A population of values has a normal distribution with \( \mu=28.2 \) and \( \sigma=97.8 \). You intend to draw a random sample of size \( n=117 \). Find the probability that a sample of size \( n=117 \) is randomily selected with a mean greater than 51.7 . \( P M>51.7)= \) Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact \( z \) - scores or 2 -scores rounded to 3 decimal places are accepted.

To solve this problem, we need to find the probability that a random sample mean \( \bar{X} \) from a normally distributed population with a mean \( \mu = 28.2 \) and standard deviation \( \sigma = 97.8 \) will be greater than 51.7, given the sample size \( n = 117 \). First, we calculate the standard error (SE) of the sample mean: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{97.8}{\sqrt{117}} \approx 9.0202. \] Next, we calculate the \( z \)-score for \( \bar{X} = 51.7 \): \[ z = \frac{\bar{X} - \mu}{SE} = \frac{51.7 - 28.2}{9.0202} \approx 2.5972. \] Using the standard normal distribution table (or a calculator), we can find the probability \( P(Z > 2.5972) \): \[ P(Z > 2.5972) \approx 1 - P(Z < 2.5972) \approx 1 - 0.9952 = 0.0048. \] Thus, the probability that a sample of size \( n = 117 \) has a mean greater than 51.7 is approximately: \[ \boxed{0.0048}. \]