Answer :
The cutoff for the highest 10% of human body temperatures, given a mean of 98.2 degrees and standard deviation of 0.73 degrees, is approximately 99.1°F. This is calculated using the z-score for the 90th percentile in a normal distribution.
To find the cutoff for the highest 10% of human body temperatures, assuming a nearly normal distribution with a mean of 98.2 degrees and a standard deviation of 0.73 degrees, we would use the concept of z-scores from statistics. The z-score associated with the top 10% can be found using standard normal distribution tables or using a statistical software or calculator that can compute z-scores for given percentiles.
In a standard normal distribution, the z-score that corresponds to the top 10% (90th percentile) is approximately 1.28. Therefore, we can use the formula for converting a z-score to an actual value in a normal distribution:
X = μ + (z * σ)
where X is the body temperature, μ is the mean body temperature, z is the z-score, and σ is the standard deviation.
Using the given mean (μ = 98.2°F) and standard deviation (σ = 0.73°F), we calculate the cutoff for the highest 10% of body temperatures as:
X = 98.2 + (1.28 * 0.73)
X ≈ 98.2 + 0.9344
X ≈ 99.1344°F
Therefore, the cutoff temperature for the highest 10% of healthy human body temperatures is approximately 99.1°F.
