So,
x0  98.2  1.96 * 0.62  99.4152  99.42
Note: This website http://duke.usask.ca/~rbaker/Tables.html can tell you the critical
value of many distributions. For instance, if Z~N(0,1) standard normal distribution, then
P( Z  Z  )   , we call Z   -critical value. If we choose   2.5%  0.025 , then you
can use this website by choosing the critical values for z-test. Then choose
  2.5%  0.025 , we get Z   1.96
Hope this helps. If you have any questions, please let me know.