Question

Assume that human body temperatures are normally distributed with a mean of

98.19 degrees Upper F98.19°F

and a standard deviation of

0.61 degrees Upper F0.61°F.

a. A hospital uses

100.6 degrees Upper F100.6°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 Upper F100.6°F

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.)

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a. The percentage of normal and healthy persons considered to have a fever is

nothing​%.

Answer 1

Solution:

Given: Human body temperatures are normally distributed with a mean of 98.19°F and a standard deviation of 0.61°F.

Part a) 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?

that is find:

Find z score for x = 100.6

thus we get:

Look in z table for z = 3.9 and 0.05 and find corresponding area.

P( Z < 3.95) = 0.99996

thus

The percentage of normal and healthy persons considered to have a fever is 0.00%

Does this percentage suggest that a cutoff of 100.6°F is​ appropriate?

No, because there is a small probability that a normal and healthy person would be considered to have a fever.

Part 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?

Find x value such that:

P( X > x ) =5.0%

P( X > x ) = 0.05

thus find z value such that :

P( Z > z ) = 0.05

that is:

P( Z < z ) = 1 - P(Z > z)

P( Z < z ) = 1 - 0.05

P( Z < z ) = 0.95

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus z = 1.645

Now use following formula to find x value: