To find the probability that the mean body temperature of a sample of eight healthy adults will be greater than 98.4°F, we can use the sampling distribution of the sample mean.
The mean body temperature of healthy adults is normally distributed with a mean (μ) of 98.2°F and a standard deviation (σ) of 0.62°F. Since the sample size is large enough (n = 8), we can assume that the sampling distribution of the sample mean is also approximately normally distributed.
To calculate the probability, we need to standardize the sample mean using the standard deviation of the sample mean (also known as the standard error). The standard error (SE) can be calculated using the formula:
SE = σ / sqrt(n)
SE = 0.62 / sqrt(8)
SE ≈ 0.2196
Next, we can calculate the z-score corresponding to the given mean body temperature of 98.4°F using the formula:
z = (sample mean - population mean) / SE
z = (98.4 - 98.2) / 0.2196
z ≈ 0.9092
Using a standard normal distribution table or a statistical software, we can find the probability associated with the z-score of 0.9092. The probability can be calculated as:
P(Z > 0.9092) ≈ 1 - P(Z < 0.9092)
Looking up the value in the table or using software, we find:
P(Z < 0.9092) ≈ 0.8188
Therefore,
P(Z > 0.9092) ≈ 1 - 0.8188 ≈ 0.1812
The probability that the mean body temperature of a sample of eight healthy adults will be greater than 98.4°F is approximately 0.1812 (or 18.12%).