To determine if it would be unusual for the sample mean to be greater than $2871, we need to calculate the probability that the sample mean is greater than this value. If the probability is very low (typically less than 5%, or 0.05), then it would be considered unusual.

Here are the key details from the problem:

• Population mean ($\mu$) = $2643
• Standard deviation ($\sigma$) = $502
• Sample size ($n$) = 102

We will use the Central Limit Theorem to calculate the probability. The Central Limit Theorem tells us that the sampling distribution of the sample mean is normally distributed with:

$\text{Standard error} (SE) = \frac{\sigma}{\sqrt{n}} = \frac{502}{\sqrt{102}}$

Next, we calculate the Z-score for the value $2871 using the formula:

$Z = \frac{X - \mu}{SE}$

Where $X = 2871$, and then find the corresponding probability.

Let me calculate this for you.The Z-score for a sample mean of $2871 is approximately 4.59. The probability that the sample mean is greater than $2871 is approximately 0.00000225, or 0.000225%.

Since this probability is extremely low (far less than 0.05 or 5%), it would indeed be considered unusual for the sample mean to be greater than $2871.

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

Here are 5 related questions you could explore:

1. How would the result change if the standard deviation were larger?
2. What happens if the sample size is reduced to 50 apartments?
3. How do you calculate the probability for a range of sample means (e.g., between $2500 and $2700)?
4. Why is the Central Limit Theorem important in this problem?
5. What is the Z-score interpretation, and how does it relate to normal distributions?

Tip: In hypothesis testing, results with probabilities under 0.05 are typically considered statistically significant or unusual.