Which of the following 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
The success-failure condition for using the normal model is met when both np and n(1-p) are greater than or equal to 10.
For the given options:
1. sample size=85; probability of success=11 percent
np = 85*0.11 = 9.35, n(1-p) = 85*0.89 = 75.65 (not met)
2. sample size=180; probability of success=95 percent
np = 180*0.95 = 171, n(1-p) = 180*0.05 = 9 (met)
3. sample size=85; probability of success=12 percent
np = 85*0.12 = 10.2, n(1-p) = 85*0.88 = 74.8 (met)
4. sample size=180; probability of success=96 percent
np = 180*0.96 = 172.8, n(1-p) = 180*0.04 = 7.2 (not met)
Therefore, the options that meet the success-failure condition when using the normal model for a sampling distribution are options 2 and 3:
- sample size=180; probability of success=95 percent
- sample size=85; probability of success=12 percent
Only one answer is correct.
Apologies for the error in my response. Let's re-evaluate the options:
1. sample size=85; probability of success=11 percent
np = 85*0.11 = 9.35, n(1-p) = 85*0.89 = 75.65 (not met)
2. sample size=180; probability of success=95 percent
np = 180*0.95 = 171, n(1-p) = 180*0.05 = 9 (met)
3. sample size=85; probability of success=12 percent
np = 85*0.12 = 10.2, n(1-p) = 85*0.88 = 74.8 (met)
4. sample size=180; probability of success=96 percent
np = 180*0.96 = 172.8, n(1-p) = 180*0.04 = 7.2 (not met)
Therefore, the correct answer is:
- sample size=180; probability of success=95 percent
Thank you for pointing out the mistake, and I appreciate your understanding.