Question

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

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

sample size=180; probability of success=95 percent
sample size=180; probability of success=95 percent

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

sample size=180; probability of success=96 percent

Answer 1

The success-failure condition states that for a sample size \( n \) and a probability of success \( p \), both \( np \) (expected number of successes) and \( n(1-p) \) (expected number of failures) should be greater than or equal to 10 in order to use the normal model for a sampling distribution.

Let's assess each option:

1. **Sample size = 85; probability of success = 11 percent (0.11):**
- \( np = 85 \times 0.11 = 9.35 \)
- \( n(1 - p) = 85 \times (1 - 0.11) = 85 \times 0.89 = 75.65 \)
- Here, \( np < 10 \), so this does not meet the condition.

2. **Sample size = 180; probability of success = 95 percent (0.95):**
- \( np = 180 \times 0.95 = 171 \)
- \( n(1 - p) = 180 \times (1 - 0.95) = 180 \times 0.05 = 9 \)
- Here, \( n(1 - p) < 10 \), so this does not meet the condition.

3. **Sample size = 85; probability of success = 12 percent (0.12):**
- \( np = 85 \times 0.12 = 10.2 \)
- \( n(1 - p) = 85 \times (1 - 0.12) = 85 \times 0.88 = 74.8 \)
- Here, both \( np \geq 10 \) and \( n(1 - p) \geq 10 \), so this meets the condition.

4. **Sample size = 180; probability of success = 96 percent (0.96):**
- \( np = 180 \times 0.96 = 172.8 \)
- \( n(1 - p) = 180 \times (1 - 0.96) = 180 \times 0.04 = 7.2 \)
- Here, \( n(1 - p) < 10 \), so this does not meet the condition.

Based on this analysis, the option that closest meets the success-failure condition when using the normal model for a sampling distribution is:

**Sample size = 85; probability of success = 12 percent.**