Question

Which of the following meets the success-failure condition when using the normal model for a sampling distribution?(1 point)
Responses

sample size=180; probability of success=96 percent
sample size=180; probability of success=96 percent

sample size=85; probability of success=12 percent
sample size=85; probability of success=12 percent

sample size=85; probability of success=11 percent
sample size=85; probability of success=11 percent

sample size=180; probability of success=95 percent

Answer 1

The success-failure condition states that for a sample to meet the criteria for using a normal approximation to the sampling distribution, both the expected number of successes and the expected number of failures should be at least 10. This can be formulated as:

- Expected number of successes (np) ≥ 10
- Expected number of failures (n(1-p)) ≥ 10

Now let's check each option:

1. **Sample size: 180; probability of success: 96 percent (0.96)**
- Expected successes = 180 * 0.96 = 172.8
- Expected failures = 180 * (1 - 0.96) = 180 * 0.04 = 7.2
- Doesn't meet the condition (expected failures < 10).

2. **Sample size: 85; probability of success: 12 percent (0.12)**
- Expected successes = 85 * 0.12 = 10.2
- Expected failures = 85 * (1 - 0.12) = 85 * 0.88 = 74.8
- Meets the condition (expected successes ≥ 10 and expected failures ≥ 10).

3. **Sample size: 85; probability of success: 11 percent (0.11)**
- Expected successes = 85 * 0.11 = 9.35
- Expected failures = 85 * (1 - 0.11) = 85 * 0.89 = 75.65
- Doesn't meet the condition (expected successes < 10).

4. **Sample size: 180; probability of success: 95 percent (0.95)**
- Expected successes = 180 * 0.95 = 171
- Expected failures = 180 * (1 - 0.95) = 180 * 0.05 = 9
- Doesn't meet the condition (expected failures < 10).

**Conclusion: The second option (sample size = 85; probability of success = 12 percent) meets the success-failure condition.**