Suppose that human body temperatures follow a normal distribution and have a mean of 98.2 degrees Fahrenheit with a standard deviation of 0.62 degrees Fahrenheit.

For a random sample of 106, compute the probability that the sample mean body temperature is 98 degrees Fahrenheit or lower.

Round your final answer to four decimal places.

To find the probability that the sample mean body temperature is 98 degrees Fahrenheit or lower, calculate the z-score using the formula Z = (X - μ) ÷ (σ/√n) and then use the standard normal distribution table to find the corresponding probability. In this case, the z-score is approximately -2.794, and the probability is about 0.0026 or 0.26%.

Explanation:

This question involves understanding how the normal distribution, specifically the standard deviation and mean, function in probability. We can define the sample mean (X) to be 98 and the population mean (μ) to be 98.2. The standard deviation (σ) is 0.62, and the sample size (n) is 106.

First, we need to calculate the z-score, the number of standard deviations from the mean. The formula is Z = (X - μ) ÷ (σ/√n). Substituting in our known values, Z = (98 - 98.2) ÷ (0.62/√106), which is approximately -2.794.

The z-score allows us to check the standard normal distribution table (Z-table) for the associated probability. The value for Z of -2.794 is about 0.0026, meaning there's a 0.26% chance that the sample mean body temperature is 98 degrees Fahrenheit or lower.

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