The proportion of homes in Florida owned by investors rather than homeowners is $$ 22 \% $$. Consider a sampling distribution where $$ n=110 $$. $$ n p= $$ $$ n(1-p)= $$ Can a normal distribution approximation be used for the sampling distribution of sample proportions? No, a normal distribution approximation can't be used .

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1. Given that the proportion is $$ p = 0.22 $$ and the sample size is $$ n = 110 $$, calculate
   - $$ n p = 110 	imes 0.22 = 24.2 $$
   - $$ n (1-p) = 110 	imes 0.78 = 85.8 $$

2. Check the conditions for using the normal approximation for a binomial distribution, which require both:
   - $$ n p \geq 10 $$
   - $$ n (1-p) \geq 10 $$

3. Since $$ n p = 24.2 $$ and $$ n (1-p) = 85.8 $$ are both greater than 10, the normal approximation is valid.

4. Therefore, a normal distribution approximation can be used for the sampling distribution of the sample proportion.
$$ n p = 24.2 $$ and $$ n (1-p) = 85.8 $$. Since both are greater than 10, a normal distribution approximation can be used for the sampling distribution of the sample proportion.

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