A population of values has a normal distribution with $u=239.5$ and $o=3$. You intend to draw a random sample of size $n=23$ Find the probability that a sample of size $n=23$ is randomly selected with a mean greater than 239.5.I

Step 1 ：The problem is asking for the probability that the mean of a sample of size 23 is greater than 239.5, given that the population from which the sample is drawn has a normal distribution with a mean (\(\mu\)) of 239.5 and a standard deviation (\(\sigma\)) of 3.

Step 2 ：We can use the Central Limit Theorem to solve this problem. The Central Limit Theorem states that the distribution of sample means approximates a normal distribution as the sample size gets larger, regardless of the shape of the population distribution.

Step 3 ：The mean of this distribution of sample means is equal to the population mean, and the standard deviation (known as the standard error) is equal to the population standard deviation divided by the square root of the sample size.

Step 4 ：We can then use the standard normal distribution (Z-distribution) to find the probability that a sample mean is greater than 239.5. The Z-score is calculated as (X - \(\mu\)) / \(\sigma\), where X is the sample mean, \(\mu\) is the population mean, and \(\sigma\) is the standard error.

Step 5 ：In this case, since we are looking for the probability that the sample mean is greater than the population mean, the Z-score is 0 (because X - \(\mu\) = 0). The probability that a Z-score is greater than 0 is 0.5 (because the Z-distribution is symmetric about 0).

Step 6 ：Final Answer: The probability that a sample of size \(n=23\) is randomly selected with a mean greater than 239.5 is \(\boxed{0.5}\).