Points: 0 of 1 Save Assume that human body temperatures are normally distributed with a mean of \( 98.21^{\circ} \mathrm{F} \) and a standard deviation of \( 0.64^{\circ} \mathrm{F} \). a. A hospital uses \( 100.6^{\circ} \mathrm{F} \) as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a fever? Does this percentage suggest that a cutoff of \( 100.6^{\circ} \mathrm{F} \) 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? (Such a result is a false positive, 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 \( \square \) \( \square \). (Round to two decimal places as needed.)

We are given that human body temperatures are normally distributed with a mean (μ) of 98.21°F and a standard deviation (σ) of 0.64°F. ───────────────────────────── Part (a) The hospital’s cutoff for a fever is T = 100.6°F. To find the percentage of healthy people who would have a temperature at or above this cutoff, we first calculate the corresponding z‐score:   z = (T – μ) / σ     = (100.6 – 98.21) / 0.64     = 2.39 / 0.64     ≈ 3.73 Next, we determine the probability that a healthy person has a z‐score greater than 3.73. For a z value of 3.73, the area to its left (the cumulative probability) is very close to 1 (approximately 0.9999), meaning that the area to the right (i.e. the probability of a temperature 100.6°F or higher) is:   P(Z ≥ 3.73) = 1 – 0.9999 ≈ 0.0001 When expressed as a percentage, this is about 0.01%. Thus, approximately 0.01% of normal and healthy persons would be considered to have a fever using a cutoff of 100.6°F. This extremely low percentage suggests that 100.6°F is set so high that virtually no healthy person would be flagged, which may not be a very practical cutoff if the aim is to detect a reasonable number of fever cases. ───────────────────────────── Part (b) Physicians want a cutoff such that only 5.0% of healthy individuals (false positives) exceed it. This means we need the 95th percentile of the healthy distribution. The z‐score corresponding to the 95th percentile is approximately 1.645. Now, we convert this z‐score back to a temperature using:   T = μ + z·σ     = 98.21 + 1.645 × 0.64     = 98.21 + 1.0528     ≈ 99.26°F So, if we set the cutoff at about 99.26°F, only about 5% of healthy people would exceed that temperature. ───────────────────────────── Summary of Answers a. About 0.01% of normal and healthy persons would be considered to have a fever using a cutoff of 100.6°F. This very low percentage indicates that the cutoff may be too high (i.e., it almost never flags healthy cases). b. To have only 5.0% of healthy people exceed the cutoff (i.e., a 5% false positive rate), the temperature should be set at approximately 99.26°F. Remember to round your answers as needed (e.g., to two decimal places).