To find the probability $P(X > 14000)$ where $X$ is normally distributed with a mean $\mu = 23943$ and standard deviation $\sigma = 3000$, you can use the following steps:

1. Calculate the Z-score: $Z = \frac{X - \mu}{\sigma} = \frac{14000 - 23943}{3000} \approx -3.3143$

2. Use the Z-score to find the probability: On a TI-83/TI-84 calculator, use the command: $1 - \text{normalcdf}(-\infty, -3.3143)$ Alternatively, find $P(Z < -3.3143)$ using: $\text{normalcdf}(-3.3143, \infty)$

3. Calculate: Using the calculator: $P(X > 14000) \approx 1 - P(Z < -3.3143) \approx 0.0005$

Thus, $P(X > 14000) \approx 0.0005$.

Would you like more details on any part of this process? Here are some related questions you might consider:

1. How do you calculate the Z-score for different values?
2. What is the significance of the standard deviation in this context?
3. How can you interpret the probability result?
4. What are other factors that can influence financial losses in a divorce?
5. How would this probability change with a different mean or standard deviation?

Tip: Always double-check your calculator settings to ensure you're using the normal distribution functions correctly!