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
(round to one decimal place as needed.)
°f.
b. using the 68 - 95 - 99.7 rule, about 95% of the body temperatures are expected to be less than
(round to one decimal place as needed.)
°f.
c. using the 68 - 95 - 99.7 rule, about 95% of the body temperatures are expected to be between
(use ascending order. round to one decimal place as needed.)
°f and
°f.
b) what fraction of people would be expected to have body temperatures above 98.6°f?
(round to two decimal places as needed.)

Explanation:

Step1: Recall the 68 - 95 - 99.7 Rule

The 68 - 95 - 99.7 Rule for a normal distribution states that about 68% of the data lies within 1 standard deviation of the mean, about 95% lies within 2 standard deviations of the mean, and about 99.7% lies within 3 standard deviations of the mean.
If the mean \(\mu = 98.2^{\circ}F\) and we assume a normal distribution, for part (a), since 95% of the data lies within 2 standard deviations of the mean. Let the standard deviation be \(\sigma\). If we assume a symmetric interval around the mean for 95% of the data (in a normal distribution), we know that \(98.2 - 2\sigma=97.4\) and \(98.2 + 2\sigma=99\) (solving \(98.2-2\sigma = 97.4\) gives \(\sigma=(98.2 - 97.4)/2=0.4\)). So most people's body temperatures (about 95% in the 68 - 95 - 99.7 Rule context for the interval part) are between \(97.4^{\circ}F\) and \(99^{\circ}F\).

Step2: For part (b)

The total area under the normal curve is 1. If the mean is \(\mu = 98.2\) and we want to find the proportion of people with body temperatures above \(98.6^{\circ}F\). First, find the z - score \(z=\frac{x-\mu}{\sigma}\), where \(x = 98.6\), \(\mu=98.2\) and \(\sigma = 0.4\). So \(z=\frac{98.6 - 98.2}{0.4}=\frac{0.4}{0.4}=1\).
Using the standard normal distribution table or the properties of the normal distribution (since the normal distribution is symmetric about the mean), the area to the right of \(z = 1\) is \(P(Z>1)\). We know that \(P(- 1<Z<1)=0.68\) (from the 68 - 95 - 99.7 Rule for 1 standard deviation), and \(P(Z < - 1)+P(-1<Z<1)+P(Z>1)=1\). Since the normal distribution is symmetric (\(P(Z < - 1)=P(Z>1)\)), we have \(2P(Z>1)=1 - 0.68\). Then \(P(Z>1)=\frac{1 - 0.68}{2}=0.16\)

Answer:

a) Most people's body temperatures are between \(97.4^{\circ}F\) and \(99^{\circ}F\) (because of the 68 - 95 - 99.7 Rule where for 95% of the data in a normal distribution, the interval is \(\mu\pm2\sigma\) with \(\mu = 98.2\) and \(\sigma=0.4\)).
b) The fraction of people with body temperatures above \(98.6^{\circ}F\) is \(0.16\) (or \(16\%\))