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.

**Step 1. Compute the Standard Error** The standard error of the sample mean is given by \[ SE = \frac{\sigma}{\sqrt{n}}. \] Here, \(\sigma = 97.8\) and \(n = 117\). Thus, \[ SE = \frac{97.8}{\sqrt{117}}. \] First, compute \(\sqrt{117} \approx 10.8167\). Then, \[ SE \approx \frac{97.8}{10.8167} \approx 9.0422. \] **Step 2. Calculate the \(z\)-score** The \(z\)-score for the sample mean is \[ z = \frac{\bar{x} - \mu}{SE}. \] With \(\bar{x} = 51.7\) and \(\mu = 28.2\), we have: \[ z = \frac{51.7 - 28.2}{9.0422} = \frac{23.5}{9.0422} \approx 2.600. \] **Step 3. Find the Probability** We need the probability that the sample mean is greater than \(51.7\), which is \[ P(\bar{X} > 51.7) = P(Z > 2.600). \] Using the standard normal distribution table or a calculator, \[ P(Z > 2.600) \approx 1 - 0.9953 = 0.0047. \] **Final Answer** \[ P(\bar{X} > 51.7) \approx 0.0047. \]