Assume that body temperatures of healthy adults are normally distributed with a mean of 98.2 degrees F and a standard deviation of .62 degrees 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 degrees F, we can use the Central Limit Theorem and convert the problem to a standard normal distribution.

Step 1: Standardize the data
First, we need to standardize the value of 98.4 using the formula:

Z = (X - μ) / (σ / √n)

where:
Z is the standard score
X is the observed value (98.4)
μ is the population mean (98.2)
σ is the population standard deviation (0.62)
n is the sample size (8)

Substituting the values, we get:

Z = (98.4 - 98.2) / (0.62 / √8)

Step 2: Calculate the probability
Once we have the standard score (Z), we can find the probability using the standard normal distribution table (also called Z-table) or a statistical software.

Looking up the Z-value in the table or using a calculator, we find that the Z-value corresponding to 98.4 is approximately 0.544. This means that the mean body temperature of 98.4 is 0.544 standard deviations above the mean.

The probability can be calculated as the area under the standard normal distribution curve to the right of this Z-value. This probability represents the percentage of values that are greater than 98.4.

P(X > 98.4) = 1 - P(X < 98.4)

Since the standard normal distribution is symmetrical, P(X < 98.4) is equal to P(X > -0.544).

Using the standard normal distribution table or a calculator, we find that P(X > -0.544) is approximately 0.705.

Therefore, the probability that the mean body temperature of a sample of eight healthy adults will be greater than 98.4 degrees F is approximately 1 - 0.705 = 0.295, or 29.5%.