suppose that the population proportion is known to be 67%. Take a sample size of 100. What is the probability that the sample proportion will be greater than 75%?

To solve this problem, we will use the **normal approximation to the binomial distribution**, as the sample size (100) is large enough for the Central Limit Theorem to apply. Here's how we can approach it:

1. **Population Proportion (\( p \))**: 67% or 0.67
2. **Sample Size (\( n \))**: 100
3. **Sample Proportion (\( \hat{p} \))**: 75% or 0.75

### Step 1: Mean of the Sampling Distribution of the Sample Proportion

The mean of the sampling distribution of the sample proportion is equal to the population proportion, which is:

\[
\mu_{\hat{p}} = p = 0.67
\]

### Step 2: Standard Deviation of the Sampling Distribution

The standard deviation of the sample proportion, also known as the **standard error (SE)**, is calculated as:

\[
\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.67(1 - 0.67)}{100}} = \sqrt{\frac{0.67 \times 0.33}{100}} = \sqrt{0.2211 / 100} = \sqrt{0.002211} \approx 0.047
\]

### Step 3: Convert the Sample Proportion to a Z-Score

Now, we convert the sample proportion of 0.75 to a z-score to determine how far it is from the mean in terms of standard deviations:

\[
z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.75 - 0.67}{0.047} = \frac{0.08}{0.047} \approx 1.70
\]

### Step 4: Find the Probability

Now, we look up the z-score of 1.70 in the standard normal distribution table, or use a calculator. The z-score of 1.70 corresponds to a cumulative probability of approximately 0.9554.

Since we want the probability that the sample proportion is **greater than 75%**, we need the area to the right of this z-score:

\[
P(\hat{p} > 0.75) = 1 - 0.9554 = 0.0446
\]

### Final Answer:
The probability that the sample proportion will be greater than 75% is approximately **0.0446**, or **4.46%**.

---

Do you want more details or have any questions?

Here are 5 related questions you could explore:
1. How does the sample size affect the standard error of the sample proportion?
2. What happens if the population proportion was 50% instead of 67%?
3. Can you explain how the Central Limit Theorem justifies the use of a normal approximation?
4. What would the probability be for a sample proportion greater than 70%?
5. How would the result change if the sample size was 500 instead of 100?

**Tip**: A larger sample size reduces the standard error, making the sampling distribution narrower and more precise.

Suppose that the population proportion is known to be 67%. Take a sample size of 100. What is the probability that the sample proportion will be greater than 75%?

Determine the probability that a sample proportion exceeds 75% given a population proportion of 67% and a sample size of 100. This problem involves concepts like the normal distribution, standard error, and z-scores, applying the Central Limit Theorem.

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