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The problem in the image involves calculating the probability that the sample mean rent is greater than $2713, using the normal distribution.

### Given dataThe problem in the image involves calculating the probability that the sample mean rent is greater than $2713, using the normal distribution.

### Given 
- Population mean ($\mu$) = \$2643
- Population standard deviation ($\sigma$) = \$502
- Sample size ($n$) = 102 apartments
- The sample mean rent we are interested in is \$2713.

### Step-by-Step Solution:

1. **Standard Error (SE):**
   Since we are dealing with a sample mean, the standard error of the mean is calculated as:
   \[
   SE = \frac{\sigma}{\sqrt{n}} = \frac{502}{\sqrt{102}}
   \]

2. **Z-score Calculation:**
   The z-score for the sample mean can be calculated using:
   \[
   Z = \frac{\text{Sample Mean} - \mu}{SE} = \frac{2713 - 2643}{SE}
   \]

3. **Finding the Probability:**
   Once we calculate the z-score, we can use the standard normal distribution table (or a calculator) to find the probability that the sample mean is greater than \$2713.

I'll calculate this for you.The Z-score for the sample mean rent of $2713 is approximately 1.4083. The probability that the sample mean rent is greater than $2713 is approximately **0.0795** (or 7.95%).

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

Here are 5 related questions to further explore:
1. How does the sample size affect the standard error?
2. What is the probability that the sample mean is less than $2713?
3. How would the probability change if the sample size were larger?
4. What does the Z-score represent in this context?
5. How can we use the Central Limit Theorem to solve this problem?

**Tip:** As the sample size increases, the standard error decreases, making the sample mean more likely to be close to the population mean.

What is the probability that the sample mean rent is greater than $2713?

Calculate the probability that the sample mean rent is greater than $2713 in a high-rent district with a population mean of $2643 and standard deviation of $502, using the Central Limit Theorem and Z-Score formula.

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