A population of values has a normal distribution with \( \mu=120.2 \) and \( \sigma=89.2 \). You intend to draw a random sample of size \( n=232 \). Find the probability that a single randomly selected value is greater than 130.7 . \( P(X>130.7)=\square \) Find the probability that a sample of size \( n=232 \) is randomly selected with a mean greater than 130.7 . \( P(M>130.7)=\square \) Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact \( z \) - scores or \( z \)-scores rounded to 3 decimal places are accepted.

To solve the problem, we will use the properties of the normal distribution. 1. **Finding the probability that a single randomly selected value is greater than 130.7:** We first calculate the z-score for a single value using the formula: \[ z = \frac{X - \mu}{\sigma} \] where \(X = 130.7\), \(\mu = 120.2\), and \(\sigma = 89.2\). \[ z = \frac{130.7 - 120.2}{89.2} \approx \frac{10.5}{89.2} \approx 0.117 \] Next, we look up the z-score of 0.117 in a standard normal distribution table or use a calculator to find: \[ P(Z < 0.117) \approx 0.546 \] Then, to find \(P(X > 130.7)\): \[ P(X > 130.7) = 1 - P(Z < 0.117) \approx 1 - 0.546 = 0.454 \] Thus, \[ P(X > 130.7) \approx 0.4540 \] 2. **Finding the probability that a sample of size \( n=232 \) has a mean greater than 130.7:** For the sample mean, we use the Central Limit Theorem. The standard error (SE) is calculated as: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{89.2}{\sqrt{232}} \approx \frac{89.2}{15.23} \approx 5.86 \] Now we calculate the z-score for the sample mean: \[ z = \frac{M - \mu}{SE} \] where \(M = 130.7\). \[ z = \frac{130.7 - 120.2}{5.86} \approx \frac{10.5}{5.86} \approx 1.79 \] Next, we find \(P(M > 130.7)\): \[ P(Z < 1.79) \approx 0.963 \] So, \[ P(M > 130.7) = 1 - P(Z < 1.79) \approx 1 - 0.963 = 0.037 \] Thus, the final results are: \[ P(X > 130.7) \approx 0.4540 \] \[ P(M > 130.7) \approx 0.0370 \] You can enter these answers as \( P(X > 130.7) = 0.4540 \) and \( P(M > 130.7) = 0.0370 \).