To determine the minimum sample size needed to use normal approximations for estimating a population proportion, we generally use the rule of thumb that both \( np \) and \( n(1-p) \) should be greater than or equal to 5, where:
- \( n \) is the sample size
- \( p \) is the probability of success
In this case, since the probability of success \( p \) is 0.10 (or 10%), we can calculate:
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\( np \) should be at least 5: \[ np \geq 5 \implies n \times 0.10 \geq 5 \implies n \geq \frac{5}{0.10} = 50 \]
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\( n(1 - p) \) should also be at least 5: \[ n(1 - p) \geq 5 \implies n \times 0.90 \geq 5 \implies n \geq \frac{5}{0.90} \approx 5.56 \]
So \( n \) must be at least 6 for this condition.
The more stringent condition is from \( np \geq 5\), which gives us:
\[ n \geq 50 \]
Thus, the minimum sample size needed to use normal approximations for an estimate about a population where the success probability is 10% is 50.
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