A population of values has a normal distribution with $u=75.9$ and $o=9.4$ Find the probability that a randomly selected sample of size $n=144$ has a mean greater than 74.8. Round your answer to four decimal places.

Step 1 ：The problem is asking for the probability that the mean of a sample of size 144 is greater than 74.8, given that the population from which the sample is drawn has a mean of 75.9 and a standard deviation of 9.4.

Step 2 ：This is a problem of normal distribution and we can use the Z-score to solve it. The Z-score is a measure of how many standard deviations an element is from the mean.

Step 3 ：The formula for the Z-score is: \(Z = \frac{X - \mu}{\sigma/\sqrt{n}}\) where: X is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and n is the sample size.

Step 4 ：We can calculate the Z-score for X = 74.8, \(\mu\) = 75.9, \(\sigma\) = 9.4, and n = 144.

Step 5 ：Then, we can use the Z-score to find the probability that a randomly selected sample of size 144 has a mean greater than 74.8. This is done by looking up the Z-score in a standard normal distribution table, or by using a function that gives the cumulative distribution function for the standard normal distribution.

Step 6 ：Final Answer: The probability that a randomly selected sample of size 144 has a mean greater than 74.8 is \(\boxed{0.9199}\).