Chapter 6, Problem 46E

a)

To determine

To find the interval that expected most people’s body temperature to be.

Answer to Problem 46E

The most people’s body temperatures are between 96.8 and 99.6.

Explanation of Solution

Given:

  $\mu$ = 98.2

  $\sigma$ = 0.7

Formula:

  $x=\mu +z\sigma$

We know, the usual values for Normal distribution lie within 2 standard deviations from the mean.

That is, z = $\pm 2$

Therefore,

  $\begin{array}{l}x=98.2-2\times 0.7=96.8\\ x=98.2+2\times 0.7=99.6\end{array}$

Hence, the most people’s body temperatures are between 96.8 and 99.6.

b)

To determine

To find the fraction in which people would be expected to have body temperatures above 98.6⁰ F

Answer to Problem 46E

The 28.43% of people would be expected to have body temperatures above 98.6⁰ F

Explanation of Solution

Given:

  $\mu$ = 98.2

  $\sigma$ = 0.7

Formula:

  $z=\frac{x-\mu }{\sigma }$

Using z-score,

  $z=\frac{98.6-98.2}{0.7}=0.57$

  $\begin{array}{l}P(Z>0.57)=1-P(Z<0.57)\\ P(Z>0.57)=1-0.7157=0.2843=28.43\%\end{array}$

Hence, the 28.43% of people would be expected to have body temperatures above 98.6⁰ F

c)

To determine

To find the body temperature that are coolest 20% of all people.

Answer to Problem 46E

The coolest 20% of all people have a body temperature of at most 97.612⁰ F.

Explanation of Solution

Given:

  $\mu$ = 98.2

  $\sigma$ = 0.7

Formula:

  $x=\mu +z\sigma$

We need to first find the z-score for which the probability of 20%.

  P(Z< z) = 0.20

  z = -0.84

Therefore,

  $x=98.2+(-0.84)\times 0.7=97.612$

Hence, the coolest 20% of all people have a body temperature of at most 97.612⁰ F.