Assume that human body temperatures are normally distributed with a mean of 98.1
ID: 3132297 • Letter: A
Question
Assume that human body temperatures are normally distributed with a mean of 98.19 degree F and a standard deviation of 0.61 degree F. A hospital uses 100.6 degree 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 degree F is appropriate? 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.) The percentage of normal and healthy persons considered to have a fever is %. (Round to two decimal places as needed.) Does this percentage suggest that a cutoff of 100.6 degree F is appropriate? No, because there is a small probability that a normal and healthy person would be considered to have a fever. Yes, because there is a small probability that a normal and healthy person would be considered to have a fever. No, because there is a large probability that a normal and healthy person would be considered to have a fever. Yes, because there is a large probability that a normal and healthy person would be considered to have a fever. The minimum temperature for requiring further medical tests should be degree F if we want only 5.0% of healthy people to exceed it. (Round to two 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 = 98.19
s = standard deviation = 0.61
Thus,
z = (x - u) / s = 3.95
Thus, using a table/technology, the right tailed area of this is
P(z > 3.95 ) = 0 [ANSWER]
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OPTION B: YES, because there is a small probability that a normal and healthy person would be considered to have a fever. [ANSWER]
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b)
First, we get the z score from the given left tailed area. As
Left tailed area = 1 - 0.05 = 0.95
Then, using table or technology,
z = 1.64
As x = u + z * s,
where
u = mean = 98.19
z = the critical z score = 1.64
s = standard deviation = 0.61
Then
x = critical value = 99.1904 = 99.19 [ANSWER]