please do NOT use excell...3 answers ... hw 6.3#8 a. Assume that human body temp
ID: 3257391 • Letter: P
Question
please do NOT use excell...3 answers ... hw 6.3#8
a. Assume that human body temperatures are normally distributed with a mean of 98.22°Fand a standard deviation of 0.64 degrees 0.64°F. A hospital uses 100.6°F as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6 degrees is appropriate?
b. Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 5.0% of healthy people to exceed it? (Such a result is a false positive, meaning that the test result is positive, but the subject is not really sick.)
1)The percentage of normal and healthy persons considered to have a fever is _____ % (Round to 2 decimal places as needed.)
2)Does this percentage suggest that a cutoff of 100.6°F is appropriate?
A.No, because there is a small probability that a normal and healthy person would be considered to have a fever.
B.No, because there is a large probability that a normal and healthy person would be considered to have a fever.
C.Yes, because there is a large probability that a normal and healthy person would be considered to have a fever.
D.Yes, because there is a small probability that a normal and healthy person would be considered to have a fever.
3. The minimum temperature for requiring further medical tests should be________degrees F° if we want only 5.0% of healthy people to exceed it. (Round to 2 decimal places as needed)
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 100.6
u = mean = 98.22
s = standard deviation = 0.64
Thus,
z = (x - u) / s = 3.71875
Thus, using a table/technology, the right tailed area of this is
P(z > 3.765625 ) = 0.0001 [ANSWER]
OPTION C:
C.Yes, because there is a small probability that a normal and healthy person would be considered to have a fever. [ANSWER]
*******************
b)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.95
Then, using table or technology,
z = 1.644853627
As x = u + z * s,
where
u = mean = 98.22
z = the critical z score = 1.644853627
s = standard deviation = 0.64
Then
x = critical value = 99.27270632
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