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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. Suppose we take a sample of eight healthy adults. What is the probability that their mean body temperature will be greater than 98.4?

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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%).
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