Answer :
A male at the 82nd percentile of the normally distributed body temperatures with a mean of 98.1 degrees F and a standard deviation of 0.50 degrees F would have a body temperature of approximately 98.56 degrees F, calculated by multiplying the Z-score for the 82nd percentile by the standard deviation and then adding the mean.
To find the body temperature corresponding to the 82nd percentile in a normally distributed population with a mean of 98.1 degrees F and a population standard deviation of 0.50 degrees F, we need to use the properties of the standard normal distribution.
Z-scores represent the number of standard deviations a data point is from the mean. To find the Z-score that corresponds to the 82nd percentile, we can use a Z-table or statistical software. After finding the Z-score, we use the formula:
Temperature = (Z-score)(Standard Deviation) + Mean
Let's say the Z-score for the 82nd percentile is approximately 0.92 (this value should be verified with a Z-table or statistical software). We can then calculate the temperature at the 82nd percentile as follows:
Temperature at 82nd percentile = (0.92)(0.50) + 98.1
Temperature at 82nd percentile = 0.46 + 98.1
Temperature at 82nd percentile = 98.56 degrees F
Therefore, a male at the 82nd percentile would have a body temperature of approximately 98.56 degrees F.
