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To solve this problem, we will need some additional information about the underlying distribution. Specifically, these key parameters are usually required:

1. **Mean (μ)** of the population rent.
2. **Standard deviation (σ)** of the population rent or the standard error of the mean.
3. **Sample size (n)**, if relevant (because the standard error $ \sigma_{\bar{x}} $ depends on sample size: $ \sigma / \sqrt{n} $).
   
Once these are available, the problem can be solved by:
- Converting the given range $[2503, 2603]$ into **Z-scores**:
  \[ Z = \frac{X - \mu}{\sigma / \sqrt{n}} \]
- Finding the **cumulative probabilities** corresponding to the Z-scores using the standard normal distribution table.
- Subtracting the two cumulative probabilities to get the required probability.

Can you provide the missing parameters (mean, standard deviation, and sample size)? Alternatively, I can walk you through the solution using placeholder values. Let me know how you'd like to proceed! 

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**Relative questions:**
1. What is the relationship between Z-scores and probability in a standard normal distribution?
2. How does the sample size affect the standard error of the mean?
3. Why do we convert raw values to Z-scores when working with probabilities?
4. What is the cumulative distribution function (CDF) used for?
5. How would this problem change if the population distribution were non-normal?

**Tip:** In most cases, increasing the sample size reduces the standard error, making the sample mean more precise.

What is the probability that the sample mean rent is between $2503 and $2603?

Learn how to calculate the probability that the sample mean rent is between $2503 and $2603 using standard normal distribution concepts. This solution covers key statistical concepts such as Z-scores, the central limit theorem, and cumulative probabilities.

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