Question

Which of the following meets the success-failure condition when using the normal model for a sampling distribution?
A. Sample size: 180; probability of success: 95 percent
B. Sample size: 180; probability of success: 96 percent
C. Sample size: 85; probability of success: 12 percent
D. Sample size: 85; probability of success: 11 percent

Answer 1

The success-failure condition states that for a normal approximation to the binomial distribution to be valid, both \(np\) (expected number of successes) and \(n(1-p)\) (expected number of failures) must be at least 10, where \(n\) is the sample size and \(p\) is the probability of success.

Let's calculate \(np\) and \(n(1-p)\) for each option:

**A. Sample size: 180; probability of success: 95 percent (0.95)**
- \(np = 180 \times 0.95 = 171\)
- \(n(1-p) = 180 \times (1 - 0.95) = 180 \times 0.05 = 9\)
- Condition not met (since \(n(1-p) < 10\)).

**B. Sample size: 180; probability of success: 96 percent (0.96)**
- \(np = 180 \times 0.96 = 172.8\)
- \(n(1-p) = 180 \times (1 - 0.96) = 180 \times 0.04 = 7.2\)
- Condition not met (since \(n(1-p) < 10\)).

**C. Sample size: 85; probability of success: 12 percent (0.12)**
- \(np = 85 \times 0.12 = 10.2\)
- \(n(1-p) = 85 \times (1 - 0.12) = 85 \times 0.88 = 74.8\)
- Condition met (both \(np \geq 10\) and \(n(1-p) \geq 10\)).

**D. Sample size: 85; probability of success: 11 percent (0.11)**
- \(np = 85 \times 0.11 = 9.35\)
- \(n(1-p) = 85 \times (1 - 0.11) = 85 \times 0.89 = 75.65\)
- Condition not met (since \(np < 10\)).

The only option that meets the success-failure condition is **C. Sample size: 85; probability of success: 12 percent.**