1. **Stating the problem:** We want to find the probability that the sample mean exceeds 130 days. 2. **Formula and explanation:** When dealing with sample means, the sampling distribution of the sample mean $\bar{X}$ is approximately normal if the sample size is large, with mean $\mu$ and standard error $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$. 3. **Calculate the standard error:** Suppose the population mean is $\mu$ and population standard deviation is $\sigma$, and sample size is $n$. 4. **Find the z-score:** $$ z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} $$ where $\bar{X} = 130$. 5. **Calculate the probability:** The probability that the sample mean exceeds 130 days is $$ P(\bar{X} > 130) = P\left(Z > z\right) = 1 - P(Z \leq z) $$ where $Z$ is a standard normal variable. 6. **Interpretation:** Use standard normal tables or software to find $P(Z \leq z)$ and subtract from 1 to get the final probability. *Note:* Since the problem does not provide $\mu$, $\sigma$, or $n$, the exact numerical answer cannot be computed here. **Final answer:** The probability that the sample mean exceeds 130 days is $P(\bar{X} > 130) = 1 - \Phi\left(\frac{130 - \mu}{\sigma / \sqrt{n}}\right)$, where $\Phi$ is the standard normal cumulative distribution function.