Question

the body temperatures of adults are normally distributed with a mean of 98.6°f and a standard deviation of 0.33°f. what temperature represents the 95th percentile? click to view page 1 of the table. click to view page 2 of the table. a. 99.02°f b. 99.25°f c. 98.06°f d. 99.14°f

Explanation:

Step1: Find the z - score for the 95th percentile

The 95th percentile corresponds to an area of 0.95 to the left of the z - score in the standard normal distribution. Looking up in the standard normal (z -) table, the z - score $z$ such that $P(Z

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the original normal distribution, $\mu$ is the mean, and $\sigma$ is the standard deviation. We want to solve for $x$. Rearranging the formula gives $x=\mu + z\sigma$. We know that $\mu = 98.6^{\circ}F$ and $\sigma=0.33^{\circ}F$ and $z = 1.645$.
Substitute the values: $x=98.6+1.645\times0.33$.

Step3: Calculate the value of $x$

$x=98.6 + 1.645\times0.33=98.6+ 0.54285=99.14285\approx99.14^{\circ}F$.

Answer:

D. $99.14^{\circ}F$