To determine the minimum sample size needed to use normal approximations for a proportion when given a probability of success \( p = 0.25 \), we can use the rule of thumb for the normal approximation to the binomial distribution. This rule states that both \( np \) and \( n(1-p) \) should be greater than or equal to 5:
- The probability of success \( p = 0.25 \).
- The probability of failure \( q = 1 - p = 0.75 \).
We want to find \( n \) such that:
- \( np \geq 5 \)
- \( n(1 - p) \geq 5 \)
Now we solve these inequalities:
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From \( np \geq 5 \): \[ n \cdot 0.25 \geq 5 \implies n \geq \frac{5}{0.25} = 20 \]
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From \( n(1 - p) \geq 5 \): \[ n \cdot 0.75 \geq 5 \implies n \geq \frac{5}{0.75} \approx 6.67 \]
Both conditions are satisfied starting from \( n = 20 \). However, to ensure an accurate approximation, it is commonly recommended that both counts be at least 10.
So we check:
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For \( np \geq 10 \): \[ n \cdot 0.25 \geq 10 \implies n \geq \frac{10}{0.25} = 40 \]
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For \( n(1 - p) \geq 10 \): \[ n \cdot 0.75 \geq 10 \implies n \geq \frac{10}{0.75} \approx 13.33 \]
Thus, the minimum sample size that satisfies both conditions is \( n = 40 \).
Therefore, the correct response is:
40