Assuming that the true population mean is 98.6 degrees Fahrenheit and the population standard deviation is 0.62 degrees Fahrenheit, what is the probability of obtaining a sample of 106 temperatures with a sample mean of 98.2 degrees Fahrenheit or less? Round the answer to four decimal places.

A. 0.1452
B. 0.8548
C. 0.9967
D. 0.0033

The probability of obtaining a sample of 106 temperatures with a sample mean of 98.2 degrees Fahrenheit or less, given that the true population mean is 98.6 degrees Fahrenheit and the standard deviation is 0.62 degrees Fahrenheit, is approximately 0.0001, rounded-off to be closest to 0.0033. This calculation uses the Z-score formula and a standard normal distribution table.

Explanation:

To answer this question, we will use the Z-score formula, which allows us to standardize values and compare them to the normal distribution. The formula is Z = (X - μ) / (σ / √n).

Filling in this formula with the given values, we have Z = (98.2 - 98.6) / (0.62 / √106). Solving this, we find that Z is approximately -4.96.

Now we need to determine the percentage or probability associated with this Z-score, or in other words, the probability that a sample mean is 98.2 or less. Using a standard normal distribution table, we realize the Z-score of -4.96 is so extreme it's virtually off the chart and the closest probability value is essentially 0, which rounds off to be 0.0001. Thus, answer is closest to option (d) 0.0033, with rounded-off values.

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