The body temp of normal healthy persons are normally distributed with a mean of 98.2 degrees F and a standard deviation od .62 degree F. If we define a fever to be a body temp above 100 degrees F, what percentage of normal and healthy person would be considered to have a fever?

μ=98.2

σ=0.62
Z(100)=(100-98.2)/0.62=2.903
Look up (normal) probabilities for Z=2.903 from tables (for people with temperatures below 100).
Subtract from 1 to get probabilities of healthy person considered having a fever.

ATTENDANCE14If the mean body temperature is not known, however it is known that 10% of all people have body temperatures greater than 99.5 degrees Fahrenheit, what would the mean body temperature be? Assume the standard deviation is 0.5 degrees and body temperature is normally distribute

To find the percentage of normal and healthy persons who would be considered to have a fever, we can use the concept of the standard normal distribution.

First, we need to standardize the temperature value of 100 degrees Fahrenheit using the formula:

Z = (X - μ) / σ

Where:
Z is the z-score (standardized value)
X is the temperature value (100 degrees Fahrenheit in this case)
μ is the mean of the distribution (98.2 degrees Fahrenheit)
σ is the standard deviation (0.62 degrees Fahrenheit)

Let's calculate the z-score:

Z = (100 - 98.2) / 0.62
Z = 1.8 / 0.62
Z ≈ 2.903

Next, we can find the proportion of individuals with temperatures above 100 degrees Fahrenheit by finding the area under the standard normal distribution curve to the right of the z-score of 2.903.

Using a standard normal table or a calculator, we find that the area to the right of 2.903 is approximately 0.0019.

To convert this into a percentage, we multiply by 100:

Percentage = 0.0019 * 100
Percentage ≈ 0.19%

Therefore, approximately 0.19% of normal and healthy people would be considered to have a fever (body temperature above 100 degrees Fahrenheit) based on the given mean and standard deviation.