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
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To determine the probability that the mean body temperature of the sampled healthy adults will be greater than 98.4 degrees F, we can use the Central Limit Theorem.
The Central Limit Theorem states that for a large enough sample size, the distribution of sample means will follow a normal distribution regardless of the shape of the population distribution.
In this case, the sample size is 8, which is relatively small. However, we can approximate the distribution of sample means as normal due to the Central Limit Theorem.
First, we need to calculate the standard error of the mean (SEM), which is the standard deviation of the sample means. The formula for SEM is:
SEM = standard deviation / √(sample size)
SEM = 0.62 / √(8) = 0.22
Next, we standardize the value of 98.4 degrees F using the z-score formula:
z = (x - μ) / SEM
where x is the mean body temperature (98.4 degrees F), μ is the population mean (98.2 degrees F), and SEM is the standard error of the mean (0.22).
z = (98.4 - 98.2) / 0.22 ≈ 0.909
Now, we need to find the probability of the mean body temperature being greater than 98.4 degrees F, which corresponds to a z-score greater than 0.909.
Using a standard normal distribution table or a calculator, we can find that the probability of getting a z-score greater than 0.909 is approximately 0.181.
Therefore, the probability that the mean body temperature of the sampled healthy adults will be greater than 98.4 degrees F is about 0.181, or 18.1%.