Human body temperatures are normally distributed with a mean.

Question

Assume that human body temperatures are normally distributed with a mean 98.20F of and a standard deviation of 0.61F.

a. A hospital uses 100.6F 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 100.6F 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.)

Answer 1

Hi MJ,

This problem uses the classic equation in introductory statistics, z= (x-mu)/sigma. Let's break this down first:

z=z-score, explained below

x=value you are given to test

mu=mean

sigma=standard deviation

Now, let's break this down specifically for part (a).

a.

x=100.6

mu=98.20

sigma=0.61

z=(100.6-98.20)/0.61

z=3.93

Now, that value is your z-score, which you use to get the proportion that you want. So, go to the z-table, which you can find online or in any statistics textbook, look up 3.9 on the left column, 0.03 in top row. This value is actually off the chart. Now, remember that the z-table gives probability that z<3.93, which is equivalent to probability that x, the temperature, is less than 100.6. So:

P(Z<3.93)=P(X<100.6)>0.99

Now, we want greater than 100.6, so we apply the complement rule or, as I call it, the "one minus trick:"

P(X>a)= 1-[P(X<a)]

P(X>100.6)=1-P(X<100.6)

P(X>100.6)< 0.01

This is a very low percentage, and statistical software will yield a more exact value, but suffice it to say that it is likely appropriate for the percentage to be that low, without me having any medical experience.

(b) We are still looking at the same formula, but this time, instead of seeking a probability, we seek a temperature that meets a probability cutoff. So, we are essentially being asked for the top 5%, which makes the percentile 95, so we go to the z-table first this time, look at its interior for 0.95, the decimal equivalent. We see 0.9495 and 0.9505 at 1.64 and 1.65 respectively, so let's assume:

z_0.95=1.645

This will be a key critical value when you look at confidence intervals. Anyway, we now have z, and we still have our population mean and standard deviation from earlier, so:

z=(x-mu)/sigma

z=1.645

mu=98.20

sigma=0.61

x=x, the temp. we are looking for

Thus:

1.645=(x-98.20)/0.61

Multiply both sides by 0.61:

1.003=x-98.2

x=99.20

I hope this helps.