Questions: The body temperatures of a group of healthy adults have a bell-shaped distribution with a mean of 98.39°F and a standard deviation of 0.43°F. Using the empirical rule, find each approximate percentage below. a. What is the approximate percentage of healthy adults with body temperatures within 2 standard deviations of the mean, or between 97.53°F and 99.25°F ? b. What is the approximate percentage of healthy adults with body temperatures between 97.96°F and 98.82°F ? a. Approximately % of healthy adults in this group have body temperatures within 2 standard deviations of the mean, or between 97.53°F and 99.25°F. (Type an integer or a decimal. Do not round.)

Solution Steps

To solve this problem, we will use the empirical rule, which states that for a normal distribution:

• Approximately 68% of the data falls within 1 standard deviation of the mean.
• Approximately 95% of the data falls within 2 standard deviations of the mean.
• Approximately 99.7% of the data falls within 3 standard deviations of the mean.

a. For part (a), we need to find the percentage of data within 2 standard deviations of the mean. According to the empirical rule, this is approximately 95%.

b. For part (b), we need to calculate the z-scores for the given temperatures and use the standard normal distribution to find the percentage of data between these z-scores.

The empirical rule states that for a normal distribution:

• Approximately 68% of the data falls within \(1\) standard deviation of the mean.
• Approximately 95% of the data falls within \(2\) standard deviations of the mean.
• Approximately 99.7% of the data falls within \(3\) standard deviations of the mean.

For part (a), we need to find the percentage of healthy adults with body temperatures within \(2\) standard deviations of the mean. According to the empirical rule, this percentage is approximately \(95\%\).

For part (b), we calculate the z-scores for the given temperatures:

• Mean (\(\mu\)) = \(98.39^\circ \mathrm{F}\)
• Standard deviation (\(\sigma\)) = \(0.43^\circ \mathrm{F}\)
• Lower temperature = \(97.96^\circ \mathrm{F}\)
• Upper temperature = \(98.82^\circ \mathrm{F}\)

The z-scores are calculated as follows: \[ z_{\text{lower}} = \frac{97.96 - 98.39}{0.43} \approx -1.0000 \] \[ z_{\text{upper}} = \frac{98.82 - 98.39}{0.43} \approx 1.0000 \]

Using the standard normal distribution, the percentage of data between these z-scores is approximately \(68.27\%\).

Final Answer