To determine the minimum sample size needed to use normal approximations for a proportion, you can use the following criteria: the sample size \( n \) should be large enough such that both \( np \) and \( n(1-p) \) are greater than or equal to 5, where \( p \) is the probability of success.
In this case, the probability of success \( p \) is 0.25. Therefore, the probability of failure \( 1 - p \) is 0.75.
You will set up the two conditions based on \( p \) and \( 1-p \):
- \( np \geq 5 \)
- \( n(1-p) \geq 5 \)
Let's compute these conditions:
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For \( np \): \[ n \cdot 0.25 \geq 5 \quad \Rightarrow \quad n \geq \frac{5}{0.25} = 20 \]
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For \( n(1-p) \): \[ n \cdot 0.75 \geq 5 \quad \Rightarrow \quad n \geq \frac{5}{0.75} \approx 6.67 \]
The more restrictive condition here is \( n \geq 20 \).
To maintain sufficient sample size and ensure that both conditions are satisfied, and in practice, it's common to round up and take a slightly larger sample size to ensure reliability.
Since none of the provided options (38, 39, 40, 41) directly relate to 20 being the minimum, you must ensure you're within a range where you can apply the approximation effectively.
Thus, based on common guidelines for normal approximation for proportions, including more than the minimum can help ensure that the sampling distribution of the proportion is approximately normal.
Among the options given, the 39, 40, or 41 options would be acceptable choices. However, using the 39 as the first valid choice given that it exceeds 20 and is the lowest may be reasonable.
So the answer by selecting the closest minimal valid option that fits could be:
39
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