To determine the minimum sample size needed to use normal approximations for a proportion, we often use the rule of thumb that both \( np \) and \( n(1 - p) \) should be at least 10, where \( p \) is the probability of success and \( n \) is the sample size.
Given \( p = 0.25 \):
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Calculate \( np \): \[ np = n \times 0.25 \]
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Calculate \( n(1-p) \): \[ n(1 - p) = n \times (1 - 0.25) = n \times 0.75 \]
We need both \( np \geq 10 \) and \( n(1 - p) \geq 10 \).
From \( np \geq 10 \): \[ n \times 0.25 \geq 10 \implies n \geq \frac{10}{0.25} = 40 \]
From \( n(1 - p) \geq 10 \): \[ n \times 0.75 \geq 10 \implies n \geq \frac{10}{0.75} \approx 13.33 \]
The more restrictive condition is \( n \geq 40 \).
Thus, the minimum sample size needed is 40.
The answer is 40.