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 .

In this scenario, we can calculate \( np \) and \( n(1-p) \) to determine the sampling distribution's viability. Here, \( n = 110 \) and \( p = 0.22 \). Thus, \( np = 110 \times 0.22 = 24.2 \). Similarly, \( n(1-p) = 110 \times (1 - 0.22) = 110 \times 0.78 = 85.8 \). Both \( np \) and \( n(1-p) \) need to be at least 10 for the normal approximation to be applicable. Since both values exceed 10, a normal distribution approximation **can** be used for the sampling distribution of sample proportions after all!