Final answer:
The probability that the sample mean of a normally distributed population with a mean of 20 and variance of 40 will be between 17 and 23, for a sample size of 16, can be determined using the Central Limit Theorem and z-scores.
Step-by-step explanation:
The question asks for the probability that the sample mean of a normally distributed variable with a population mean (mu) of 20 and a population variance (sigma^2) of 40 will fall between 17 and 23, given a sample size of 16. To solve this, we use the Central Limit Theorem which states that the sampling distribution of the sample mean will also be normally distributed with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size (standard error).
The standard error (SE) can be calculated as sqrt(sigma^2/n), which in this case is sqrt(40/16), leading to a standard error of 1.58. We then find the z-scores for the sample means of 17 and 23 using the formula z = (x - mu) / SE. For x=17, z = (17 - 20) / 1.58 = -1.90, and for x=23, z = (23 - 20) / 1.58 = 1.90.
Finally, we look up these z-scores in a standard normal distribution table or use a calculator to find the probabilities associated with each and subtract the lower probability from the higher one to find the probability that the sample mean will fall between 17 and 23.