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To solve this problem, we need to calculate the probability that the sample proportion \( \hat{p} \) is between 0.50 and 0.60, given the population proportion \( p = 0.54 \) and sample size \( n = 111 \).

### Step-by-Step Solution:

1. **Determine the Standard Deviation of the Sampling Distribution of \( \hat{p} \):**
   The standard deviation of \( \hat{p} \), also known as the standard error, is calculated using the formula:
   \[
   \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}
   \]
   Substituting \( p = 0.54 \) and \( n = 111 \):
   \[
   \sigma_{\hat{p}} = \sqrt{\frac{0.54 \times (1 - 0.54)}{111}}
   \]

2. **Calculate \( \sigma_{\hat{p}} \):**
   \[
   \sigma_{\hat{p}} = \sqrt{\frac{0.54 \times 0.46}{111}} = \sqrt{\frac{0.2484}{111}} \approx 0.0473
   \]

3. **Convert the Bounds to Z-Scores:**
   To find \( P(0.50 < \hat{p} < 0.60) \), convert 0.50 and 0.60 to z-scores using the formula:
   \[
   z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}
   \]

   - For \( \hat{p} = 0.50 \):
     \[
     z = \frac{0.50 - 0.54}{0.0473} \approx -0.8452
     \]

   - For \( \hat{p} = 0.60 \):
     \[
     z = \frac{0.60 - 0.54}{0.0473} \approx 1.2676
     \]

4. **Find the Probability Using the Z-Scores:**
   Now, find the probability between these z-scores, \( P(-0.8452 < Z < 1.2676) \).

   Using the z-table or a calculator:
   - The probability corresponding to \( z = -0.8452 \) is approximately 0.1992.
   - The probability corresponding to \( z = 1.2676 \) is approximately 0.8975.

   So,
   \[
   P(0.50 < \hat{p} < 0.60) = 0.8975 - 0.1992 = 0.6983
   \]

### Answer:
\[
P(0.50 < \hat{p} < 0.60) \approx 0.6983
\]

Would you like more details or have any questions on this solution?

### Related Questions:
1. How do we interpret the z-scores in this context?
2. Why is the standard error of \( \hat{p} \) used in the calculation?
3. How does sample size impact the standard error of the sample proportion?
4. Can this method be applied to different confidence intervals?
5. What assumptions are required for the Central Limit Theorem to hold in this context?

#### Tip:
When using the Central Limit Theorem for sample proportions, ensure the sample size is large enough to approximate a normal distribution, generally \( np \geq 10 \) and \( n(1 - p) \geq 10 \).

Below, n is the sample size, p is the population proportion and \( \hat{p} \) is the sample proportion. Use the Central Limit Theorem and the TI-84 calculator to find the probability. Round the answer to at least four decimal places.

n = 111
p = 0.54

\( P(0.50 < \hat{p} < 0.60) \) = ?

Solve for the probability that the sample proportion is between 0.50 and 0.60 using the Central Limit Theorem. Given a population proportion of 0.54 and sample size of 111, this solution involves calculating the standard error, converting bounds to z-scores, and determining the probability. Suitable for college-level statistics students.

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