To solve $P(x > 14000)$, we need to find the probability that the financial loss $x$ is greater than $14,000. Given:

• The mean $\mu = 23943$
• The standard deviation $\sigma = 3000$

We can calculate the $z$-score using the formula: $z = \frac{x - \mu}{\sigma}$

Step 1: Calculate the $z$-score for $x = 14000$

$z = \frac{14000 - 23943}{3000}$ $z = \frac{-9943}{3000} \approx -3.3143$

Step 2: Find $P(z > -3.3143)$

On the TI-83/84 calculator, go to:

1. 2nd -> DISTR -> normalcdf(.
2. Enter the bounds for $z$: normalcdf(-3.3143, ∞). Since the calculator doesn't allow ∞, use a large positive number (e.g., 10).
3. Input the mean $0$ and standard deviation $1$.

The probability $P(z > -3.3143)$ should return approximately 0.9994.

Thus: $P(x > 14000) \approx 0.9994$

Would you like details on the steps, or do you have any questions?

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Related Questions

1. What is $P(x < 20000)$ for this distribution?
2. How would we find $P(x > 25000)$?
3. What is the probability that a randomly selected person spent between $20,000 and $30,000 on their divorce?
4. What does the probability $P(x > 14000) \approx 0.9994$ imply in practical terms?
5. How would changes in the standard deviation affect the probability $P(x > 14000)$?

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Tip: When using the TI-83/84, remember that normalcdf(lower bound, upper bound, mean, standard deviation) is the format for finding probabilities under a normal curve.