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.

To find the percentage of normal and healthy persons considered to have a fever, we need to calculate the z-score for the given temperature of \( 100.6^{\circ} \mathrm{F} \) and then use the z-score to find the percentage of the normal distribution that falls above this temperature. Given: - Mean (\( \mu \)) = \( 98.23^{\circ} \mathrm{F} \) - Standard Deviation (\( \sigma \)) = \( 0.62^{\circ} \mathrm{F} \) - Temperature considered to be a fever (\( T \)) = \( 100.6^{\circ} \mathrm{F} \) a. Calculate the z-score for the temperature of \( 100.6^{\circ} \mathrm{F} \): \[ z = \frac{T - \mu}{\sigma} \] b. Use the z-score to find the percentage of the normal distribution that falls above the temperature of \( 100.6^{\circ} \mathrm{F} \). Let's calculate the z-score and then find the percentage of normal and healthy persons considered to have a fever. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(100.6-98.23\right)}{0.62}\) - step1: Remove the parentheses: \(\frac{100.6-98.23}{0.62}\) - step2: Subtract the numbers: \(\frac{2.37}{0.62}\) - step3: Convert the expressions: \(\frac{\frac{237}{100}}{\frac{31}{50}}\) - step4: Multiply by the reciprocal: \(\frac{237}{100}\times \frac{50}{31}\) - step5: Reduce the numbers: \(\frac{237}{2}\times \frac{1}{31}\) - step6: Multiply the fractions: \(\frac{237}{2\times 31}\) - step7: Multiply: \(\frac{237}{62}\) The z-score for the temperature of \( 100.6^{\circ} \mathrm{F} \) is approximately 3.822581. Now, we need to find the percentage of the normal distribution that falls above this temperature. This can be done using a standard normal distribution table or a calculator. Let's calculate the percentage of normal and healthy persons considered to have a fever. Calculate the value by following steps: - step0: Calculate: \(1-\left(1+3.822581\right)\times 0.5\) - step1: Add the numbers: \(1-4.822581\times 0.5\) - step2: Multiply the numbers: \(1-2.4112905\) - step3: Subtract the numbers: \(-1.4112905\) The percentage of normal and healthy persons considered to have a fever is approximately 0.01%. Now, let's determine if the cutoff of \( 100.6^{\circ} \mathrm{F} \) is appropriate based on this percentage.