Question

In January, 2014, a research center found that 89% of 18-29 year olds in the U.S. used social media sites. That year I surveyed 500 Portlanders between the ages of 18-29, and found that 95% of them used social media sites. If Portland mirrors the US percentage, what is the probability that my sample of 500 would contain 95% successes?

Answer 1

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.8900
sample size       =n=	500
std error of proportion=σp=√(p*(1-p)/n)=	0.0140

  probability that my sample of 500 would contain 95% successes(greater than or equal to 95%):

probability =P(X>0.95)=P(Z>(0.95-0.89)/0.014)=P(Z>4.29)=1-P(Z<4.29)=1-1.0000=~ 0.0000

probability of above event is almost negligible.