I apologize for the oversight. Let's solve it and provide the answer.
Step-by-step explanation :
1. **Standard Error Calculation :
\[ SE = \frac{\sigma}{\sqrt{n}} \]
\[ SE = \frac{25.3}{\sqrt{215}} \]
\[ SE \approx 1.7292 \]
2. **Z-Score Calculation:**
\[ z = \frac{X - \mu}{SE} \]
\[ z = \frac{197.2 - 195.5}{1.7292} \]
\[ z \approx 0.9836 \]
3. Finding the Probability :
To find the probability that the sample mean is greater than 197.2, we need to find the area to the right of the z-score in the standard normal distribution table.
For \( z \approx 0.9836 \), the area to the left is approximately \( 0.8374 \).
Since we want the area to the right (the probability the sample mean is greater than 197.2), we need to subtract that value from 1 :
\[ P(X > 197.2) = 1 - 0.8374 \]
\[ P(X > 197.2) = 0.1626 \]
Answer : The probability that the sample mean is greater than 197.2 is approximately \( 0.1626 \) or \( 16.26\% \).