Question

assume that human body temperatures are normally distributed with a mean of 98.21°f and a standard deviation of 0.62°f.
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? does this percentage suggest that a cutoff of 100.6°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.)
click to view page 1 of the table. click to view page 2 of the table.
a. the percentage of normal and healthy persons considered to have a fever is
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the z - score for part a

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 100.6$, $\mu=98.21$, and $\sigma = 0.62$.
$z=\frac{100.6 - 98.21}{0.62}=\frac{2.39}{0.62}\approx3.85$

Step2: Find the proportion of values above the z - score

We want to find $P(Z>3.85)$. Since the total area under the standard - normal curve is 1, and $P(Z < z)+P(Z>z)=1$. Looking up $P(Z < 3.85)$ in the standard - normal table, we find that $P(Z < 3.85)\approx0.9999$. So $P(Z>3.85)=1 - 0.9999 = 0.0001$.
Converting to a percentage, we get $0.01\%$.

Step3: Calculate the z - score for part b

We want to find the z - score $z$ such that $P(Z>z)=0.05$. Then $P(Z < z)=1 - 0.05=0.95$. Looking up 0.95 in the standard - normal table, the z - score corresponding to a cumulative probability of 0.95 is approximately $z = 1.645$.

Step4: Find the temperature for part b

Using the z - score formula $z=\frac{x-\mu}{\sigma}$ and solving for $x$, we get $x=\mu+z\sigma$. Substituting $\mu = 98.21$, $z = 1.645$, and $\sigma=0.62$, we have $x=98.21+1.645\times0.62=98.21 + 1.0199=99.23^{\circ}F$.

Answer:

a. $0.01$
b. $99.23^{\circ}F$