To find the percentage of students who have SAT scores greater than 1731, we can use the properties of the normal distribution. We will first convert the SAT score of 1731 into a z-score using the formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
where:
- \(X\) is the score we are interested in (1731),
- \(\mu\) is the mean score (1465),
- \(\sigma\) is the standard deviation (295).
Plugging in the values:
\[ z = \frac{(1731 - 1465)}{295} = \frac{266}{295} \approx 0.900 \]
Now that we have the z-score, we can use the standard normal distribution table (or a calculator) to find the probability associated with \(z = 0.900\).
Looking up the z-score of 0.900, we find the corresponding cumulative probability \(P(Z < 0.900)\).
The cumulative probability for \(z = 0.900\) is approximately \(0.8159\) (or 81.59%).
Since we want the probability of students scoring greater than 1731, we calculate:
\[ P(X > 1731) = 1 - P(Z < 0.900) \approx 1 - 0.8159 = 0.1841 \]
Now, to express this as a percentage:
\[ 0.1841 \times 100 \approx 18.41% \]
Rounding to one decimal place, we conclude:
\[ \boxed{18.4} \]