To find the probability $P(X < 13,500)$ for a normally distributed variable, we'll calculate the z-score and use it to find the probability. Here are the steps:

Step 1: Calculate the z-score

The z-score formula is:

$z = \frac{X - \mu}{\sigma}$

where:

• $X = 13,500$ (the target debt amount),
• $\mu = 16,061$ (the mean debt amount),
• $\sigma = 2,011$ (the standard deviation).

Substituting the values:

$z = \frac{13,500 - 16,061}{2,011} = \frac{-2,561}{2,011} \approx -1.2736$

Step 2: Find the probability using the z-score

Now, using the z-score of $-1.2736$, we can look up the probability from a standard normal distribution table, or use the normalcdf function on a TI-83 Plus/TI-84 Plus calculator as follows:

1. Press 2nd then DISTR.
2. Select 2:normalcdf.
3. Enter -1E99, -1.2736 (where $-1E99$ represents negative infinity).
4. Press ENTER.

The result should be approximately:

$P(X < 13,500) \approx 0.1019$

Final Answer

$P(X < 13,500) \approx 0.1019$

Let me know if you want further clarification or details on using the calculator.

---

1. What is a z-score, and why is it useful in probability?
2. How would the probability change if the debt limit was increased to $14,000?
3. Why is the normal distribution appropriate here?
4. How would you find the probability of a debt above $20,000?
5. How does standard deviation affect the shape of a normal distribution?

Tip: For accuracy in real-world applications, rounding to four decimal places is a standard practice in statistics.