Assume that human body temperatures are normally distributed with a mean of \( 98.23^{\circ} \mathrm{F} \) and a standard deviation of \( 0.62^{\circ} \mathrm{F} \). a. A hospital uses \( 1006^{\circ} \mathrm{F} \) as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a fever? Does th \( 100.6^{\circ} \mathrm{F} \) 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? meaning that the test result is positive, but the subject is not really sick.) Click to view page 1 of the table. Click to view page 2 of the table. a. The percentage of normal and healthy persons considered to have a fever is \( 0.01 \% \). (Round to two decimal places as needed.) Does this percentage suggest that a cutoff of \( 1006^{\circ} \mathrm{F} \) is appropriate? A. Yes, because there is a small probability that a normal and healthy person would be considered to have a fever B. No, because there is a small probability that a normal and healthy person would be considered to have a fever. O C. No, because there is a large probability that a normal and healthy person would be considered to have a fever. OD. Yes, because there is a large probability that a normal and healthy person would be considered to have a fever.
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To find the percentage of normal and healthy persons considered to have a fever at \( 100.6^{\circ} \mathrm{F} \), we first calculate the z-score using the formula \( z = \frac{(X - \mu)}{\sigma} \). Plugging in the values, we get \( z = \frac{(100.6 - 98.23)}{0.62} \approx 3.82 \). Consulting the z-table, this gives an area of about \( 0.00003 \), which translates to roughly \( 0.01\% \). So yes, it suggests that \( 100.6^{\circ} \mathrm{F} \) is an appropriate cutoff (Option A). Next, for the minimum temperature where only \( 5.0\% \) of healthy people exceed it, we look for the z-score that corresponds to the \( 95\% \) cumulative area, which is approximately \( 1.645 \). We can rearrange the z-score formula to find \( X \): \( X = \mu + (z \times \sigma) = 98.23 + (1.645 \times 0.62) \approx 99.34^{\circ} \mathrm{F} \). Thus, if we want only \( 5\% \) of healthy people to exceed the threshold, it should be set at around \( 99.34^{\circ} \mathrm{F} \).
