**Credit Card Debt** The average credit card debt for households that have credit cards is 

$16,061

. If the distribution is approximately normal and the standard deviation is 

$2011

, what is the probability that a randomly selected household has one or more credit cards with the debt of each card less than 

$13,500

? Use a TI-83 Plus/TI-84 Plus calculator. Round the final answer to at least four decimal places.
 
P(x<13500)=

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.

The average credit card debt for households that have credit cards is $16,061. If the distribution is approximately normal and the standard deviation is $2011, what is the probability that a randomly selected household has one or more credit cards with the debt of each card less than $13,500? Use a TI-83 Plus/TI-84 Plus calculator. Round the final answer to at least four decimal places. P(x<13500)=

Learn how to calculate the probability of a household having credit card debt below $13,500, given an average debt of $16,061 and a standard deviation of $2,011. This solution explains the use of the z-score formula and normal distribution properties, suitable for high school statistics students.

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