To determine the minimum sample size needed to use normal approximations for a proportion, we typically apply the rule of thumb that requires both \( np \) and \( n(1-p) \) to be at least 5, where \( n \) is the sample size and \( p \) is the proportion of success.
Given a probability of success \( p = 0.10 \), we calculate as follows:
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Calculate \( np \): \[ np = n \times 0.10 \]
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Calculate \( n(1-p) \): \[ n(1-p) = n \times (1 - 0.10) = n \times 0.90 \]
Now we set up the inequalities:
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For \( np \) to be at least 5: \[ n \times 0.10 \geq 5 \quad \Rightarrow \quad n \geq \frac{5}{0.10} = 50 \]
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For \( n(1-p) \) to be at least 5: \[ n \times 0.90 \geq 5 \quad \Rightarrow \quad n \geq \frac{5}{0.90} \approx 5.56 \]
Since the second inequality yields a minimum sample size of approximately 6, the more restrictive condition is the first one. Thus, the minimum sample size required to ensure both conditions are satisfied is:
\[ \text{Minimum sample size } n = 50 \]
In conclusion, the minimum sample size needed to use normal approximations with a probability of success of 10 percent is 50.