Assume that body temperatures of healthy adults are normally distributed with a mean of 98.2°F and a standard deviation of 0.62°F. Find the 90th percentile of the sample mean body temperatures for samples of size 16.

98

To find the 90th percentile of the sample mean body temperatures for samples of size 16, we need to use the concept of the sampling distribution of the mean.

The sampling distribution of the mean refers to the distribution of all possible sample means that can be obtained from a population. For large sample sizes, the sampling distribution of the mean is approximately normally distributed, regardless of the shape of the population distribution.

Now let's use the following formula to find the standard deviation (standard error) of the sampling distribution of the mean:

Standard Error (SE) = (Standard Deviation of the Population) / √(Sample Size)

SE = 0.62°F / √(16)
SE = 0.155°F

Next, to find the Z-score corresponding to the 90th percentile, we need to use a Z-table (also known as a standard normal distribution table) or a calculator that can provide Z-scores.

The Z-score represents the number of standard deviations a particular value is from the mean. Since we want to find the value corresponding to the 90th percentile, the Z-score will give us the number of standard deviations from the mean that corresponds to this percentile.

Using the Z-table or a calculator, the Z-score corresponding to the 90th percentile is approximately 1.28.

Finally, we can find the 90th percentile of the sample mean body temperatures by multiplying the Z-score by the standard error and adding it to the population mean:

90th percentile = (Z-score) * SE + Mean
90th percentile = 1.28 * 0.155°F + 98.2°F
90th percentile ≈ 98.4°F

Therefore, the 90th percentile of the sample mean body temperatures for samples of size 16 is approximately 98.4°F.