To determine if the success-failure condition is met for using the normal model in a sampling distribution, we check whether both \(np\) (expected number of successes) and \(n(1-p)\) (expected number of failures) are greater than or equal to 10.
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sample size = 180; probability of success = 96% (0.96)
- \(np = 180 \times 0.96 = 172.8\)
- \(n(1-p) = 180 \times (1 - 0.96) = 180 \times 0.04 = 7.2\)
Here, \(n(1-p)\) is less than 10. This does not meet the condition.
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sample size = 85; probability of success = 11% (0.11)
- \(np = 85 \times 0.11 = 9.35\)
- \(n(1-p) = 85 \times (1 - 0.11) = 85 \times 0.89 = 75.65\)
Here, \(np\) is less than 10. This does not meet the condition.
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sample size = 180; probability of success = 95% (0.95)
- \(np = 180 \times 0.95 = 171\)
- \(n(1-p) = 180 \times (1 - 0.95) = 180 \times 0.05 = 9\)
Here, \(n(1-p)\) is less than 10. This does not meet the condition.
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sample size = 85; probability of success = 12% (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 \(np\) and \(n(1-p)\) are greater than or equal to 10. This meets the condition.
Therefore, the last option \( \text{sample size = 85; probability of success = 12%} \) meets the success-failure condition.