Answer :
Answer: 0.7752
Step-by-step explanation:
Given : The proportion of all homes purchased in 2004 were considered investment properties estimated by NAR: p = 0.23
Sample size : n= 800
Required sample proportion : [tex]\dfrac{175}{800}=0.21875[/tex]
Now , the probability that at least 175 homes are going to be used as investments will be :
[tex]P(p\geq 0.21875)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\geq\dfrac{0.21875-0.23}{\sqrt{\dfrac{0.23(1-0.23)}{800}}})\\\\=P(z\geq-0.756)\ \ [\because\ z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}]\\\\=P(z<0.756)\ \ [\because\ P(Z>-z)=P(Z<z)]\\\\=0.7751754\approx0.7752[/tex] [using p-value calculator or z-table]
Hence, the required probability = 0.7752