Question

A body temperature of 96.7°F is given, considering that human body temperatures have a mean of 98.2°F and a standard deviation of 0.62°F. Find the z-score corresponding to the given value and use the z-score to determine whether the value is unusual.

Answer 1

The z-score for a body temperature of 96.7°F, given a mean of 98.2°F and a standard deviation of 0.62°F, is -2.42. This indicates that the temperature is unusual, as it falls more than two standard deviations below the mean.

Step-by-step explanation:

To calculate the z-score for a body temperature of 96.7°F, given that the mean human body temperature is 98.2°F with a standard deviation of 0.62°F, use the formula:

Z = (X - μ) / σ

Where:

X = the value in question (96.7°F)

μ = the mean value (98.2°F)

σ = the standard deviation (0.62°F)

The formula yields:

Z = (96.7 - 98.2) / 0.62 = -2.42

A z-score of -2.42 means the body temperature is 2.42 standard deviations below the mean. In general, a z-score lower than -2 or higher than 2 is considered unusual. Therefore, a temperature of 96.7°F is unusual for human body temperature.