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Let's go through each part of this problem step-by-step to provide a comprehensive answer.

### Given Data:
- Population proportion of satisfaction, \( p = 0.80 \)
- Sample size, \( n = 100 \)

### Part (a)
The question asks if the response to "Are you satisfied with the way things are going in your life?" is qualitative or quantitative.
- **Answer**: The correct choice is **(D)**. This is because satisfaction is a qualitative response (categorical) that can be classified into "satisfied" or "not satisfied".

### Part (b)
The question explains why the sample proportion, \(\hat{p}\), is considered a random variable and asks for the source of its variability.
- **Answer**: The correct answer is **(A)**. The sample proportion \(\hat{p}\) is a random variable because its value varies from sample to sample due to the differences in individual responses. Each random sample of 100 people may give slightly different satisfaction proportions.

### Part (c)
This part asks to describe the sampling distribution of \(\hat{p}\), the sample proportion of people satisfied with their lives, and verify the model requirements.
  
1. **Conditions for Normal Approximation**:
   - Since the sample size \(n = 100\) is less than 5% of the population (assuming a large population), this condition is met.
   - For a binomial distribution to be approximated by a normal distribution, we check if \( np \geq 10 \) and \( n(1 - p) \geq 10 \):
     - \( np = 100 \times 0.80 = 80 \)
     - \( n(1 - p) = 100 \times 0.20 = 20 \)
   - Both values are greater than 10, so the normal approximation is appropriate.

2. **Sampling Distribution**:
   - The mean of \(\hat{p}\), \( \mu_{\hat{p}} = p = 0.80 \).
   - The standard deviation of \(\hat{p}\), \( \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.80 \times 0.20}{100}} = 0.040 \).
   - **Answer**: \(\hat{p}\) is approximately normal with \( \mu_{\hat{p}} = 0.80 \) and \( \sigma_{\hat{p}} = 0.040 \).

### Part (d)
This part calculates the probability that the sample proportion \(\hat{p}\) exceeds 0.84.

1. **Z-score Calculation**:
   - We calculate the Z-score for \(\hat{p} = 0.84\):
     \[
     Z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.84 - 0.80}{0.040} = \frac{0.04}{0.040} = 1.0
     \]
   
2. **Probability Calculation**:
   - Using the Z-table, the probability that \( Z \leq 1.0 \) is approximately 0.8413.
   - Therefore, the probability that \( Z > 1.0 \) (or \(\hat{p} > 0.84\)) is \( 1 - 0.8413 = 0.1587 \).

   - **Answer**: The probability that the proportion of people satisfied exceeds 0.84 is approximately **0.1587**.

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### Summary of Answers:
- **(a)**: (D) The response is qualitative because it can be classified based on satisfaction.
- **(b)**: (A) \(\hat{p}\) is a random variable because its value varies from sample to sample.
- **(c)**: The distribution of \(\hat{p}\) is approximately normal with \( \mu_{\hat{p}} = 0.80 \) and \( \sigma_{\hat{p}} = 0.040 \).
- **(d)**: The probability that \(\hat{p}\) exceeds 0.84 is approximately **0.1587**.

Would you like more details or have any questions about any part?

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### Related Questions:
1. What would happen to the standard deviation of \(\hat{p}\) if the sample size increased?
2. Why is it important to verify the normal approximation conditions in sampling?
3. How would the probability change if the threshold was 0.85 instead of 0.84?
4. What is the probability that the sample proportion is below 0.76?
5. Can we use this normal approximation if the population proportion was unknown?

### Tip:
When working with sampling distributions, always check the conditions for normality, especially when dealing with proportions. This helps ensure accurate probability calculations.

According to a study conducted by a statistical organization, the proportion of people who are satisfied with the way things are going in their lives is 0.80. Suppose that a random sample of 100 people is obtained. Complete parts (a) through (d) below.

Explore how to calculate the probability of sample proportions exceeding a threshold in a satisfaction survey. This problem demonstrates concepts such as sampling distribution, normal approximation, and the Central Limit Theorem. Suitable for college-level statistics.

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