Reword the following answer: Let's break down the solution to each part of the question step-by-step: **First Part: P(X < 223.1)** 1. **Identify the Parameters:** - Population mean () = 228.8 - Population standard deviation () = 24.1 - Value of interest (X) = 223.1 2. **Calculate the Z-score:** The Z-score formula for a single value is: \[ Z = \frac{X - \mu}{\sigma} \] Substituting the given values: \[ Z = \frac{223.1 - 228.8}{24.1} = -0.2365 \] 3. **Find the Probability:** Using a Z-table or statistical software, find the probability corresponding to Z = -0.2365. This gives: \[ P(X < 223.1) = 0.4345 \] **Second Part: P(M < 223.1)** 1. **Identify the Parameters:** - Sample mean (M) = 223.1 - Population mean () = 228.8 - Population standard deviation () = 24.1 - Sample size (n) = 22 2. **Calculate the Standard Error:** The standard error (SE) of the sample mean is calculated as: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{24.1}{\sqrt{22}} \approx 5.139 \] 3. **Calculate the Z-score for the Sample Mean:** The Z-score formula for a sample mean is: \[ Z = \frac{M - \mu}{SE} \] Substituting the given values: \[ Z = \frac{223.1 - 228.8}{5.139} = -1.4284 \] 4. **Find the Probability:** Using a Z-table or statistical software, find the probability corresponding to Z = -1.4284. This gives: \[ P(M < 223.1) = 0.3557 \] These calculations show how to determine the probabilities for both a single value and a sample mean being less than a specified value in a normally distri