Question

most people think that the
ormal\ adult body temperature is 98.6°f. in a more recent study, researchers reported that a more accurate figure may be 98.2°f. furthermore, the standard deviation appeared to be around 0.4°f. assume that a normal model is appropriate. complete parts a through c below.
a) in what interval would you expect most peoples body temperatures to be? explain. select the correct choice below and fill in the answer box(es) to complete your choice.
a. using the 68 - 95 - 99.7 rule, about 95% of the body temperatures are expected to be at least
°f.
(round to one decimal place as needed.)
b. using the 68 - 95 - 99.7 rule, about 95% of the body temperatures are expected to be less than
°f.
(round to one decimal place as needed.)
c. using the 68 - 95 - 99.7 rule, about 95% of the body temperatures are expected to be between
97.4°f and 99°f.
(use ascending order. round to one decimal place as needed.)
b) what fraction of people would be expected to have body temperatures above 98.6°f?
15.87%
(round to two decimal places as needed.)
c) below what body temperature are the coolest 20% of all people?
°f
(round to one decimal place as needed.)

Explanation:

Step1: Recall the 68 - 95 - 99.7 Rule

The 68 - 95 - 99.7 Rule states that for a normal distribution:

• Approximately 68% of the data lies within \( \mu\pm\sigma\)
• Approximately 95% of the data lies within \( \mu\pm2\sigma\)
• Approximately 99.7% of the data lies within \( \mu\pm3\sigma\)

We are given \( \mu = 98.2^{\circ}F\) and \( \sigma=0.4^{\circ}F\)

Step2: Calculate the interval for part a

For part a, we want the interval where most (about 68% - 99.7% is not most, 68% is a single - sigma interval, 95% is a two - sigma interval. But when we say'most' in the context of the 68 - 95 - 99.7 Rule, we consider the 95% interval).
The lower bound \(L=\mu - 2\sigma\) and the upper bound \(U=\mu + 2\sigma\)
Substitute \( \mu = 98.2\) and \( \sigma = 0.4\)
\(L=98.2-2\times0.4=98.2 - 0.8=97.4\)
\(U=98.2+2\times0.4=98.2 + 0.8=99.0\)

Step3: Calculate the value for part b

We know that the total area under the normal curve is 1. If we want the proportion of people above \(x = 98.6\)
First, calculate the z - score \(z=\frac{x-\mu}{\sigma}\)
Substitute \(x = 98.6\), \( \mu = 98.2\) and \( \sigma=0.4\)
\(z=\frac{98.6 - 98.2}{0.4}=\frac{0.4}{0.4}=1\)
The area to the left of \(z = 1\) is about 0.8413 (from the standard normal table). So the area to the right (proportion above \(x = 98.6\)) is \(1 - 0.8413=0.1587 = 15.87\%\)

Step4: Calculate the value for part c

We want to find the value \(x\) such that \(P(X\lt x)=0.2\)
Looking up the z - score in the standard normal table for an area of 0.2. The z - score \(z\approx - 0.84\)
Using the formula \(z=\frac{x-\mu}{\sigma}\)
Substitute \(z=-0.84\), \( \mu = 98.2\) and \( \sigma = 0.4\)
\(-0.84=\frac{x - 98.2}{0.4}\)
Multiply both sides by 0.4: \(x-98.2=-0.84\times0.4=-0.336\)
Add 98.2 to both sides: \(x=98.2-0.336 = 97.9\)

Answer:

a) \(97.4^{\circ}F\) and \(99.0^{\circ}F\)
b) \(15.87\%\)
c) \(97.9^{\circ}F\)